 So as you can see here, we are talking about the pair of straight lines, which is based on a real life story of straight line couple. That's not to be taken seriously. Okay, we'll start this session. Now what do I mean by a pair of straight lines? So let us first understand the basic concept, pair of straight lines. So let us say we have two lines, L1 and L2, whose general equation looks like AX plus BY plus C equal to 0, and let's say A1X plus B1Y plus C1 equal to 0, okay? Now pair of straight lines is nothing but a joint equation of these two lines. And how do we join them? Something like this. So we just multiply, you can say the left hand side of these equations like this. So this joint equation is basically called the pair of straight lines equation. Okay, now there is a rule by which we do this multiplication, when you are performing the multiplication of the two lines, it cannot be done in any fashion. There is a rule for it. You should have brought these equations into their general form. Now don't ask me what is the general form. In class 11, they had done different forms of a straight line, didn't we? Right? Slow form, intercept form, normal form, they realized two point form, slope point form, and there was something called general form. General form is your AX plus BY plus C equal to zero form. So when you're performing the operation of formulating a pair of straight lines equation, you should have brought the two lines in their general form, very, very important. Else, your answers will not come out to be as expected. Let me give you an example. Let me give you an example for the same. Let's say there are two lines going to a X equal to Y, and let's say another line is X equal to minus Y, correct? If I say, give me the equation of the pair of straight lines whose constituent lines are these two. How will you do that? The first step is you have to bring these two lines in their general form, something like this, okay? And then perform the multiplication, okay? And you'll end up seeing X square minus Y square equal to zero. If you do this, let me write it in red because this is not a correct way. If you just multiply them literally, let us X square is equal to minus Y square. Remember, you'll end up getting some different answer altogether. By the way, this is a point circle, isn't it? Point circles satisfied by only zero comma zero, okay? So this is not the right way to do it. The right way to do it is this approach. Are you getting my point? So basic ground rules are clear, okay? So if you type such an equation on GeoGebra, you'll see that. GeoGebra will show you two straight lines. Let me show you how it works. Okay, hope you can see my GeoGebra software. Yeah. So if you see this point, this line, X square minus Y square equal to zero. Let me plot it and show you. X square minus Y square, X square minus Y square equal to zero. You see that the graph automatically shows you two straight lines, right? So primarily we're dealing with a combination or you can say a joint equation of two straight lines. Now they may be intersecting, they may be parallel, they may be coincident. They may be both imaginary also. But imaginary part, we are not taking into our account, okay? So that is beyond our discussion of this chapter. Okay, any questions so far? So good morning to everybody who has joined in. Say something here, good morning. Good morning sir. I know Corona virus has taken a huge impact. Good morning guys. Good to see you back, my first class with you in class 12th. Okay, so let me give you a question to begin with. You know how to find the equation of the pair of straight lines given the two lines, right? Now I'll give you an opposite question. Let's say this is a equation, a combined equation. By the way, I will not use the word combined equation every time. I'll just say it's a pair of straight lines equation. Can you tell me which are the two straight lines which constitute this pair? You can type it privately to me. Which are the two straight lines? Or you can say find the lines which constitute this pair. That constitute this pair. By the way, nobody has problem in audio or video. Everybody is able to hear me properly. If not, please type it out that I'm not able to hear. Because first time join is they have a slight issue with. I can see everybody has joined in. Krishna, are you there? Absolutely Ajay, very good. Anybody else guys? Reply privately to me. Correct, Arpita. Arpita, by the way, equation should be something equal to something, right? You're just written expression. Same goes with Viva. Right, right. Correct, Rupan. So this is very simple. All of you probably have answered this correctly. You need to first factorize this. You know you have to split the middle term over here. So we can split the middle term over here as x square minus 4xy minus 2xy plus 8y square equal to 0. Take x common, x minus 4y. Take a minus 2x minus 2y common. You'll end up getting x minus 2 4y. And it just means that these are the two lines which actually made this pair of straight lines. So one is this line and other being this line. Is that fine? Now guys, I would like you all to pause and ponder over here for some time. What I see here is that the two lines which I got from this particular pair of straight lines were both passing through origin. They're both passing through origin. Passing through origin means what? Intersecting each other at origin. Correct? So these two lines graph, if you draw, you realize that, okay, I'll just draw a close to real graph. So this is y is equal to x by 2. So y is equal to x by 2 is something like this. Let's say y is equal to x by 2. I'm talking about this line. Let's say L1 line and L2 line is y is equal to x by 4. So it would be, let's say I draw it in yellow. So it would be a lesser slope line like this. You realize that these lines are passing through origin. So these two lines are passing through origin. And this fact has a straight impact on this pair. As you can see, the pair does not contain any x term. It doesn't contain any y term. When I say x term, just x term, I'm talking about something like 5x. It is containing x square, agreed. It is containing xy, agreed. It is containing y square, agreed. But there will be no x term kind of a thing. There will be no only y term kind of a thing. There will not be any constant term also. Do you see that? So what I claim here is that if you have any two lines, y is equal to m1x and y is equal to m2x. So we know that these are two lines which are passing or intersecting, whatever you want to call it. Intersecting at origin. And you make a combined equation out of it. Let's say I want to make a pair of straight lines from it. You would realize that this will never have an x term. So there will be no x, y or constant terms coming in this multiplication. What will be only a present? What will be the only terms present? x square, y square and xy. So this is something which I am going to talk about in my next slide. But meanwhile, any question so far? Very simple exercise. Any question anybody? No. Should we move on? Okay. So I am now going to talk about something called second-degree homogeneous equations in two variables. Okay. By the way, let me explain these terms. Second-degree, everybody knows what is second-degree. What is second-degree? The highest power of the variable or the variables combined should be not more than 2. That is the meaning of second-degree. What is homogeneous equation? That is something which needs to be understood. Two variables everybody knows. So that means your equation will only contain x, y or any two variables. Okay. Not more than that. So let me explain to you what is homogeneous equation. But before that, we need to understand homogeneous function into variable. Okay. Guys, try to understand this concept clearly because you will be using this again in the chapter differential equations. So it is useful. Let me write it here. Useful in differential equation chapter, which we will be doing in the month of September. Okay. It is actually the last chapter of calculus. By the way, all of you would be knowing that 12th is a calculus heavy year. Limits, continuity, differentiability, differentiation, application of derivatives, integration, that is indefinite, definite, area under curve, differential equations. So almost 10 to 11 chapters are just based on calculus. Okay. Yeah. What is a homogeneous function? So let me write the formal definition for it. So a function f x comma y. Okay. Why x comma y? Because this function is in two variables is said to be a homogeneous is said to be a homogeneous function said to be a homogeneous function in two variables. Of degree n degree n. If it satisfies this condition f lambda x comma lambda y is lambda to the power n f x comma y. Now don't be scared. This is not something like rocket science or something which you cannot understand. See, I'll explain you with a simple example. Let's say I have a function in two variables, which is homogeneous of degree two. So let's say a simple example like this that the word itself means homogeneous homogeneous in what homogeneous in its degree. So every term in this expression will be of same degree that is degree two. You see that x square degree two y square degree two x y degree two. Okay. This is just a homogeneous equation of degree two. That's what you need to understand nothing else. This is just a mathematical jargon. So what it says it says that if you replace your x with lambda x and if you replace your y with lambda y, then something like this will happen. So x square will become lambda square x square y square will become lambda square y square and this will become lambda x lambda y which could be written as lambda square in common multiplied to the original function. So that becomes lambda square f x comma y. Isn't it? That is what this guy is saying. So you can take lambda one lambda to the power n, n being the degree. Okay. So here degree is two. So lambda two has come out. This is a way to identify whether a function is homogeneous or not. Okay. But I'm sure with this example you are very clear. What's the homogeneous function? So all of you will type one homogeneous function on your chat box. Everybody should type. Don't copy from each other. I mean try to give your own function. Type a homogeneous function on your chat box which is of degree two. For square you can write x2. So I can understand your writing x square. Everybody, I'm waiting. Nobody is typing. What happened guys? My question was everybody should type a homogeneous function of degree two in two variables. Come on Aditya. In two variables. Very good Tripan. Very good Anjali. Very good Kirtanam. In two variables, Arimand. That's also correct. Very good. Very good. Sana. Very good. Just xy also you could have typed. That is also correct. Yes. Can I say zero is also acceptable? Zero can be of any degree right? Zero degree is not defined. Yes or no? Correct. Okay. Very good guys. So what I'm going to claim here my dear is that a homogeneous function like this. A homogeneous function like this when equated to zero becomes a homogeneous second degree equation. It becomes a homogeneous second degree equation. Many people ask me sir what is this zero doing over here? Doesn't it make the you know a homogeneity doesn't it lose out on the homogeneity of the entire expression? Let me tell you no. Zero can also be treated as zero into square of any variable. So zero can acquire any degree it wants. It doesn't have a degree of its own. Okay. But if I put a one over here then it would no longer remain a homogeneous second degree equation. Remember that. Getting my point guys. So what am I claiming here my dear? My claim is let's say right claim. I claim that such a equation will always represent always represent a pair of straight lines. A pair of straight lines passing or intersecting at origin. Now I do not claim that the lines will be always real. I have just said that it would be some pair of straight lines. Now they may be real. They may be imaginary also. Okay. So this may not give you any point at all also. This may not give you any two lines at all. It may just give you a point. For example, remember I had taken this example in the beginning of the chapter. Isn't this also a homogeneous equation? This will not represent a line. It will just represent a point. Okay. In this case my lines would become imaginary in nature. My lines will become imaginary in nature. Okay. So my claim here is such a equation which we call as homogeneous second degree equation would always represent a pair of straight lines passing through the origin. Okay. Now let us say the pair of straight lines that constitute this are y is equal to m1x and y is equal to m2x. Now what I'm trying to do is I'm trying to know what is the relationship between m1 m2 and these coefficients that you see. Okay. So let us say these lines constitute this pair of straight lines. So I'll let it down. Let this and this constitute this particular pair of straight lines equation. Now what is my agenda? My agenda is to know the relationship between m1 m2 a h and b. Okay. So let's try to work on this. So what I'll do here is that if you see these two lines, it suggests that y by x could acquire two values m1 and m2. Isn't it? If you bring the x down, you can see y by x could take two values. What are those two values in my present assumption? Those are m1 and m2. Correct. So what I'm going to do is I'm going to make a quadratic in y by x from here. See how I'm going to do it. It's very easy. I'm going to first flip the positions of these terms. I'm going to take b common. Guys, don't be too much worried about copying here. Understand because notes will be anywhere shared on the group. So you can sit and fulfill your notebooks. All of you should be just glued to the screen just trying to understand what I am saying. Don't be busy scribbling, scribbling, scribbling. Else you'll miss out on important things. I will send this PDF on the group. I'll send the recording also on the group. Then you can listen to the video n number of times and copy down the notes also. Okay. Next is I'm going to divide throughout with x square. So if I divide throughout with x square, this is what you are going to see. Correct me if I'm wrong. So what I did here, I let it down. I divided by x square. Okay. This operation I did. Now you would see here that it has resulted into a quadratic in y by x, right? And this quadratic has two roots m1 and m2. Agreed or not? Is it making sense? Is the chronology making sense? Now y by x has two roots m1 and m2. So this guy should have two roots m1 and m2. So if this has two roots m1 and m2 by using our fact that the sum of the roots is b by a, see x square is a variable. It's not a constant. So variables can take any form it wants. In case x is zero, y will also be zero and thrippen it will take an indeterminate shape. Because we know that it has to pass through zero zero, right? Okay. So if x is zero, y will simultaneously become zero. Then it will form an indeterminate scenario. And we know that for indeterminate scenario, a real value can exist for that. Getting this thrippen. Does this answer your question? It's a very good question you asked. Okay. So what is the sum of the roots? Can anybody tell me? Minus b by a. Everybody knows the Vitas relation? Vitas relation. In any quadratic, sum of the root is minus b by a. Product of the root is c by a. That is Vitas relation. Okay. This is just a fancy name. You know, after the name of the Italian mathematician who discovered this. Okay. So some of the roots will be minus b by a. So this is playing the role of your b. One is playing the role of your a. Please note it's a quadratic in y by x. So in your mind, you should have a picture of x in place of y by x. Getting my point. One is playing the role of a. 2h by b is playing the role of your b of your quadratic. And a by b is playing the role of your c of the quadratic. So sum of the roots, that is m1 plus m2 is minus 2h by b. And product of the root is a by b. Let me write it in yellow. Okay. This two relation is going to be very useful guys. I'm telling you, you are going to use this at endless number of places. So remember it. This is a symbol which normally I use for you to remember it. I'm not saying memorize it. Remember it so that while problem solving, we can recall these results. Make sense. Okay. Now. Something very interesting related to this quadratic. Don't waste time copying. If you're fast enough to copy, then good and good and well, but don't waste time copying it. Okay. We'll go back to the same quadratic which we just not talked about. y by x square plus 2h by b y by x plus a by b. Okay. Now we know that the roots are what are the roots m1 and m2. Okay. Now we can use our quadratic equation formula. What we call as the Sridharacharya formula to get these roots. What is the formula of Sridharacharya formula? Minus b plus minus under root b square minus 4 ac by 2a. Correct. Now remember a little while ago we were saying that it represents a pair of state lines. But those state lines can be real also. The state lines can be imaginary also. So now the situation has come that we know under what condition or under what relationship between H and B, I would end up getting real lines or I will end up getting coincident lines or end up getting imaginary lines. So this guy is a very, very important decision maker. So you can say that this is playing the role of your discriminant. So if it's square 4 x square by b square minus 4 a by b. If this is positive. Okay. Let me write it in a proper way. You can take 4 b square common that is x square minus a b. If this is positive. Now 4 and b square are already positive. That means it further implies if x square minus a b is positive, you will end up getting real and distinct values of m1 and m2. That in turn means you will get two real and distinct lines. That means your lines will look like this. That means you can differentiate between them. That's why distinct. Getting my point guys. Any question here? Stop me. Feel free to ask me. Anywhere you feel that I have jumped something and there is a conceptual void. If your x square minus a b is zero, what does it mean? m1 and m2 both will be equal. That means your line would be coincident. Coincident in this case. That means when you are dealing with a homogeneous second degree equation, your lines will be coincident. Many people ask why coincident? Why not parallel? Because when you are saying x square minus a b is zero, m1 will be equal to m2. That means they could be parallel also like this. One could be y equal to mx. Other also y equal to mx. But can you see that they cannot be parallel because both of them have to pass through? Origin. That's why I didn't write parallel. I directly wrote coincident. So this situation can never occur that two straight lines and both of them are passing through origin are parallel. They can always be coincident. Are you getting my point guys? What happens when x square minus a b is less than zero? In this case, you will end up getting imaginary lines. But nevertheless, imaginary lines passing through origin. So imaginary lines may pass through a real point. However, our discussion agenda is not to talk about imaginary lines in this course. You may do a further reading of this to different sources. Math stack exchange is one of the sources. Wikipedia also you can look for it. So we are primarily going to talk about these courses. So we will not deal with the imaginary lines. Any questions so far? If no, then I have a question for you. So Vibhav has a question. Sir, can there be a case where one line is imaginary and the other line is real? See, in that case, what will happen? Your coefficient should be such a way that it should end up giving your a, h and b both as real. All of them as real. It may happen. It may happen when one of them is real other is imaginary. Your classical example is your argon plane. Your y axis is your imaginary line and your x axis is your real line and both of them meet at origin. So that's a very classic example of such a scenario to happen. So it can happen Vibhav. Is that all right? Does it satisfy your question? Okay. So now I have a question for all of you. If the slope of one of the lines represented by a x square to h xy plus by square equal to 0 b n times the slope of the other, then which of the options is correct? Single option correct. So I'm writing down the options. 4n square h square is equal to a b 1 plus n the whole square. Option b 4n h square a b 1 plus n. Option c 4n h square is equal to a b 1 plus n the whole square. Option d 4n square h square is equal to a b 1 plus n. I'll be running a poll for this. So once you've solved it, I'll allow you three minutes for this. Once you've solved it, please press on the poll button so that I know what is the percentage of people answering which option. Time starts. Time has already started, by the way. If you're not able to see the poll, don't worry, you can type it privately to me also. So Pratham, I would also request you to press on the poll button so that we have an idea of how many people are giving the right answer to it. So as a teacher, I would come to know how well the concept has been understood by you. So two people have responded so far. Three have responded so far. Guys, 50 seconds more, then everybody should press on some button. Six people have voted so far. Time is running out. Guys, fast, fast, fast, fast. Okay, time is up but I'll still give you 30 seconds more. I'll end the poll in exactly 30 seconds. Please press on some button. Guys, time is up. Please press, please press something. Only... Yeah, yeah guys, come on, come on, press it. Good, good, good. Okay. End of poll. Now see the result in front of you. 39% people have said option C and then option B. Okay, so between B and C there is a kind of competition. So let's see, let us see what is the actual answer. Okay. So what is given to us is that the slope of one of the lines is M and the slope of the other is N times M. Okay, so let these be the two slopes. So can I say first that from our previous relation, the sum of the slopes that is M1 plus M2 is minus 2x by B. So I'm using that. Okay. And product of the slopes that is M into NM, that's going to be A by B. So let me call this as my first equation, let me call this as my second equation. Correct? Now, from the first equation I can say, from one I can say M times 1 plus N is equal to minus 2x by B. So M is nothing but minus 2x by B1 plus N. Okay. Let us substitute this. Let us substitute this in 2. Okay. So when we do that, we end up getting NM squared. Instead of M squared, I will write minus 2x by B1 plus N the whole square. And that is equal to A by B. Let's expand this. So we'll have N4H squared by B squared 1 plus N the whole square. B goes off. Let me take the term in the denominator to my RHS. So I'll end up getting 4NH squared is AB1 plus N the whole square. Okay. So this is my final result. Let's see which option says that. Which option says that? Which option says that? Option C says that, my dear. Option C is the right option. So Janta was correct. So those who replied with C, well done. Your answer was correct. So I can see Pratham was the first guy to get this answer. Pratham Shushan. Well done Pratham. Very good. Okay. We'll move on to the next one. Let me pull out a question from my repository. Okay. Let's take this up. Right. Hope you are able to see this question properly. This is a prove that question. Guys, let me tell you one thing. Please do not underestimate prove that question. In fact, please do not ignore prove that question because you can always, the question setter can always objectify or prove that question also. So many of us while practicing, we think, this is a prove that question. Why are we wasting time on it? Now the exam is happening in objective way. No. Prove that questions are equally important because they can be objectified any day. Let's have three and a half minutes for this question. And once you are done, please type done so that I know how many of you are done. Read the question properly. If the slope of one of the lines represented by that second degree homogeneous equation be the nth power of the other, then prove that whatever is written there. Done. So one person, Anjali is done. Guys, just a casual suggestion. Please do not roam around. Stay indoors. These holidays are not meant for you to socialize. Isolate yourself in your study room. Sorry to say that. But this is not a time for socializing because it may lead to a huge outbreak later on. So please stay isolated as far as possible. Just one person done so far. What about others? I don't have to name all of you. Okay, let me give you some hints on this. See, what it says is that, let's say if the slope of one of the guys is m, the slope of the other is m to the power n. Isn't it? And we already know that the sum will be minus 2h by b and product will be a by b from our previous relations. Isn't it? That's why it's so important that we remember these. In the exam time, we'll not get time to derive it. Okay. So let me call this as 1. Let me call this as 2. So from 2, I can say m to the power n plus 1 is a by b. That means m is a by b to the power of, or you can say n plus 1th root of a by b. Nice. Okay. Now we'll substitute this in 1. We'll substitute this in 1. Okay. So when we do that, what do we end up getting? This will become a by b to the power of 1 by n plus 1. m to the power n will become a by b to the power of n by n plus 1. And this is equal to a by b, sorry, minus 2h by b, minus 2h by b. Correct. Any questions so far? Now, I'm coming over here at the top. Multiply the root with b. So b times a by b to the power of 1 by n plus 1. And b times a by b to the power of n by n plus 1 is equal to minus 2h. Now minus 2h also. Let me bring it to the left-hand side. Okay. Now this is just a game of exponential laws. Nothing else. So you can see b to the power 1 over here, and you can see b down over here with the power of 1 by n plus 1. So it will become b to the power 1 minus 1 by n plus 1. Guys, nothing. I'm using exponential laws. Okay. Everybody knows it. Here also, b is power 1, and denominator will have b whose power is n by n plus 1. It will become b to the power 1 minus n by n plus 1. Okay. And a to the power of n by n plus 1 plus 2h equal to 0. Now let's try to simplify these powers. So here on top, I'll end up getting n by n plus 1 a to the power 1 by n plus 1. Plus, here I will get b to the power 1 by n plus 1 into a to the power n by n plus 1 plus 2h equal to 0. It still doesn't seem to be like this. What we'll do is we'll take a b to the power n common from these two ways. Okay. So a b to the power n, if you take common, let me write it properly. Then this entire thing is as if you are taking n plus 1th root of it. Isn't it? So if you multiply these powers on these variables or these numbers a and b, you'll end up getting this term. In the same way, this I can write it as a to the power n into b whole base to the power 1 by n plus 1 plus 2h equal to 0. This is what we wanted to prove. Is that fine? Simple question. You are just playing with the exponent guys. Nothing else. Nothing else. Any question, any confusion anywhere, please feel free to stop me. Clear? Very good. I can see Anjali was the first person, then Pratham, Aditya, Kirtan. Guys, let's move on to the next problem. I'm going to put forth a lot of problems for you. Okay. Again, here we'll have a question which I'll make objective. See how an objective question is framed. Okay. The question was find the find the condition that one of the lines given by this may be perpendicular to one of the lines given by this. So what I'll do as a question setter, I will scratch out this find. Okay. See how an objective question can be made from this and I'll say the condition that one of the lines may be perpendicular to the one of the lines here is. So I just put an is over here and I'll give you options. So here are your options. Okay. I'm giving you options. Meanwhile, you can start thinking over it. So options are for h dash a dash h dash b dash plus b b dash option B is for h dash a dash h dash b dash h a plus h b plus b b dash minus a a dash square is equal to 0. Option C for h a dash h dash b h dash a h b dash plus b b dash minus a a dash square is equal to 0. And option D is for h a dash h dash b h dash a h b dash plus b b dash plus a a dash square equal to 0. Okay. And here comes your poll and your time also begins simultaneously. Let's have three and a half minutes for this because if I give you three and a half minutes, you'll end up taking four. So let's have a three and a half minutes. So I've got one response so far. If you have no idea, you can type that out also. Okay. I don't want you to just sit and wait for three minutes and do nothing about it so that I can know that many of you have no idea how to begin with. Let's be honest. Cross multiplication formula. Okay. Aditya is asking me how to use cross multiplication formula. Very good question Aditya. Many of you would have learned how to solve simultaneous equation. Correct. There are three methods to solve simultaneous equation. One is by substitution. Other is by elimination. Another is by a cross multiplication. I think that is what Aditya is asking me. Okay. How it works Aditya. Let me show you. So let us say you have something like this. There is one equation in two variables. Linear equation like this. There is another one like this. Okay. Then cross multiplication says you just have to write. Let me write it in different pen. X, Y, 1. Okay. Below X whatever you have to write is basically obtained by hiding this. So hide this guy. Okay. Whatever you see the constants B1, C2 minus B2, C1. So here B1, C2 minus B2, C1 will come. So there is a cross multiplication that you have to do. Okay. For Y you have to do the opposite here. C1, A2 or you can say A2, C1 minus A1, C2. You can note down this formula if you want. And for 1 you have to do A1, B2 minus A2, B1. A1, B2 minus A2, C1. Okay. So only in case of Y you have to switch from, you have to take the multiplication of right top with left bottom. So C1, A2 minus A1, C2. Okay. But for others it is always left top with right bottom first. Okay. A1, B2 minus A2, B1 for the last one. And for this guy B1, C2 minus B2, C1. Is this clear? Aditya. So this is how we can solve for X and Y. So if you bring the denominator of X on top over here, you will get X. If you bring the denominator of Y on top over here, you will get a Y. So this is another way by which we can solve simultaneous equation in two variables. It is, it is actually vector cross product. It is related to that. Analog. Later on when we do vectors or later on when we do Cramer's rule in determinants, you would realize that it is actually related to vectors. You're actually using vectors to solve this. When you're cross multiplying, you know what you're doing? You're actually trying to find out the area of a parallelogram. Okay. I'll talk about all these things when the vector counts. So in mean, meanwhile, time is up guys. How much time will you take? It's already five minutes. Oh my God. Please, please type something. Please press the poll button. I want to see even if you take a guesswork, how many of you has guesswork is correct? Guys, wait a minute. Before it is six minutes, press something. Press something, press something, press something. I'm going to stop now. Look at the poll result. 69% of you have said option B. Okay. Maximum Janta has gone with option B. Okay. Only 23% says option C. So it's B versus C again. Let's discuss it. Let's discuss it. Okay. Now try to understand here. Let's say the two lines that make up the first set of equations. So this set of equation, let's say the two lines are these L1 and L2. Okay. The second set of equations, let me show it in some different color. Let me show it in blue is something like this. But what we know from the scenario is that, and let me call this as L1 dash and L2 dash. But what we know here is that one of the lines, let's say L2 is perpendicular to the other line of this blue one. So let's say L1 dash and L2 are perpendicular. This is what is known to us. Correct? Has everybody understood the problem itself properly? This is what you all understood, correct? Anybody who had a different understanding? So one of the lines of this, which I'm calling as L2, is perpendicular to the lines of this guy. And let's say that line be L1 dash. So you can see this line and this line is what we are referring to. Okay. Now, since this is a homogeneous second-degree equation, I can assume that let L2 have an equation Y equal to MX. Correct? Now, the second equation is also a homogeneous equation and this line is perpendicular to this. So what should be the equation for this L1 dash? Y equals minus 1 by MX. Absolutely. Y is equal to minus 1 by X. Okay. Now, see how am I going to use this? First, this line, it will satisfy this equation. Sorry for writing in wide, on wide. Okay. So to see the arrow on the screen, Y equal to MX will satisfy this guy. Okay. So let me replace wherever I see a Y with MX. So AX square will remain as it is. In face of Y, I'm writing MX. BY square means BM square X square. Correct? As you can see, X square is present everywhere. Drop it. Okay. So write it like this. Let me call it as 1. So what I'm saying is that M is basically satisfying this relationship. In the same way, this guy should satisfy this equation, which I'm circling with a blue. Yes or no? So let me substitute that. So I will end up getting A dash X square to H dash X. In face of Y, I'll put minus one by MX. Okay. Let me just shorten the length of this guy. And plus B dash minus one by MX square equal to zero. Is that clear? Now let me do one thing. Let me cancel out X square from both the sides of this equation or let me drop X square and multiply with M square. So it will become A dash M square minus two H dash M plus B dash equal to zero. Okay. Let me call this as two. Now, if you look at the options, the options actually wants you to eliminate M from one and two. Eliminate M from one and two. And that's where I think Aditya had asked for the cross multiplication method because he wanted to treat M square as X, M as Y. So basically end up seeing a simultaneous equation in two variables. Okay. So let me show how this cross multiplication can help us over here. So let me write these equations once again. B M square two H M plus A equal to zero A dash M square minus two H dash M plus B dash equal to zero. Right. So first I will write M square. I will write M and I will write one below that. I will put these dashes for M square. Hide this. You just have to cross multiply this guy. Two H B dash minus minus. So it'll become plus two H dash A. Correct. Same for M. You have to do a dash minus B B dash. So only in case of the middle term, the order will switch. So you'll start with right top multiplied with left bottom right top with left bottom rest for both the extreme cases. You have to start with left top multiplied with right bottom first. So for one, you will do minus B into minus two H dash. So you can say minus two H dash B minus two H A dash. Guys, I'm not doing any rocket science. This is just your cross multiplication method, which you had already learned in your class 10. Okay. Already learned in class 10. Okay. Now see how I'm going to eliminate M from it. So my, from my these two equations, I can say M square is two H B dash plus two H dash A by negative of two H dash B plus two H A dash. And from these two, I can say M is equal to A A dash minus B B dash by negative two H dash B plus two H A dash. Is that correct? Now, since this is M square and this is M, can I say this guy is square of this guy? Am I right? Am I right or not? So I can say this fellow that is two H B dash plus two H dash A by negative two H dash B plus two H A dash is square of this guy. So I'm squaring it individually. Okay. By the way, negative sign here will become irrelevant to us because we are squaring it. Okay. So this factor will cancel off with one of the factors here cross multiply and bring it to one side. So when you cross multiply, you'll realize two two two two here can be taken common. So it becomes four and you'll end up getting H B dash H dash A times H dash B H A dash plus A A dash minus B B dash square equal to zero. So this should be your final answer. Now check it out which option matches with this option. Let me scroll up a bit. So you should have B B B dash minus A A dash squared. Doesn't matter even if you flip the position. So this and this can never be my answer. Okay. Secondly, it should have H A dash H dash B H dash A H B dash. So let me see which B or C which is correct. Yes. So option C is going to be the correct option. This is also ruled out option C is the right option. Okay. And the poll result was B is correct. Oh my God. So Janta has gone wrong here. Okay. Sorry. Okay. The poll result was B, right? But it is actually C that is correct. Okay. Pratham. Got your mistake. Okay. So this was on a slightly different note. Okay. Now I'll not leave you. I'll ask you one more question. Next question is listen to this question very, very, very carefully. If two of the three lines, if two of the three lines represented by a X cube B X square Y C X Y square and D Y cube equal to zero are at right angles are at right angles. Okay. Then which of the option is correct. Single option options are a square plus AC plus BD plus D square is equal to zero. Option B B square plus AC plus CD plus a square is equal to zero. Option C C square plus AB plus BD plus D square is equal to zero. Option D a square plus, sorry, a square plus. Let me write it again. A square plus AC plus CD plus D square equal to zero. But guys, before you start working it out, let me explain the question to you. Two of the three lines. Remember, it's a cubic homogeneous equation. That means it would represent how many lines? All passing through origin. Are you getting my point? That means if you formulate a cubic in, if you formulate a cubic in Y by X, then there would be three roots to it. Let me call it as M1, M2, M3. Okay. Getting my point. Now two of them are at right angles. Then which option is correct. Understood? Okay. Running a poll on this, time starts now. Let's have three and a half minutes for this as well. In case you're not able to see the poll, don't worry too much. Just type out the answer privately to me. Guys, we are about to finish three minutes. Just one, two people have answered so far. Okay. Three of you have answered. Four of you. Come on. Come on guys. More response. Don't worry. Yes. We'll discuss about it. How to approve these kinds of questions. I'm there for that only. Don't worry. Seven responses guys. Five seconds, four seconds, three seconds. Come on. End it. Okay guys. Three and a half minutes over. Now please press on some button. No worries Aniludha. Just put forth your response on the poll button. We'll discuss. I'm going to end it in next 10 seconds. 20 of you have responded. 10 of you still, I'm waiting for you guys. Fast, fast, fast, fast. And closed. Let's have the results shared. 65% Janta is for A. Then the next response is B. So now there's a clash between A and B. Cello will see it. Let's have a look at it. Don't worry Aniludha. Okay. Now all of you please pay attention. As I discussed with you, I actually given you a big hint. So we have to first formulate a cubic in Y by X. So the first step that I would do is I would divide this throughout by X cube. When I do that, I end up getting Y by X whole cube. Okay. You'll get C Y by X whole square. You'll get B Y by X and you'll get an A. Okay. For the time being, I'll name this expression to be M. Okay. I don't want to write Y by X, Y by X, Y by X every time. So let's have something like this. Okay. So M is basically just like your variable X that you write in a cubic equation. Okay. So this is basically a cubic in M. This is basically a cubic in M. And how many roots it has? M1, M2, M3. Right. Now we'll use beta's relation once again. Beta's relation once again. What is beta relation say in case of a cubic, my dear? That the sum of the roots would be minus B by A. See the motion of my pen. Minus B. This is your B now. By A. This is your A. So it'll be minus C by D. Correct. Let me call this as one. Then product, sum of product two at a time. That is M1, M2, M2, M3, M3, M1 would be C by A. I think most of you have not learned how is that same relation which you had learned in quadratic scaled up to a cubic. In fact, it can be scaled up to any degree polynomial equation. Okay. It's not a rocket science. You can just Google this out or you can just get hold of any standard J book and look into the chapter theory of equations. But anyways, I'm going to do that with you in the next session. And the last one is the product of all of them that is M1, M2, M3 is equal to so minus B, C and then minus D by A. So it's actually the naming is ultra over here. So minus A by D. Is that fine everybody? Are you guys happy with this? Okay. Because I'm going to use this now in order to crack this problem, especially I'm going to use the third one over here. I'm going to use this third one. What does the question say? Question says two of them out of this three are perpendicular. Isn't it? So if two of them out of three are perpendicular, let's say the two of them have the slopes M1 and M2 without the loss of generality. I can say M1 M2 is minus one. So if I put it over here, I get minus one times M3 is equal to minus A by D, which means M3 is A by D. Correct? Now M3 is the root of this guy. Isn't it? It's roots are M1, M2 and M3. Correct? So this M3 must satisfy this first equation in M. Isn't it? So let me write this here. M3 must satisfy. M3 must satisfy this guy, which is one. Isn't it? So let me substitute A by D in place of M over here. This will become D times A by D cube C times A by D square B times A by D plus A equal to zero. Okay? Let me just give it a final shape. This will become A cube by D square. This will become A square C by D square. This will become AB by D and this is A. Okay? Do one thing. Multiply with a D square. You end up getting A cube A square C ABD and AD square. Drop a A. Drop a A from everywhere. If you drop an A, it becomes A square plus AC plus BD plus AD, sorry, plus D square equal to zero. Let's see, is this present in my options? A square plus AC plus BD plus D square. A square plus AC plus BD plus D square. Yes. Yes. It is very much present in A. So Janta was correct. Well done. Okay. I'm not asking who answered with what, but yes, your approach is correct. Your answer is correct. Sorry. Yeah. Now we'll talk about a different subtopic. Let's talk about now the angle between the lines which constitute this homogeneous second degree equation. So basically, let's say if somebody provides me with a homogeneous equation like this, homogeneous second degree equation like this and asks me, boss, tell me what is the angle between the two lines which make this pair of straight lines? Okay. If I have been given an equation, the equation actually doesn't give me M1 and M2 explicitly. Right? M1 and M2 is hidden in this equation. Correct? So I don't know M1 and M2 directly and I have to find this particular angle. Yes or no? Yes or no? Now if M1 and M2 were known directly and let's say I'm finding for the acute angle, then I'm sure each one of you know the formula for finding the angle between two lines whose slope is known. What is that? M1 minus M2 by 1 plus M1 M2. Very important formula guys. Is there anybody who has never come across this formula? I don't think so. Okay. The problem is M1 M2 and these guys are not known to me directly. However indirectly I know that M1 plus M2 is what? Minus 2H by B and M1 into M2 is A by B. Can I use this to crack this? If yes, please tell me how. Somebody please unmute yourself and tell me how. I'm waiting guys. Anybody? You can get M1 minus M2 by M1 plus M2 whole square minus 4M1 M2. Okay. So Siddhartha is suggesting me that see there is no problem with the denominator. Nominator already has M1 into M2. Correct? And we know that over here. The problem is with the numerator fellow. Okay. Yes. So what I'm going to do is I'm going to write that as M1 plus M2 square minus 4M1 M2 under root. Guys, let me tell you this expression is giving you mod M1 minus M2. If you doubt that, I will explain you further. But if you're fine, I'll move on. Anybody who is not able to understand how did I transform? How did I transform mod M1 minus M2 as under root of M1 plus M2 square minus 4M1 M2? Anybody who wants to know how I got this? It's a simple identity which you have learned in class seventh. A minus B square is A plus B square minus 4AB. Isn't it? That's what I've written. Okay. Now this actually helps me to fit in these values. So M1 plus M2 put it over here, square it. So let me write it under the root sign. So 4H square by B square minus 4M1 M2 that is 4AB. Okay. In denominator, I'll get mod of 1 plus AB. Okay. Let me further simplify this. Let me further simplify this. So in the numerator part, I can take a B square common. I can take a 4 out and I can write it as H square minus AB. Correct? By the way, when a B square comes out of the under root symbol, what will it come out as? When a B square comes out of the under root sign, what will it come out as? So it won't be 2. I'm sorry. Yeah. So my question is the B square of the denominator, if I'm pulling it out of the under root sign, what will it come out as? Type it out. B. B. A. Mod B. Mod B. Come on guys. How many times I explained this? Under root of something square is mod of that quantity. This is a blunder which people do when they don't use it. So please remember this. Very, very important. Under root of B square will come out as mod B. Here also I can write it as mod A mod B by mod B. So this fellow denominator, I can take mod B as LCM. Mod B mod B gone. So finally we are left with the expression that tan of theta is 2. By the way, this theta is acute angle by this. Numerator doesn't need a mod because under root already takes care that it is positive. This formula is very important. Please remember this. Now we'll do some analysis of this before we move on. I have a very bad habit of analyzing every result which I get. Now if h square is AB or you can say h square minus AB is 0. Do you realize that theta will also be 0? 0 degrees, 0 radians. Which means your lines will be coincident. People are wondering why not parallel? 0 means parallel also. Again my same argument holds here. It is a second degree homogeneous equation. It has to pass through origin. It has to pass through origin means they will be coincident, not parallel. Did you come across this guy before? Did you come across this guy before? Should I move back some slides? This came in the discriminant part, right? So if this guy is 0, remember m1 is equal to m2. And if m1 is equal to m2, the angle between them have to be 0. Hence the lines will become coincident. Try to relate. Go back couple of slides. If h square minus AB is positive, that means theta is a positive quantity. And the lines will be making some acute angle. So lines are real and distinct. And if h square minus AB is negative, anyways we know theta is going to become imaginary because the lines are imaginary. Because the lines are imaginary. So I am only interested in the first two cases. This is one thing. Other thing is if your A plus B becomes 0, that means if your denominator becomes 0, what does it mean? It means theta is 90 degrees. That means the lines that constitute that pair are perpendicular to each other. Are perpendicular to each other. Guys, let me tell you this is a very important and useful concept in solving locus questions. You will see this concept coming a lot in locus type of questions. Especially when you are solving conic sections. So as a matter of thumb rule, if a pair of straight lines shows this characteristic that the coefficient of x square, that is your A plus the coefficient of y square, that's your B, adds up to give you 0. That means it represents perpendicular pairs. Perpendicular pair is there. That means the lines would be at 90 degrees to each other. Are you getting this point? Simple example would be your x square minus y square. As you can see, coefficient of x square is 1. Coefficient of y square is minus 1. So 1 plus minus 1 is 0. And we all know that it represents y equal to x and y equal to minus x. Angle here is 90 degrees. Getting my point guys. Any question with respect to this, please let me know. Sir, I have a doubt. Yes sir, please tell me. Sir, but which pair of angles do you take? There can be a pair of acute angles and there can be a pair of obtuse angles also. Why do you call a pair of acute angle? Because the angle here and the angle here will be the same, right? You can take the other pair of angles also, right? Didn't I say it's an acute angle over here? They can be only one acute angle possible? Sir, but if you take the other angle then what will be the case? Other angle, basically we'll remove the mod and we'll allow it to have a negative value. Okay, so one angle will be obtained as from this formula, other will be automatically pi minus theta. This will be your obtuse. Okay, yes sir. Let's have a question based on this. Let's have a question based on this. Okay, now again I will objectify it. Instead of find the angle, I'll remove it and I'll say the angle between the lines given by this pair of lines equation is which of the following and your options are? And your options are option A, alpha, option B, to alpha, option C, beta, option D, to beta. If you're wondering what is alpha and beta, look at the question. Question has alpha and beta in them. Okay, let us take a poll on this. Three minutes, can I have three minutes for this? Time starts now. Guys, one more minute. Only one person has responded so far. 30 seconds more. Here angles means the acute angle is what I want. It's time up, but still I'm going to give you 30 seconds more. Sorry, I'm very particular about the time you're taking because we have to work under time constraints. Nobody is giving us infinite amount of time to solve these questions. Three and a half, four, max four and a half is what we should be eyeing for. In next 10 seconds, I'm going to close the poll. Press in any button that you feel you're going to take a chance. Next 10 seconds, I'm going to switch off the poll and the poll. Press on something. Guys, fast, fast. Four, three, two, one, stop. And here is the result. 65% Janta has given B as their options. And then the next heavy weight is your D. 22% of you say D is the option. Okay. Let's discuss this and see who was correct. Okay, Pratap, let's check. So now, first of all, we need to understand is this a homogeneous second-degree equation? I think so. It is because every term over here is a second-degree because second-degree term, second-degree term, and even if you square this up, it's only going to give you a second-degree term. So let us write it down first in our normal form that we have known it as. That is AX square to HXY and BY square. Okay. So let me collate all the terms which are associated with X square. Or let me collate the coefficient of X square, so as to say. So you'll have sine square alpha coming from here and you'll have minus cos square beta coming from here. Okay. Let me collate all terms containing XY. So XY can only be obtained from this guy, isn't it? So if I bring it to the left-hand side, it's going to give me 2XY, sine beta cos beta. Okay. Let me collate Y square terms. Y square terms will be having sine square alpha here and from this guy, I'll get sine square beta also. So first and foremost, you should have got your expression like this. Now, this is playing the role of my A. This is playing the role of my H. And this is playing the role of my B. Any doubt regarding this, guys? All good so far? Great. So now let's use the formula tan theta is 2 under root. Guys, don't miss out on this too. I've seen many people forgetting this too. Okay. This too is very important. So 2 under root X square minus AB by mod A plus B. Let's not forget mod also. Okay. So now let me fit this A, H, B, et cetera into this. So 2 under root of H square. H square will be sine square beta cos square beta minus AB. AB. Okay. All divided by mod of A plus B. A plus B. Okay. This is just a game of trigonometry. It should be very good in your trigonometric identities to solve this problem. Now, in the denominator, if you see these two guys, they will leave minus one, isn't it? So at least denominator, let me simplify this. Mod of 2 sine square alpha. That means collect these two and these two will be minus one. Okay. On the numerator, you'll end up getting sine square beta cos square beta. You'll end up getting minus sine to the power 4 alpha. You'll end up getting plus sine square alpha sine square beta. And again, plus sine square alpha cos square beta. I don't have my space over here so I'll drag it. And you will have a negative sine square beta cos square beta. This term, this term goes for a toss. From these two terms, take sine square alpha common and you'll end up getting sine square beta and cos square beta. That will give you one anyways. So let me show you what you finally see. You see sine square alpha minus sine to the power 4 alpha. Okay. And down in the denominator, we have the same term. We shorten this up. Okay. Now take, in fact, include everything under the root side. Let's do one thing. We have everything under the root side. So this 2 will become a 4. And here you will have sine square alpha if you take common. 1 minus sine square will become a cos square. I can. And down over here, I can write this as 2 sine square alpha minus 1 the whole square. Isn't it? Because mod when introduced within the root will become a square. Okay. Now, numerator is nothing but sine square 2 alpha. Check it out. And this is basically coming from your identity that cos 2 alpha is 1 minus 2 sine square alpha. Yes or no. So isn't this square of cos 2 alpha? So ultimately, you are trying to simplify this guy, which is very simple tan 2 alpha. And that means your angle is 2 alpha. That means your option B is going to be correct, my dear. So Janta was right. Janta was right. Sure Aditi. See, this step you want. See, this step I took sine square alpha common. By the way, 2 I introduced inside that will become a 4. So if I take a sine square alpha of here common, I'm just writing it down here short. Won't it look like this? And won't it look like this then? Okay. That's what we have. And isn't this... Now we have 4 sine square alpha. Isn't it sine 2 alpha square? Because sine 2 alpha square is nothing but 2 sine alpha cos alpha square. Yes or no? So from here to here, I'm writing this guy as sine square 2 alpha. And the denominator part was nothing but... The square of this term, isn't it? See, it doesn't matter the position of 1 and 2 sine square alpha because it is subjected to square power. Okay. So it is as good as saying cos square 2 alpha. So denominator I replaced with cos square 2 alpha. Now sine square 2 alpha, cos square 2 alpha is tan square 2 alpha. Under root of that is mod. And since we are already taking acute angle, mod becomes irrelevant because tan of 2 alpha will be positive. So those who answered with 2 alpha, brilliant. Absolutely correct. Well done. Time to move on to the next concept guys. This concept is equally important. Equation of angle, bisectors of a pair of straight lines whose equation is a homogeneous second degree. Now people must be wondering. Sir is only talking about homogeneous, homogeneous, homogeneous. What about the general cases? Guys, we'll come to it. First we have to master the homogeneous part. Your general equation is basically evolved out of this homogeneous thing. So first we have to master this homogeneous thing. Don't worry, we'll come to the general form also. Okay. So look at the concept that we are talking about. We are talking about concept of angle bisectors which means pair of bisectors. Guys, we are only dealing with pairs, pairs, pairs, pairs, pairs. So pairs of bisectors of these lines, the lines which constitute this pair. Okay. So let me draw this out for you. So let's say ax square plus 2hxy plus by square are these two blue lines. Okay. Where will they meet? Where will they meet? Origin, needless to say that. Now we are looking for the equation of these angle bisectors, these yellow ones. Let me name them. So L1 and L2, let it be your original lines and B1 and B2, let it be your bisector equations, bisector lines. So how will I get the equation of these B1 and B2 combined? How will I get the equation of these angle bisectors combined? Any idea? By the way, all of you know angle bisectors, it bisects the angle between the lines. So this angle and this angle will be equal, this angle and this angle will be equal. So kindly unmute yourself, anybody, and help me how do I resolve this situation? How would I get the equation of the pair of bisectors? Any idea? Say something. Okay. Have you all done the concept of bisectors of straight lines in your class 11th straight line chapter? What was the core principle? The core principle was if you choose any point on the bisector, let's say I choose this point to be h, k. Okay. Can I say the distance, the perpendicular distance of this point from the two given lines will be equal, this will be equal to this and it doesn't matter whether you choose it on B1 or B2. So even if you choose it at B2, I can say that if I drop a perpendicular from B2 on to this line or to this line, okay, these two lengths will be still equal. Okay. So any generic point h, k, if you drop, if you drop a perpendicular on to the two given lines L1 and L2, the two lengths would be equal. This is the core principle and we're treating this question as if we are solving a locus question and I'm trying to find locus of all such points whose perpendicular distances from the two given lines L1 and L2 are equal. Is this clear? Is this argument, is this statement of mind clear? So the basis of getting the equation is I'm trying to find out the locus of all such points, all such points, whose perpendicular distance, whose perpendicular distance or whose distance, if you don't say perpendicular, it is automatically perpendicular, whose distance from L1 and L2, where L1 and L2 are the lines which constitute this pair, are equal. This is my guiding principle, correct? Any question with respect to this? Good. So now you tell me, let's say I call this y equal to m1x and I call this blue line as y equal to m2x. What is the perpendicular distance of h, k from y is equal to m1x? Everybody knows the perpendicular distance formula? What is the perpendicular distance of h, k from this line? k minus m1h by under root of 1 plus m1 square mod of it because distance is positive. In the same way, what is the perpendicular distance of h, k from y equal to m2x? But let me write it like this, y minus m2x equal to 0. Everybody knows distance formula, right? Hello friends, can you hear me? Yes sir. Don't give me a shock sir, where did this come from? So you must all be knowing this formula by this time. So it will be k minus m2h by under root of 1 plus m2 square mod. So by this locus condition, they both must be equal, correct? Now we'll resolve this, we'll square it and we'll do all kind of simplification to get your combined equation, combined equation, combined equation of b1, b2. Okay. So guys, please allow me to replace k with y and h with x because sooner or later I have to do it. So can I do it now only with your permission? So let me square and cross multiply and replace my k with y, h with x. Is that fine? Okay. Now, a lot of simplification has to go in here because m1 and m2 is something which I have assumed. The question never said m1, m2 separately. The question just gave me the equation of this pair of straight lines, right? The question only mentioned this. So this m1 and m2 need to go. How will they go? We'll use our old relation that we have been using so far. What is that? m1 plus m2 is, m1 plus m2 is, say in your mind, minus 2h by b. m1 into m2 is what? a by b, yes. So I'll use this here. Okay. So before that, I'll simplify. First of all, I'll collect all the y square coefficients. Guys, I have a habit of collecting it without expanding it. I can handpick. So y square coefficient will be this fellow. And by the way, let me bring it to the other side. So let me remove this and write a negative here and put a zero on the other side. Okay. So y square coefficient will be 1 plus m2 square minus 1 plus m1 square. Isn't it just m2 square minus m1 square? Correct me if I'm wrong. Okay. What about minus x square coefficient? See, minus x square coefficient will have m2 square 1 plus m1 square minus m1 square 1 plus m2 square. Am I correct? Guys, let me know if I missed out on anything. Okay. Can I resolve this here itself and write it as again m2 square minus m1 square? Okay. This is something which you can do it at your end also. You don't have to see what I'm doing. You can do this. Try it yourself also. Let me write it. Try simplifying yourself. Try simplifying yourself. You don't have to watch me here. Okay. And now I'm going to simplify for xy term. So xy term, if I'm not mistaken, I can take 2xy term in fact. 2xy term. 2xy term will have minus minus m1 into this fellow. Okay. And from here also I will get plus m2 into this fellow. So let me simplify that also. And write here m2 minus m1. So I'm just taking care of the ones here. And minus m1 m2 m2 minus m1. Okay. Again, as I told you, try it yourself. I'm giving you one minute. Everybody just verify whatever I've got. Is it matching with what you have got? Then we'll move on. One minute everybody, check your calculations. Only when I've got a green signal from you, I'll move on. Just type matching on your chat box if it is matching with my result. Trippan sits matching. Aditya is also matching. Siddhartha also matching. Okay. Okay. Now we'll do one thing. I can clearly see that m2 minus m1 factor can be dropped from everywhere. Now I'm dropping it under the assumption that the slimes will not be coincidental. Okay. So let me write it here. So m1 is not equal to m2, which means m1 minus m2 is not equal to 0. So I can drop that factor from everywhere. So I'll end up getting m2 plus m1. Okay. Here also I'll get m2 plus m1. So I'll act smart over here. I will group y square minus x square also. Okay. From here I'll get 2xy. This will become a 1 and this will become a m1 minus m2. Wow. This brings a smile on my face because I have been able to successfully convert it into m1 plus m2 and m1 m2, whose results are already known to me. And what is that? Minus 2h by b. And here also 1 minus a by b. Getting my point. Okay. Let's simplify this further. Let me take one of the terms to the other side. Let me write it as b minus a by b. I think b could be dropped. I think 2 could be dropped. Okay. And now we can write this as y square minus x square by b minus a is equal to xy by h. Or many books will write it like this. x square minus y square by a minus b is equal to xy by h. The formula is your equation of the bisectors. Combine the equation of the bisectors and this also has to be memorised. Not memorised. You have to keep in mind because we can't sit and derive this every time. Is that fine? Note it down. Guys, again, I have a habit of analysing my result. The first analysis that I would like to do here is what is the coefficient of x square in this particular bisector? What is the coefficient of x square? 1 by a minus b. Agreed? See. This is the coefficient of x square, isn't it? What I am showing with this boomerang structure. Isn't it? In a similar way, what is the coefficient of y square? Coefficient of y square is negative of 1 by a minus b. What happens when you add them? You get a 0. And remember when we were talking about angle between pair of straight lines, I told you when the coefficient of x square and y square add up to 0, that means the lines are perpendicular. That means your bisectors are perpendicular. Which we all know is true. Our bisectors are always at right angles to each other. This is 90 degree always. See further it justifies our formula also. Second analysis that is very important. What if a is equal to b? What will happen if a is equal to b? Will hell break loose? What will happen? Will the bisectors be undefined? Because here the denominator here will become 0, isn't it? Look at the motion of my pen. This guy will become 0 if a is equal to b. Yes or no? So what will happen in that case? Will the bisectors cease to exist? What will happen? Anybody? See we can have a pair of straight lines whose a is equal to b, right? That means all those straight lines, this guy will become 0, isn't it? Now in that case, guys don't panic. In that case, your equation will just become this is equal to 0. This will be your bisector equation because this will just become a symbolic representation. It is not an operation that you are dividing by 0. So this 0 can go on the other side and that will leave you with x square minus y square equal to 0. Guys let me tell you when you are writing an equation, it is a symbolic representation of how the points on that particular line behave. Getting my point. Later on in 3D geometry, you will come across such weird equations. x plus 1 by 0 is equal to y minus 1 by 2 is equal to z plus 3 by minus 1. So don't think like, oh my god, how did somebody divide by 0? This is not possible in maths. No, it is just symbolic. Symbolic means it doesn't have any operational power. Just like the president of India. He is the symbolic head of our country. He doesn't take any decisions. It is taken by the prime minister, Mr. Modi. In the same way, when your aim is equal to b, it just becomes x square minus y square equal to 0. That's it. Don't panic in this case. Similarly, if your h becomes 0, your equation just becomes x y is equal to 0. By the way, who will tell me x y equal to 0? Which two lines does it represent? x axis and y axis. Getting my point. Are you getting this? So in any situation where your h is 0, let me just take a simple example. y square minus 3x square. Let me just show you on GOG also. y square minus 3x square equal to 0. You see your bisectors are your x and the y axis. Let me type that also. x y equal to 0. In fact, it's not showing over here. I don't know why it is taking this, but x y equal to 0 is nothing, but your x and the y axis. I know why it is taking this. Anyways, x is 0 means you're this line. I'll show you where x is 0 means you're this line. y axis. And y equal to 0 is your x axis. So you can see your y axis is bisecting this angle. x axis is bisecting this angle. Getting the point. Let me move back to this scenario. Can we begin with a question? If you have understood this, let's have a question on this. I'll give you a break. Let's have this question first. In case you're not able to read this hidden part, I'll write it down for you. This is 3 x square minus 5 x y plus 4 y square equal to 0. So the question is, find the equation of the bisectors. Bisectors means pair of bisectors of the angle between the lines represented by this. Do it just one minute for it. After one minute, please type done so that I know you're done. Yes, it's hardly a 10, 20 seconds question because it's just use of the formula. Done, done, done, done, done, done. Okay guys, let's discuss quickly. a is 3, b is 4, h is minus 5 by 2. And we know the bisector equations. Bisectors equations are x square minus y square by a minus b is equal to xy by h. So just have to put the values of a and b. That's 3 minus 4 and h is minus 5 by 2. Let's simplify this. So 5 by 2 will go up. Minus sign will go off because of the minus in the denominator. So that will give you 5 x square minus y square is 2xy. And this becomes your final result. As you can see coefficient of x square and y square will add up to 0. That means these two lines will always be perpendicular. Everybody happy with this? Look out easy question. Let's move on to the next one. Okay, look at this question. I'll read it out also for you. If pairs of straight lines given by this and this be such that each pair bisects the angle between the other pair. Prove that pq is minus 1. Again, this can be objectified. If you approve that question, that doesn't mean you will not take it seriously. It can always be objectified. Now what is the situation? I'll draw it for you. See, let's say this pair of straight lines. You can see the motion of the plus symbol going on. By the way, this is perpendicular lines, isn't it? Because x square coefficient and y square coefficient add up to give you 0. So these are two perpendicular lines. Let me show you by these green lines. So this green line is your this pair. So this is your green lines. Okay, and there's another pair. Let me show that with blue, which is also perpendicular. Oh my God, these guys are also perpendicular. Okay, so let me show it like this. And at the same time, they're bisecting the angle also between each other. Got it. So I've drawn in such a way that you will see that the green lines are bisecting the angle between the blue ones. And the blue ones are bisecting the angle between the green ones. Okay, then in such condition prove pq has to be equal to minus one. Two minutes. Once you're done, type done so that we can discuss it. Okay, done, done, done for most of you. Should we discuss this? This is very easy. First, let us focus on the first pair. Let's write down the equation of the bisectors. So what will be the bisectors for the first pair? Use the formula x square minus y square by a minus b is minus one. So we'll end up getting one minus of minus one is equal to xy by h. H is minus p. Hope you have not done a mistake in identifying the coefficients. So this gives me minus px square plus py square is equal to 2xy. That is minus px square minus 2xy plus py square equal to zero. This is your bisector. Okay. Now the question claims this equation is your blue lines. That means these two lines are basically representing the same pair of state lines. Isn't it? That's what my question claims. So if these two are the same, I can solve this by comparing coefficients. Let's compare coefficients of x square, xy, y square in both these lines. Okay. Because my question is claiming that these two guys are same equations. So if you compare the coefficients, you can say minus p by one. See minus p and there's a one over there minus p by one is minus two by minus two q. Is equal to p by minus one. Okay. By the way, the first and the third are basically the same things. So nothing will happen from these two. What is going to happen is from the middle two. So these two says minus p by one is equal to one by q. Just cross multiply. So p q is equal to minus one proved. Is that clear? Okay. Any question you have, please ask me. If no, we'll move on to the next question and post that I will give you a break for 10 minutes. Read the question carefully. If the lines represented by x square minus two p x y minus y square equal to zero are rotated about the origin. The angle of theta one in clockwise and the other in anti clockwise direction. Find the equation of the bisectors of the angle between the lines in the new position in the new position. Is it done guys? I don't think so. It should take a lot of time. This is a trick question. Tell me frankly, do you believe that the bisectors of this line and the bisectors of the new line will be any different? Correct Sana. It would be the same. See, it's very simple. Let's say this is the white line and this is your bisectors. Okay. The dotted yellow lines are your bisectors. Okay. If I move this line by theta, let's say I bring it to a new position. Let's say something like this. Okay. So I moved it by theta in a clockwise sense. This I moved by theta in an anti clockwise sense. Okay. Do you think that the bisectors are going to change? No, it will not. The yellow ones would still remain to be your bisectors. Correct. So this question was actually just to trick you. Nothing else. So your final answer would be the answer that you obtained for the bisectors of these guys. What is that? Let's quickly write that x square minus y square by one a minus b is equal to xy by h. Okay. Let's simplify this. So minus px square plus py square minus 2xy equal to zero. This is your answer. Okay. Now let's have a break. Okay. Let's have a break for 10 minutes. Okay. We'll resume at... Look at the time here. It's 11.08. We'll resume at 11.18 am sharp. Okay. See you in 10 minutes. Eat something, drink something, come back. So guys, now we are going to talk about the heavy part of this chapter. When I say heavy part, it is slightly more challenging as what we did before the break. I would request everybody's full attention over here. Now, instead of homogeneous second degree equation of pair of straight lines, we are now going to talk about general second degree equations of pair of straight lines. Of pair of straight lines. Yes, my dear. Now, a pair of straight lines can also acquire the shape of a general second degree equation like this. When I say general second degree equation, I mean to say, apart from x square, xy, y square, there could be other terms also like 2gx, 2fi and c equal to zero. That means it can resemble the equation of a conic. I'm sure most of you would recall from our previous year that a conic also has a general second degree equation as its equation. If you take a circle, circle also has an equation somewhat like this. It's a different thing that h is zero in a circle. Ellipse can also have an equation like this. Hyperbola can also have an equation like this. But Abula can also have an equation like this. And so can a pair of straight lines also. That's why pair of straight lines is sometimes treated as a special case of a conic where your eccentricity of a hyperbola has been made infinity. If your hyperbola's eccentricity goes infinity, the hyperbola will actually, the two arms will coalesce and become a pair of straight lines. This structure that you know of a hyperbola, this will slowly come together and this will become a pair of straight lines like that. Now, it's very obvious from the fact that these three terms have come because of the presence of constant terms in the equation of the lines which constitute this. See, earlier when we were talking about ax squared plus 2hxy plus by squared equal to zero, we know that this was actually made up of two such lines which did not have any constant term, isn't it? As you can see, y minus m1x and y minus m2x doesn't have any constant term. If I introduce a constant term in these two lines, that would give birth to these three terms. That means if such a line is to be factorized, it would be of this nature y minus m1x minus let's say c1 and y minus m2x minus c2. So because of these constants c1 and c2, these three terms have come into picture. First of all, do you all appreciate this? Second of all, I want you to appreciate the fact that these three terms actually carry the information about their slopes. So these three terms, they carry the information info, let me write it in shortcut. They carry the info about the slopes of the line. As you can see, m1 and m2 terms, so first this two term and this two term if you multiply, you'll end up getting only the second degree terms. So the constant terms do not have any role in building up these three terms. They have only the roles building up in the last three terms. So I will just summarize what I said so far. The last three terms come because of the presence of constants that is c1 and c2. So if you remove your c1 and c2 from here, you'll only get the first three terms. Second thing, what I said was the information about the slope of the two lines, which actually make up this pair of state line is only carried in the second degree terms, not in the linear terms or the constant term. Is that clear? Is that clear everybody? If it is not clear, I'm going to ask you a question. I'm going to ask you a question. Okay, let me put a question for you. I'm looking for the right one. Yes, let's have this question. Hope you can read this question. Yeah, Aditya, I'll repeat once again. See, whenever you see a general second degree equation that is of this nature, the first three terms that is this guy, this guy and this guy, they hold the information about the slope of your lines. And the last three terms, which I'm tickmarked with yellow, they come because of their constant terms in the line. If you multiply this, you'd realize y2, xy and x2 terms would be your first three terms. Other terms would be leading to your x, y and constant. Can I go to that? Do you need it? Yeah? Is that fine? Okay, so let's talk about this question. Find the combined equation of the straight line passing through 1, 1. This point is 1, 1. And parallel to the lines represented by this equation. Let's see from this particular information, how many of you are able to crack this question? The idea is there are two lines which are given by this equation. I have to get another set of lines. Let me make it blue, yellow ones like this. Says that the yellow lines are parallel to your blue ones. So this is parallel to this and this is parallel to this. And they are passing through 1, 1. So I want this equation. What is the equation of these guys? This equation is already known. You can type it out on your screen. Okay, so Shamik has given an answer. Shamik, your expression doesn't contain any x or y or constant term. Are you sure? That means your answer is suggesting that the yellow lines will pass through origin. Which is not the case. They are passing through 1, 1. Okay, change your answer then. Anybody else who would like to answer this? Okay, let me handle this. See what did I say a little while ago? These terms which I am marking with blue, they carry your information about the slope. Yes or no? These carry the information about the slopes of the line. Something like the DNA. So first let us factorize them. So if I were you, I would first factorize this. Okay? So it can be easily be factorized, split the middle term. Not a very difficult thing to factorize here. So if you factorize that, you will end up getting x minus 4y. And let me put a bit of gap over here. And x minus 5. Okay, that means you come to know that the slopes of these lines will be 1 by 4 and 1 respectively. Okay, Shamik. Okay Pratam, we will check it out. Cool Shabris. Okay, now why did I leave a gap over here? I left a gap over here because there were other terms also like x, 2y and minus 2. And these terms came because of some other constants present along with this. That is what I was trying to convey my dear. Now is it clear why I was saying that the last three terms come because of the presence of constant terms in the equation of the line? Okay. Now how do I find p and how do I find q is something which I'll take up little later on also. In fact, I'll take here also. So in order to find p and q, by the way, we don't need them to solve this question. I'm just doing it for the sake of your full understanding. So how do you find p and q? So you must be wondering slopes are fine, but I need to know the constants also of the line. Okay, for that you compare the coefficients of x and coefficient of y from this and this. So what will the coefficient of x from here? Can anybody tell me x coefficient? What will it be? So when x multiplies with q or when p multiplies with x. So p plus q or q plus p is your coefficient of x and that should be compared with one over here. See what is the coefficient? Getting my point. How will I get my other equation? That is by comparing the coefficient of y. So why term here? How will I get this check minus four multiplies with q to minus four q and p multiplies with minus one. So minus p. And this will be your coefficient present over here, which is two. So two equations, two unknowns, we can solve it. Let's add them. So if you add them minus three q is equal to three. So q is negative one. If q is negative one, p will become minus, sorry, p will become plus two. You can also check that when p and q multiply, it should give you minus two. Is it giving us? Yes, it is giving us minus one into two is giving us this. So this is correct. So this line that you see over here on your question was actually made up of these two lines. x minus four y plus p equal to zero and p was two. And the other line was x minus y minus one equal to zero. Okay, even though it is not required here, but I'm just telling you so that in future if you require it, you should be able to split the pair of straight lines into their constituent lines. Is it clear? Yeah, sure, Aniruddha. So what I did was, see Aniruddha, what I'm trying to say is these three terms, they help us to get these two. And because there are other terms also, that is x, y and two, it has come because of the presence of constants over here. So I assumed it to be p and q. Okay. Then in order to find p and q, what I did was I first found out what is the coefficient of x that we will get from this entire expansion. That you can do it at your end also. So p into x will be one coefficient and q into x will be another coefficient. So p plus q will be a coefficient. And that is compared to the coefficient over here, which is one. Okay, see in the equation we have been given, this is the coefficient one over here. Sorry, I'm writing in yellow there. We're writing in black. One over here, isn't it? So this I'm comparing with one. In the same way I compared the coefficient of y also with two. This guy. Getting my point. And these two are simultaneously solved. And I got my p and q. Clear? I'll come to that. I'll come to that, Siddhartha. I have not solved the question. I'm just saying that I'm trying to split the line into two constituent lines. Wait, Siddhartha, wait. I'm not solved it. Okay. Now my question is I need to get two such lines which are passing through one comma one and parallel to L1 and L2. Okay. So I will say let a line parallel to, let a line parallel to L1 be x minus 4y plus let's say lambda equal to zero. Correct? As we know that when lines are parallel, coefficient of x and coefficient of y doesn't change or may not be changed. And but constant term should be different. Now this line should pass through one comma one. This line should pass through one comma one. So put a one and one in place of x. So what does lambda become? Three. So the line which I'm looking out for let's say L1 dash that will be x minus 4y plus 3 equal to zero. Okay. In a similar way we can say let the line parallel to L2 that is your this line be x minus y plus some beta. Again this should pass through one comma one. So one minus one plus beta should be zero. So beta is zero again. That means my L2 dash will be x minus y equal to zero. Now the question setter is asking me to give the combined equation of this. So all you need to do is multiply them. Job done over. Let me expand it also for you x square plus 4y square minus 5xy. And you'll end up getting plus 3x minus 3y equal to zero. This is your final answer. Have I made sense here everybody? Is that clear? Aniruddha understood? Siddhartha understood? Ananya. Ananya is it clear? Okay. Yes sir. Awesome. Now guys a very very serious part of this chapter. I would like you all to pay attention. It's not that whatever I was saying so far was not serious. But here is the slightly difficult part. So for the next half an hour I want you to be very very attentive. Here and there if you miss out, gone. Okay. So stay glued to your seats. Don't look around. Okay. Guys do you remember last year I had given you that if such an equation. If this equation were to represent a pair of straight lines. Then it must satisfy that is the coefficients a, h, b, g, f and c would be related to each other by this relation. a, b, c plus 2f, gh minus af square minus bg square minus cs square equal to zero. Do you remember this? If you don't or if you have not come across this expression so far in your life. Time is now to remember this. So this has to represent a pair of straight lines. This expression must be zero. This expression is what we call as delta also for in shortcut. By the way, this delta can also be written as the determinant. Hope you guys have worked with determinant before. Okay. So as a determinant, you can write it as a, b, c, h, g, h, g, f, f. So this guy, what you have written over here is actually a determinant. Anybody over here who doesn't know how to deal with determinants. Anybody. Please type. I have never, please type not dealt with determinants in any form, whether in physics or in maths. chemistry. I really doubt you have seen a determinant. Okay. I do not know determinants. No worries. I'll just tell you how a determinant is expanded. Okay. No worries. Anita and just look at this. How do you expand the determinant? A determinant is expanded with respect to any one row or any one column number one. A determinant is expanded. So if I have to expand it, I'll have to, I'll get this isn't, I'll show you how it works. A determinant is expanded. There's a separate chapter for this. So anyways, we'll do that chapter. Any determinant is expanded with respect to any one row, any one row or any one column. Okay. Let me give somebody a chance to choose a row or a column. Let's have Aditi. Aditi Balaji. Aditi. Which row or which column you want to choose? Just write R dash the number or C dash the number, whichever you want to choose. R dash what? Okay. R dash 2. So she's choosing R2. This is your R2. By the way, this is row number one, row number two and row number three. Okay. So Aditi has chosen R2. So I'll expand this determinant with respect to R2. Now how it works. Now first element H. It comes in the second row first column position. Isn't it? Okay. So write second row first column as powers of minus one. So minus one to the power two plus one. Okay. Then write the H. Okay. Put a bracket. Okay. Now hide the row and the column in which H falls. So hide this from your fingers. You can hide it. What do you see? H, G, F and C, correct? Just cross multiply. HC minus GF. So right here. HC minus GF. Clear? Next. Next element in row number two, which Aditi has chosen is B. Okay. By the way, let me erase this so that you're not confused which rows we are hiding. So I'm erasing this. It's very, very attentive you should be because there's something which is new for you. The next element is B. B comes from second row second column position. So write minus one. Right up plus over here minus one to the power two plus two and write B. Okay. Then hide the row and the column in which belongs. Then hide the row and the column in which B belongs. What do you see? HCGG, right? So multiply it. AC minus G square. So cross multiply and subtract. In the same way do for F also, which is minus one to the power second row third column into F. Hide the row and the column. Let me show you on the figure. So hide the row and the column in which F falls. So this will be hidden. So AF minus GH. Is that fine? Correct? Now, if you write this in a proper form, you will have minus H times HC minus GF. You will have plus B into AC minus G square. And you will have again minus F into AF minus BG. Sorry, GH. Expand it. So you have minus CH square. You have plus FGH. You have plus ABC. You have minus BG square. You have minus AF square. You have plus FGH. Collect the terms together. So ABC, I'll write it first. Then I'll write FGH, FGH, which is 2 FGH. Then I'll write AF square. Then I'll write BG square. Then I'll write CH square. So this is the expansion of this determinant, which is actually the same thing which I have written over here. Check it out. This is the same thing as what we have written over here. Trust RO. Okay. Is it understood roughly how a determinant is expanded? Don't worry too much. I will take up a separate chapter on this altogether. Yes, S, whatever is your full name. The determinant will remain the same irrespective of whichever row you are choosing or whichever column you are choosing to expand it. Trust me, the result will not vary. Try this exercise after the class today. Pick up any other row or any other column and try expanding it. Your result will still remain the same. Good enough? Okay. Now, normally we write it like this because it is easy for us to remember. Okay. This expression many people forget. So determinant, why is many people remember it? Okay. Now the question comes, prove it. Isn't it? How does this expression come when you say this particular equation should represent a pair of straight lines? So what I'm going to do is I'm going to prove it. So I would need everybody's attention over here. See, that is something which is related to cofactors. Aditi, right now I just told you a rule. Just follow that rule. How that rule comes, that we will discuss when the chapter is taken up. So if I start teaching you determinant now, it will be like another class in itself. Okay. So don't worry about how this minus one and all comes. It is just a rule which says take the element, whichever row you have selected or whichever column you have selected. Let's say, had you selected your third column, then you would have written minus one to the power three plus one over here. G, then hide the column in the row in which G falls, then cross multiply the other elements. Okay. This is how the signs are governed. How it comes, what is the theory behind it? We will take that up in the determinant chapter. Okay. Don't worry too much about it Aditi right now. Method should be clear. Is that okay? Aditi? Happy? Okay. Don't worry. I'm not going to use determinant in this chapter. It is just for your extra knowledge. Okay. Now, proof. How do we prove that when AX square plus BY square plus 2HXY plus 2GX plus 2F5 plus C equal to zero represents a pair of straight lines. Then that delta thing must always be zero. And what is the delta thing? In your mind, ABC plus 2FGH minus AF square minus BG square minus CS square. Very important. This is a very, very difficult expression to remember. We will forget this if you don't practice enough questions. Okay. So now I'm trying to prove that this must be zero for it to represent a pair of straight lines. So prove number one. Let us begin with our equation of the straight lines. Okay. Now I will try to give it a shape of a quadratic in X. How do I do that? See here. Okay. Do you see that I have given it a structure or a shape of a quadratic in X? Okay. Guys, you have to be very attentive over here because this is a difficult kind of a proof. So here my, this term is your like B term of your quadratic and this term is like your C term of your quadratic. Think like this. Okay. So it is a quadratic in X where coefficient of B and constant term themselves are functions of Y. Very strange, right? Because we can, we have never seen quadratic whose constants themselves are some functions of other variable. But yes, we can always take it like a quadratic in X. Now, if you have taken it as a quadratic in X, I can also apply my quadratic equation formula to solve it. That is minus B plus minus B square minus 4 AC by 2A formula. Shridharacharya formula. So let us apply that. So if I do that minus B, this will be a minus B, isn't it? So minus, let me write it as 2 common. So minus B plus minus B square. B square will be 4. HY plus G the whole square minus 4 AC. Okay. Whole divided by 2A. Is this clear so far? Any doubt regarding this? Step number one, I structured my pair of straight lines as a quadratic in X. Step number two, I have used my quadratic equation formula. Now step number three, very important. Listen to me very, very carefully. I'm going to slightly factorize the term in the under root sign. So this term, which is the under root sign, I'm going to first simplify it. So first Y square terms, I will write it. So Y square terms would be, by the way, 2 also you can expand. Very good. Somebody was suggesting me that, remove the 2 outside. In fact, 2, 2 can be cancelled off from everywhere. Okay. Okay. If you take the coefficient of Y square from everywhere, it will be H square from here and it will be AB minus AB from here. Okay. So I'm just writing it properly. Next Y term would be, Y term would be coming from this guy 2HG or you can say 2GH and from here it will come 2AF. Okay. So 2GH minus A to AF. Okay. And constants would be G square minus AC. Correct me if I'm, I've missed out anything. So step number 3, I just simplified this term. Okay. Nothing else. Now guys, listen to my claim here. I claim that this term, which is under the under root sign, which I'm showing with the curly bracket, this must be a perfect square. If it has to represent a pair of straight lines, can somebody tell me why? Anybody, please unmute yourself. I have made a claim over here that the term in Y, the quadratic in Y, which is under the under root sign must be a perfect square. Why? You must be thinking, sir, you only have made this claim, so you should justify why and you're asking us. No, I want you to see whether you are able to crack that, whether the same feeling which I have, you have the same feeling as well. Anybody, unmute yourself. Talk. Why that should be a perfect square? Nobody, anybody? Vibhup says because two lines must be obtained. You are almost 80% correct Vibhup, but a better way to justify it is, see, by this equation, you are trying to write a line X in terms of Y, isn't it? So let's say I have a line AX plus BY plus C equal to zero and I'm writing X the subject of the formula. Okay, so something like this. So I'm writing X as minus C minus BY by A. Okay. Do you see that when you're writing X in terms of Y, there cannot be any under root sign over Y or Y square because if there is, let's say, I give you an equation like this, okay? Y square plus Y plus one by let's say A. And ask you, does it represent a line? Does it represent a line? You will say no, sir. It cannot represent a line because there is an under root over a quadratic. These are not how your lines look like, isn't it? So in an equation of a line, if you are expressing X in terms of Y, you should never get an under root of a quadratic. If you're getting it, it should actually convert it to a linear term in Y. It can only happen when this term is the perfect square. Now does it make sense to you? Y should be a perfect square, okay? Nobody has a question on why it is a perfect square. So a quadratic must under root of a quadratic in Y. As you can see, it's a quadratic in Y. It must give you linear answer. That can only happen when that quadratic is a perfect square, my dear. So I'm claiming here that Y square X square minus AB plus Y, or you can say too common. This term must be a perfect square. In other words, I'm stating the same thing in a different language. Try to see whether you have understood that. In other words, this guy when equated to zero must have equal roots. Yes or no? Because equal roots come from perfect squares. See, I give you a simple example. X square minus 2X plus 1. Isn't it not a perfect square? We all know it's a perfect square, correct? If you equate it to zero, doesn't it give you two equal roots, one and one? The same thing I've applied over here. If this is a perfect square, you would agree that this equation will have equal roots. Any doubt regarding this? And if this has equal roots, my dear, can I say that the discriminant of this, the discriminant, now don't get confused with determinant and discriminant. Both are different things. I've seen many people misusing the word discriminant as determinant. The discriminant of this, which is your B square minus 4 AC term, should be what? Zero because it has equal roots. Yes or no? So far so good. I'm taking a pause for 10-15 seconds. Any question here? Please ask. The chronology is clear. From where am I coming? Okay, now B square, what is the B over here? This guy is your B, so 2 into GH minus AF is your B. So B square will be 4 GH minus AF square minus 4 AC. Let me bring the 4 AC on the right-hand side and say equal to 4 AC. AC, A is this guy, C is this guy. Agreed? I've reached a very critical step now. In some time, you will see the result in front of you. 4, 4 gone. Expand muddy. G square, H square, A square, F square minus 2 AF GH. From here also, G square, sorry, H square, G square minus ACH square minus ABG square plus A square BC. Cancel? Cancel. Drop A. Drop A from everywhere. We'll end up having AF square minus 2 F GH is equal to negative C H square, negative BG square plus ABC. Bring these two terms to the left side. Sorry, right side. We'll end up getting ABC plus 2 F GH minus AF square minus BG square and minus C H square equal to 0. This is what we wanted. This is our result. Happy face. Go at the top of the page. This is what we wanted to prove. See, this is what we wanted to prove here. The same thing we have got. We wanted to copy the proof or anything. Let it stay with you as a document over here. But this is the idea behind it. Getting my point. Remember last year I didn't prove it. I told you wait for the right time for me to prove it. So now the time has come that you appreciate the proof. But many people raise a question regarding this proof, including myself. What would happen if this guy A were 0? Then everything will go for a toss. But I say even if A is 0, the result would not change and let me defend my argument. So let me go back to the next page. Anything you want to copy here or can I move on to the next page? Anything that you would like to copy here, I would recommend not to copy anything. I'll send this PDF to you on the group. Then you can easily copy whatever you want. What else? Should we go to the next slide? Now some people say what will happen if A is 0? Let me defend my argument that nothing will happen if A is 0. The result would still persist to be the same. How? Let's see. If you A is 0, your equation would reduce to BY squared plus 2HXY plus 2GX plus 2FY plus C equal to 0. A term will be disappearing. There was an AX squared term. Now this term will be no more because your A is 0. So now this is your pair of state lines equation. Again, let me treat this as a quadratic in Y. So I'm treating this as your A of the quadratic. I'm treating this as the B of the quadratic. I'm treating this as the C of the quadratic. Again, can I use my Shridhar Acharya formula? Thanks to this guy who has made our life very easy. So minus B plus minus under root B squared. By the way, I'm just pulling out a 2 out. So HX plus F the whole square minus 4AC by 2A here. A here is my B. So I like 2B. But the very same argument which I gave in the previous slide, can I say this guy must be, must be, must be perfect square. Because I've never come across a straight line equation which has got a under root of a quadratic. Isn't it? So in other words, let me write this separately for you. In other words, this guy must be a perfect square. Correct? In other words, this equal to 0 must have equal roots by the very same logic. Correct? You may draw 4 factors from everywhere. A factor of 4 you can remove. Because that is just a number multiplied to the whole quadratic. And let me write it in a proper way. So X squared coefficient will be H squared here. And that's it. And X coefficient will be 2FH minus 2BG. And constants will be F squared minus BC. So this must have equal roots. This must have equal roots. Correct? So if it is having equal roots, again, can I say discriminant must be 0? So B squared, that is your 4 times FG minus BG. Sorry, FH. I'm so sorry. FH. Thank you, Krishna, for that. FH. I'm so sorry. So B squared minus 4AC. So let me take the 4AC term on the other side. So 4AC. Which means 4, 4 will go off. Square it. F squared H squared plus B squared G squared minus 2BFGH is equal to again F squared H squared minus BCH squared. Drop F squared H squared and drop a B from everywhere. Drop a B from everywhere. So you end up getting BG squared minus 2FGH is equal to minus CH squared. In other words, you have ended up getting 2FGH minus BG squared minus CH squared equal to 0. Now guys, let me tell you that this expression is the same as this. Same as this. Where your A value has been put as 0. So if you put A as 0, this term will vanish. This term will vanish. And only these three guys will be left off, which you can see clearly over here. Getting my point. So what is the final conclusion? That your expression of the condition for it to represent two straight lines does not change. Now there are some people, some sadist people who don't want to see me happy. They say what if B was also 0? Because here I wrote B is equal to B in the denominator. So what if this guy is 0? So for those people, I have another argument. Let me justify that also. So let us say now I have a situation where A is also 0 and B is also 0. So they don't want to see me happy. So let's go back to a situation where A is also 0 and B is also 0 in the line. That means your equation is only reduced to 2hxy plus 2gx plus 2fy plus c equal to 0. In this case, how do you justify or how does your condition still hold to be true? So I claim that my condition will not change. Let's see how. First of all, I will multiply throughout with h by 2. So multiply by h by 2. So if you do that, it becomes h square xy. This will become hgx. This will become hfy. This will become ch by 2. Am I right? Bring this ch by 2 on the other side. So I'll do it here itself. Okay, fine. Now I will add a fg term to both the sides. So plus fg plus fg. Okay. Now in this case, you will see if you take these two terms together and these two terms together, see what will I get? So h square xy plus hfy and hgx and fg term together. I can actually factorize this by taking hy common. Okay. So it will give me hx from here and I will get a f. Okay. Here if I take a g common, I again get hx plus f. Okay. And on the right hand side, it still remains to be fg minus ch by 2. So this is factorized as hy plus g into hx plus f. Okay. Now my idea, I'm claiming that if this has to represent a line, this fellow must be zero. My question to you is again, why? Can anybody justify that? Anybody unmute yourself and say why? If it is not zero, what will happen? What will it represent then? It will not represent a pair of straight lines through origin because if you expand this, if you expand this, you'll still get a constant terminal. See again, hint number two. Hint number two here, all of you see. Take an h common from both the terms. So you have hx plus f by g, sorry f by h and y plus g by h is equal to fg minus ch by 2. Okay. So h square will come out now. Okay. So bring this h square down. Doesn't it give you a structure, something like xy is equal to some constant. It will represent a hyperbola, my dear friend. And if you wanted to become a straight line, this guy must become a zero. I'll show you on the GeoGebra tool. Let me take, let me just assume a situation like this. Let me just assume a situation like this. So let's say x minus 1 times y plus 2. I'm just randomly picking up some values. Okay. Don't judge me for why I picked up 1 and 2. And here I'll pick up any unknown value. Let's say a, but let me first put a slider for it. I'll vary it from 10 to 10. Okay. I'm putting this as equal to a. Okay. Now look at the graph. Why are you showing me a region? Let me do it again. Sorry. Yeah. So let me pick up a slider. Slider is something which can vary. So I'm calling that's very varying term as a. And I'm choosing that a to be from minus 10 to 10. Those who have used GeoGebra before would know this. Okay. So this is my next I'm writing an equation x minus 1 times y plus 2 equal to that same a. Okay. You can see it's a hyperbola in this case. Now if I start varying my a, see the hyperbola, how it is varying. And the moment I reach a equal to zero, see what does it become? What does it become? A pair of straight lines. If I again move away from zero, it again becomes a hyperbola. See this. This is the only situation when it is a pair of straight lines. So that's why I used to say a pair of straight lines is a very special case of a rectangular hyperbola where your eccentricity becomes infinity. So let me run this thing so that we can enjoy this. It's going positive. Positive. See how hyperbola is still there. Now the hyperbola branches are coming, coming, coming, coming, coming, coming straight line. And then again hyperbola. Do you appreciate that? Do you appreciate this movement? Everybody? Yeah. Okay. So I'll give it a stop here. So that's why my argument here was this guy is doesn't become a zero. It will represent a hyperbola. So this fellow has to be zero. In other words, FG minus CH by two must be zero. In other words, two FG must be CH. In other words, two FGH must be CH square. In other words, two FGH minus CH square must be zero. Let me tell you this is the same condition as what we had started with. It's just that my A is zero. So this term you can't see. B zero also you can't see this term. So two FGH minus CH square is what you see. So it doesn't matter whether you're A zero or B zero. You will still end up getting the same condition for the pair of straight lines. Is that clear? Is that clear? Now guys, proof number two. This is a slightly different proof which comes from your understanding of your second degree homogenous equation. So the principle which I'm going to use is that such an equation will always represent a pair of straight lines passing through. Passing through. Origin. Origin. I will use this principle over here. How? Let me explain. So let's say this is my general second degree equation. This is my general second degree equation. Again, no need to write anything. Just listen to me. So this general second degree equation is not a homogenous equation. So they must be meeting at some other point which is not origin. So they'll meet at some x1 y1 point, let's say. Let's say they meet at x1 y1 and x1 y1 are not zero. Okay. Fine. Now, later on we'll talk about how to find these points also from the equation itself. Now listen to this carefully. If I have to make this a homogenous second degree equation, how will I make it a homogenous second degree equation? How will I make this as a homogenous second degree equation? That means I want to convert it to this form. What should I do? The g and f should be zero. That is fine. That will come out to be zero automatically. But what event will you do to make it happen? I've written it on the board. Shift the origin. So if I put my origin to this point, that means I made the intersection point origin. Can I say then this equation will transform to a second degree homogenous equation? Do you agree with me? Do you agree with me that if let's say this equation, I call this as equation one, if I shift my origin from zero zero to x1 y1, then one will get transformed to a homogenous second degree equation. Correct? Yes or no? That means it will lose out on g, it will lose out on f and it will lose out on c. That means only the second degree terms will remain. Yes or no? Now you guys tell me, you have already done the shifting of origin with me a lot of times in the class. So if you shift your origin from zero zero to x1 y1, how does this equation change? That means what do you replace your x with and what do you replace your y with? Either you would replace it with capital X plus x1 or capital Y plus y1 or you would replace with, I mean I'm giving you one more option because I have seen in past people still getting confused between this. So which is your answer? Option A or option B? B. Okay, we'll take a poll here. Now you just have to press A and B over here. Okay, 30 seconds poll. Come on, everybody should answer. Either A or B. Last three seconds. Poll stop. Please press something. Nine people are still there to be voting for. Okay, guys, very disappointed. B is wrong and you guys have chosen, 61% of you have chosen B. Oh my God. Guys, you need to revisit. Revisit shifting the origin concept. Your x will be replaced with a new x plus x1 and y will be replaced with y plus y1. This is something very, very important. A is the right option. See, y is the right option. Very simple. Let us say there was a curve like this. Okay. This curve was y is equal to x2. Okay. If I shift my origin to, let's say, a point 1,2. Okay. How would this graph change my dear? Now the origin has come over here. So I will, I'll draw another graph. This origin has come over here. So earlier this point was 1,2. How would this graph change? Won't this graph come down and left something like this? Yes or no? If origin goes to right and top, the graph goes to left and down. Right. So what would be the equation of this parabola B? It will become y plus 2 is equal to x plus 1, the whole square. Getting my point. So in the bridge course, class 11, remember the first few classes when we were talking about shifting of graphs. So if a graph is moving left and down, x will be replaced with x plus how much it is shifting to the left and y will become y plus. Isn't it? So your x has to be replaced with x plus x1 and y has to replace with y plus y1. So why are you making this mistake again and again? Okay. Fine. So now you have realized that B is wrong. So I am removing this. So now I'll make these substitutions over here. A, x plus x1 square, 2h, x plus x1, y1, B, y plus y1 square, 2g, x plus x1, 2f, y plus y1, plus c. Okay. Now, I've used capital X and capital Y. You can bring it back to small x and small y. So it's just a repetition of the same thing because we normally never write any equation in capital alphabets. Have you ever seen any equation written in capital alphabets? No, right? You always use small alphabets. So capital X was only used for you to make a distinction between the new x and the new and the old x. Now bring it back to the old one. Okay. Now guys, what am I claiming that this will represent a homogeneous second-degree equation? What is the meaning of homogeneous? Will this contain any x term? Should this expression contain any x term? No, right? Because homogeneous will never contain any x term. It will never contain any y term also. And it will never contain any constant also, right? So can I say the coefficient of x, the coefficient of y, and the constants must all be 0 in this particular equation agreed or not? Agreed or not? Because without that we will not be able to move forward. Just write agreed on your chat box. Okay. Ananya, Pratam, Krishna, Sana all are agreeing with me. Okay, good. Good. So let us bring out those coefficients. Let's handpick them out. Okay. I have a habit of not expanding things. You may call me lazy, but I feel that if you expand things, you'll unnecessarily waste a lot of time and do a lot of mistakes also. So let me get the coefficient of x from the first term. So from first term it will be 2ax1. From the second term it will be 2hy1. Okay. From the third term there is nothing. From the fourth term we'll get 2gx1 again. Okay. Will there be any other term which will contribute to an x? No I guess. So this coefficient must be 0. Okay. Let me call it as 1. In the same way let's handpick coefficient of y from every term. This will not give me any y term. This will give me 2hx1. This will give me 2by1. This will not give me 2f. Sir it should be 2g. It should be 2g right? Not 2gx1. Where sir? First equation. First one. Oh I'm sorry sir. My mistake I don't know. Sorry. Thank you sir. Next one. Coefficient or sorry constant term. What will be constant term over here? Ax1 square from here. From here you'll get 2hx1y1. See x1y1 are numbers. See RMR I would just request you to hold on for next 10 minutes. After that you can leave. This is also a part of j main only. I wanted to finish off this chapter. So just hold on. Next is the constant from here will be by1 square. The constant from here will be 2gx1. The constant from here will be 2f y1. And plus c equal to 0. Let me call this as the third equation. Now all of you please pay attention to this simplification. First of all 2 and 2 can be dropped from the first two equations. So I can write this as ax1hy1 plus g equal to 0. Let me call this as fourth equation. From here also I'll drop it 2. So hx1by1 plus f equal to 0. Let me call this as the fifth equation. Now from third equation I'll do something very interesting. I will break these terms like this. Just pay attention. ax1 square this 2hx1y1 I will break it up as hx1y1. And this 2gx1 I'll break it up as gx1. The remaining hx1y1 again I'll write it. See they were 2hx1y1. So one is here another is here. There was by1 square so I'll copy it as such. And this 2fy1 again I'll break it up as fy1. So if you recall 1gx1 still remain. 1fy1 still remain and a c remains. So the equation number 3 I have broken it up like this. You must be wondering why I did that. You'll come to know when I start grouping terms. So let's group these three terms. Let's group these three terms and let's group these three terms. Take x1 common. You'll end up getting this. Take from here y1 common. You'll end up getting this and let this term be as it is. Now all of you go back to your fourth equation and fifth equation. Do you realize this is coming from your fourth equation only and that is 0? So this you can put as 0. This is coming from your fifth equation and this can also be put as 0. Correct? What does it leave you with? It leaves you with the fact that gx1 plus fy1 plus c is equal to 0. Let me call it as a sixth equation. Now guys the time has come that we now summarize things. We have got three major equations with us. Ax1, hy1 plus g equal to 0 that is your fourth one. Hx1, by1 plus f equal to 0 that is your fifth one. And gx1 plus fy1 plus c equal to 0 that is your sixth one. Now if you look at it very very carefully what does it tell you? It tells you that there were three lines like this and all of them pass through which point? All of these pass through which point? x1, y1 isn't it? See x1, y1 is satisfying all these conditions isn't it? Isn't it? That means these three lines are concurrent. See it is something like this. These are the three lines and all of them are passing through a single point x1, y1. So aren't they concurrent? Yes or no? Now all of you have Aadi Sharma at home right? Open your Aadi Sharma. If you have you can bring it now on the table. I am giving you 30 seconds to get your book on the table. Open up that book and see in the straight lines chapter what is the condition for? What is the condition for concurrency? Guys if you have the book next to you, pull it out. Open two straight lines chapter, 30 seconds I am giving you. I will write down here. If you don't have a book with you I will write down over here. If A1x, B1y plus C1 equal to 0, A2x plus B2y plus C2 equal to 0 and A3x plus B3y plus C3 equal to 0 are three lines which are concurrent. Concurrent means all passing through the same point. Then the determinant formed by these coefficients over here would be equal to 0. This is another way of saying if you solve these two then the solution will satisfy the third one. Getting my point. If you have the book I am sure you would have seen that expression. From this we can say the determinant formed by A, H, G, H, B, F, G, F, C must be equal to 0. This is what we had actually discussed while we were learning how to write delta in a determinant form. Which actually again gives you A, B, C, 2, F, G, H minus A, F squared minus A squared minus C, F squared equal to 0. Is that fine? So this is your proof number two. Hence proof. Are you getting the roadmap? What all I did? Right? First I shifted the origin. I converted the two lines which were passing through X1, Y1 as a homogeneous second-degree equation. From homogeneous second-degree equation we got these three criteria. These three criteria we converted it as if there are three lines all of them passing through X1, Y1. Then we use condition for concurrency and finally, finally, finally our result is there in front of us. Nice proof, no? Difficult or easy? Okay. A for, sorry, I was just sharing with you, sorry. Guys, whenever this thing is blocking your view please let me know. Okay. A is for difficult, B for easy. Ten seconds. A difficult, B for easy. Oh my God, everybody found it difficult. Okay, some of you are finding it easy. Okay. Thanks for your response. Yes, it was a challenging proof. So 68% of you have said it was a challenging proof. No worries. Okay. Now guys, our story has not ended in the proof. I have some takeaways from that proof. As you already know, I have a habit of analyzing my, you know, proofs. I'll go back to the previous slide momentarily. If you see this, especially focus on these two guys. It says that your X1, Y1, which is actually your point of intersection satisfies these two equations. Isn't it? Isn't it? Or you can say your X1, Y1 is satisfying these two equations. So X1, Y1 satisfies this equation. X1, Y1. Satisfies this equation. Doesn't it? Okay. Well, there's something very interesting with respect to that. What is that? Let me go back to the next slide and explain you. Let me go back to the next slide and explain you. So we learned that X1, Y1 satisfies AX plus HY plus G equal to 0. And HX plus BY plus F equal to 0. Basically, I'm just copying the two equations which I wrote. Okay. So from here, can we solve for X1, X1 and Y1? Let's do that. Again, I'll use my cross multiplication. So thanks Aditya for asking that question because anyways, we need to revisit this concept of cross multiplication over here. So tell me what should I write within X? So hide this HF minus BG. Under Y, what should I write? Tell me. F minus. GH minus AF. Now don't make that mistake. If there was minus Y, you would write AF minus GH. Okay. So if you're getting confused, you can better write a minus and then switch the position of it. Okay. For one, what will I write? AB minus X squared. In other words, I can say my X value would be take this to the numerator over here. HF minus BG by AB minus X squared. And my Y value will be GH minus AF by AB minus X squared. Many books will write it like this. So X1, Y1 basically will be your BG minus HF by X squared minus AB. I just turned the position of the numerator and denominator. So don't get confused. Okay. Okay. This again has to be, again has to be, again has to be memorized. But people say, sir, this is too difficult to memorize. I'm getting confused between BG, HF, AB and all. Okay. So I'll tell you a simple method to remember it. So memory, memory aid for this. Do you remember we had written a determinant little while ago? If I'm assuming that you remember this, then your life will be simple. So how to remember this formula from this? Copy the first two rows like this, AH, HB, GF. Okay. Then copy the first column after this. So make it AH. So this I have copied over here. That's it. Okay. Now see what I'm going to do. Cross, cross, cross, cross, cross, cross. So AB minus X squared. Write it down. AB minus X squared. Here HF minus BG. HF minus BG. And here you write GH minus AF. Okay. Divide by AB minus X squared throughout. AB minus X squared throughout. AB minus X squared throughout. Okay. This will give you a one. Okay. So remove this. This becomes your X1, Y1. Check it out. See, is this here? And this fellow is here. So this is a memory aid. By this you can remember your points of intersection. Okay. Now, let us say it was a very, very bad day for you and you forgot this form of method also. Okay. Don't worry, there is another way out. This is something which is very interesting also. Let us go back to our second degree equation that we had started with. General second degree equation. Okay. Now, all of you know what is partial derivative. If I say do a partial derivative of something with respect to X. What do you mean by it? Let's say I give you an expression X squared plus Y squared plus 2XY. I say do a partial derivative of this expression with respect to X. What would you get? What do you get? Partial derivative means differentiating. Partial derivative means differentiating with respect to X keeping Y as constant or keeping constant as constant. What will you get? 2X and this derivative will be 0. This derivative will be 2Y. Everybody knows this? Partial derivative with respect to X. Same way if I say what is the partial derivative of this with respect to Y. What will you do? Write down the answer for partial derivative with respect to Y on your chat box. Partial derivative with respect to Y means treating Y only as the variable. X and any constant will be treated as constant. Correct Aditi. 2Y plus 2X. Very good. Very good. Very good. Okay. Now, I am going to use a similar concept here also. Let me just erase this off. Let me call this expression. Let me call this expression. See, partial derivative signifies that if there is a function which is in two variables. Let's say I write a function. Let's say this is a function S. This is in two variables. If I do do S by do X, it means keeping Y as constant. If I vary my X, how will the function vary? Or what is the change in the function with respect to X? Getting my point. If I do do S by do Y, it means keeping X as constant. If I vary my Y, how will my function vary? So normally the things that we deal around with our day to day life, they are multivariate functions. Price of a food item. It is dependent upon the price of water. It is dependent upon the price of petrol. It is dependent upon the price of the seed. So so many factors. Price of the fertilizer. So let's say as an economist, I want to see how does the price of my food change if I just change the price of water? Or if I just change the price of petrol? If I just change the price of the seed? If I just change the price of fertilizer? So in that way I have to do partial derivatives. A very beautiful concept. Guys, I would recommend you after your class 12th, when you have free time, I'll suggest you some books on calculus. One of them is Shanti Narayan. You can read through partial derivative concept from there. But meanwhile, let's pay attention over here. Now let us say I call this term on the left-hand side as S. S stands for second degree. That's why we call it as S. Remember, we were using S in conic sections also? Right? So in pair of straight lines, I'll call this term as S. Now I would request all of you to give me do S by do X expression. Everybody. Okay, I'll also work along with you. So the derivative of this with respect to X will be 2AX. The derivative of this with respect to X will be 2HY. The derivative of this with respect to X will be 0. The derivative of this will be 2G. Okay? That's it. We will not get anything else. Very good. Now equate this to 0. You will end up seeing something very, very strange over here. Isn't this equation this fellow? Check it out. So what do you have got is this guy? Okay. Interesting, right? If you do do S by do Y. Let's see what will I get. I'll get 2HX. I will get 2BY. I will get 2F. If you put it to 0, you'll end up getting HX plus BY plus F equal to 0. Another interesting thing. It is actually your second equation. Correct? So what do I want to say here? I want to say here is that if you want to find out the point of intersection, if you want to find out the point of intersection, okay? Solve do S by do X equal to 0 and do S by do Y equal to 0 simultaneously and you'll get your answer. Okay? So you don't have to rely on these memory aid. You don't have to rely on this cross multiplication formula. Getting my point. Beautiful concept, no? Yes, it is independent. It is very good to observation. It doesn't depend upon your C. It doesn't depend upon the constant of there. Okay? Now guys, many people ask me, sir, does it mean that these two lines where the two lines which made up this pair of straight lines? No. Okay? Even though these two intersect at the very same point where the lines making this pair will also intersect, that doesn't mean they are the same lines. Okay? See, let's say this line, this is basically these two lines. Okay? L1 and L2. And let's say these two lines, see basically you are getting a line only, no, from here. These two lines can be something like this. Okay? Even though they meet at the same position where L1 and L2 meets, it doesn't mean they are the same lines. So don't be under the impression that oh, by using do s by do x and do s by do y equal to 0, I could get the two straight lines which make that pair of straight lines. No, you will not. You will not get that. Getting my point, even though it will give you such straight lines which will meet at the same point, it doesn't mean it is the same lines. No, no, no, no. No triple. From my experience of 12 years, I have seen they do not give you the same line. I don't know if I missed out on any case. Getting my point. If you perform such an operation on any other conic, guys, if you do it for any other conic, that means if you do it for circle, if you do it for ellipse, hyperbola, then this will actually give you the coordinates of the center. This is also another thing which you need to note down. I'll show you a simple example. We all know this is the equation of a circle, correct? Tell me what is the center here? All of you have done circle in your class 11. You have already given your semester exam. What is the center? Quickly type. Center, center, center. Type it out. This is a circle. What is the center? Center, center, center. 3 comma 2. Very good. Now I'll get the same 3 comma 2 by using these guys. So think as if this is your S. What is dou S by dou X equal to 0 give you? Dou S by dou X means 2X minus 6 equal to 0. X becomes 3. What is dou S by dou Y equal to 0 give you? Y is equal to 2. 3 comma 2. So this gives you the center. So just get the summary of the entire situation. If you have been provided with a pair of straight lines, solving dou S by dou X equal to 0 and dou S by dou Y equal to 0 gives you the intersection of the lines. If you have been given any conic, solving dou S by dou X equal to 0 and dou S by dou Y equal to 0 will actually give you the center of the conic. Are you getting my point? Let's have a quick question on this. Find the point of intersection of the lines whose pairs are given by whose pair of equation is given by. Guys, even if I don't mention anything over here, let's say if I put lambda also here, you should still be able to find out the answer. Should I give you options? Options 3 by 2, 5 by 2, minus 3 by 2, 5 by 2, 3 by 2, 3 by 2, minus 5 by 2, and finally minus 3 by 2, minus 5 by 2. Let me run a poll. Can I just have two minutes for this? In two minutes, will you be able to wrap this up? Okay, I've got a response. It's time up. Time to press on the button. If I can press something, I'm going to stop the poll. Okay, enough poll. So 44% of you say option D. Okay, let's check. So first of all, I'll use my dou S by dou X equal to 0 concept and dou S by dou Y equal to 0 concept to solve this. So that will give me 24 X minus 10 Y plus 11 equal to 0. And dou S by dou Y will give me minus 10 X plus 4 Y minus 5 equal to 0. Okay, let's check it out, Ananya. Okay, so one way to solve this is multiply. We can multiply something with 2 and we can multiply this to the 5. Correct? And add them. 48 X minus 20 Y plus 22 equal to 0. And we have minus 50 X plus 20 Y minus 25 equal to 0. Adding it gives you minus 2 X minus 2 X minus 3 equal to 0. So X has to be negative 3 by 2. Okay, these two options are ruled out and put it in any one of them. Let me put in the second equation. So minus 10 into minus 3 by 2, that will give me 15 plus 4 Y minus 5 equal to 0. That means 4 Y is equal to minus 10. So Y is equal to minus 5 by 2. So minus 5 by 2 and minus 3 by 2 only D is there. So correct. Janta was right. Option D is the right option. Okay, if you have not used this method, you would unnecessarily factorize this equation. Okay. Now many people ask me, sir, if this was lambda, then how would I find the point of intersection given that I don't know this method? Very easy. First find out lambda. How will you find out lambda? How will you find out lambda? How will you find out lambda? Anybody? By using the fact that it represents a pair of straight lines, isn't it? That condition a, b, c. So let me write it. Find lambda using a, b, c plus 2 f g h minus a f square minus b g square minus c s square equal to 0. Then you factorize it by splitting the second degree terms. Remember, I had discussed a similar question earlier, so I'm not going to repeat it. And then get the two lines and solve them simultaneously and get the answer. But isn't it like, oh my God, too much thing, too many things to do? So this is always recommended for comparative exams, so I recommend this. Okay. Save your time. Save your energy. Let's move on to the next concept now. That is the concept of angle between, angle between the two lines whose pair of equations or whose pair of straight lines equation is a general second degree. Remember, I'm repeating the very same concepts which I did for homogeneous, but now I'm applying it to a general second degree equation like this. Okay. Guys, the answer for this is the angle theta between the lines, even though let's say your line is not passing through origin. So let's say this is the two lines and let's say theta is your acute angle. The answer still remains to be 2 under root at square minus AB by mod A plus B. Can you tell me why? That means G, F and C have no role to play in your angle. Why? Sir, because the slope of the line is only decided by the first three. Absolutely, Siddharth. Bang on. The slope's information is only carried by the first three. Or so as to say, even if let's say I shifted my origin to this position and I converted this to a homogeneous one, let's say, you would have used this formula. So if you shifted it back, does it change any angle? Shifting, does it change the angle between the two lines? You tell me. No, right? If I'm shifting two lines, let's say if I take the same two lines somewhere else, let's say I draw it like this. Does this angle change? Does shifting change the angle? No. No. So formula will also not change. And remember, A, H and B will not change on shifting. So if you bring your origin over here, A, H and B still remain to be A, H and B for the new line. Only these terms are going to get vanished. Try it out. Try it out. It doesn't change. Okay. So the concept here is pretty easy. So even if you have a general second degree equation, nothing is going to change as per the angle between the lines. Okay. I have again the same question for you which I had taken a while ago. Tell me what is the angle between the two straight lines? This is a plus, minus sign, minus xy plus 2y squared plus 11x minus 5y plus 2 equal to 0. Tell me what is the angle between them? What is the angle between the lines? Just 30 seconds for this. I want the answer on the chat box. Your answer will be in terms of tan inverse something. Absolutely correct, Aditi. Very good. Correct, Sharmik. Correct, Viver. Okay. So all you need to do is apply the formula tan inverse of 2x squared minus AB by mod A plus B. Okay. So it is tan inverse 2. Who's paying the rule of H? Minus 5. So minus 5 squared is 25. AB is 24 by mod A plus B. That is mod of 14. So that gives you tan inverse of 2 by 14 which is 1 by 7. Okay, so before we close one last concept I will take up. That is the equation of the bisectors of the angle for the pair of lines given by a general second-degree equation. Here also I will not do a lot of hard work. I will use my shifting of origin concept. Let's go back to our homogeneous equation. So let's say this is our homogeneous equation of second-degree. Okay. And this were my bisectors, the blue lines. What are the equations of these bisectors? So these were the lines. What were the equations of my bisectors? X squared minus Y squared by A minus B is equal to XY by H. Remember these formulas. You will be using them very, very frequently. Okay. Now guys, let me ask you this question. If I want this particular line to become this line, where should I shift my origin? Where should I shift my origin? Origin will get shifted to which position? Type it out or answer it out? Minus X minus Y. Who is that? Absolutely correct. Guys, if you want your origin to become, if you want this point to become X1, Y1, you should have taken your origin to minus X1 minus Y1. Is that obvious or not? So this will become minus X1 minus Y1 point. So how would this equation, how would this equation change? It will become this. How would this equation change? How would this equation change? You will replace your X with, you will replace your X with capital X plus this point. That is plus minus X1. That is minus X1 itself. And you will replace your Y with Y plus Y1. Here plus Y1 is actually minus Y1. Remember the same formula I am using. I am not using a different formula. See the formula is this. Your origin shifts to H comma K. Your old X gets replaced with capital X plus H. Your old Y gets replaced with capital Y plus K. And then you put your capital X and capital Y back as small x and small y. This is something which you need to keep in mind. Please remember this. We will not be discussing this over and over again. So what I am doing is, this is my H, this is my K. So that is what I have used over here. So H is negative X1 and K is negative Y1. So how would this equation change? This equation will become capital X minus X1 square minus capital Y minus Y1 square by A minus B is equal to capital X minus X1, capital Y minus Y1 by H. I mean I am leaving the simplification part up to you. That is not required. But this is what, by the way, write it back in small x because it is not a good practice to write any equation in capital alphabets, isn't it? So let me put it back in small alphabets. Is that fine? Is that clear? Okay. One quick question on this. The same old question that we had. A 12X square minus 10XY plus 2Y square. Why I am choosing this question is because we already know where they intersect. X1, Y1 is already known to us. See, we need X1, Y1, right? That is already known to us. For this equation, find the equation of the bisectors. Find the pair of bisector equation. Don't simplify it. Just write down the expression on your notebook. Just 30 seconds for that. So your answer should look like this. Remember your X1, Y1 points where minus 3 by 2 minus 5 by 2. So your answer should look like this. X minus X1, that is plus 3 by 2. Whole square minus Y plus 5 by 2 whole square by A minus B, that is your 10. I'll write it down directly. X is equal to X plus 3 by 2 Y plus 5 by 2 by H. Simplify it. I am not going to waste time doing the simplification. Just simplify it. Is that fine? So guys, we learned a lot of things in this class but still we are not able to complete this chapter. I don't want to start a new concept. Given that we have very less time left. But the agenda for the next class, I am making it very clear over here, we'll talk about, number one, the concept of homogenization. This is a very important concept that has been recently asked in JEMain a number of times. We'll talk about solving problems based on these concepts. So probably I'll take around one hour more for this in the next class. And the third thing that we are going to begin with is theory of equations. Happy birthday, Siddhartha. Thank you, sir. You are living for thousands of years. You are 50,000 years old. May God give you a lot of happiness. Of course, hard work is also there. Have a great year ahead. Your party is pending. Once we meet, we'll have a small get together for this. So this is our agenda for the next class. So signing off now, over and out. Thank you. Have a good day. Study hard. Don't go out. Thank you, sir. Thank you, sir.