 So almost everybody has come in now. Yeah, so pair of straight lines So let me start with this concept. So if you take two straight lines, let's say I have a line L1 Whose equation is ax plus by plus c equal to 0 and you have another line L2 Which is a1 x plus b1 y plus c1 equal to 0 and you multiply these two equations in In this way That means you are formulating a joint equation of these two lines in this way Then this joint equation is basically referred to as the equation of pair of straight lines pair of straight lines I Would request everybody to mute their mic in case you want to speak something you can unmute yourself and speak out Or you can type it on the chat Okay Yeah, now so There's a rule by which we actually create this pair of straight lines in order to create a pair of straight lines The rule is we must have a zero on one of the sides of the equation That means you convert it to general form as you all know this is called the general form of a equation of a line Ax plus by plus c equal to 0 so before finding out the pair of straight lines You should have converted it to general form and then perform this multiplication If you don't do that There's something very unusual that will happen. Let me give you an example Let's say I take a line l1 Whose equation is x is equal to y Okay, I take another line l2 whose equation is x is equal to minus y Okay, you can say y is equal to x or y is equal to minus x okay, if you have to create a pair of straight lines from it will first have to Bring both the equations to their general form like this Okay, and then perform the multiplication that would give you the pair of straight lines as x square minus y square equal to 0 so this would be a pair of straight lines Okay, as you can see, it's the second degree equation. That's why we study this chapter along with Koenig sections Now there is another student who decides to not follow this way and Directly multiplies l1 and l2. So let me write that in red because he's creating a mistake here So when you multiply this directly, you'll end up getting x square is equal to minus y square Correct, and in that case you'll end up getting something very different and Remember, this is not going to represent a pair of straight lines. In fact, it represents a dot It represents just 1.0 comma 0 Getting my point. So please whenever you are formulating a pair of straight lines, make sure you have converted them to the general form right Any questions with respect to this? May well if you are not able to hear me rejoin again and select dial in via audio device dial in via the The device whichever you are Are you seeing this on okay? Why I'm speaking to him. He's not able to hear me Rejoin Dial in via device audio Okay, guys any any question with respect to this if no, I have a very simple question for you all I'm giving you Sorry, let me write it in yellow once again So I'm giving you a equation of a pair of straight lines very easy question just as a warm-up question Hope you can see the screen x square minus 6 x y Plus 8 y square equal to 0 Tell me what are the separate equations of the line represented by this equation? I Think they're very simple once you can type it out on the chat box, but type it privately not for everyone to see Let's see who gives the answer for this first That's correct good my Russian that's one of the lines correct You should completely write down the equation with equal to sign also, okay? So just like x minus something y is equal to zero whatever you want to write correct correct correct So everybody who's answering me Rahul, that's correct. I think correct Very nice. So all you need to do was just factorize it. Okay. Now while factorizing Don't think there are two variables You may treat this as a single variable function also if you divide throughout with x square Okay, so the multiple ways one is directly you can just split the terms like minus 4 x y minus 2 x y plus 8 y square But many people they don't find this convenient to use so I'll tell you another method after this So just take x common x minus 4 y take minus 2 common x minus 4 y is equal to 0 So that gives us two factors Which is these two so basically it corresponds to two lines You can say y is equal to x by 2 and y is equal to x by 4 so these two lines jointly have this equation You have another way to do that You can actually divide throughout with x square So let's say I call this as my first method second method is divide throughout with x square. We'll end up getting something like this Okay, you may treat y by x to be some other Variable, let me call it as m Now you all know why I'm calling it as m because slope. It is something to do with the slope Well, I'll connect the picture completely little later on So when I do that it gives me a plain and simple quadratic equation Where you can apply the spitting of the middle terms just like we did it in the Approach 1 and finish it off So take 4m to m minus 1 take a minus 1 to m minus 1 equal to 0 So that gives us m equal to half and m is equal to 1 fourth Yes, so this this can be completed and said that we are dealing with two equations y by x is equal to half That means y is equal to x by 2 and y by x equal to 1 by 4. That is y is equal to x by 4 Now a slightly complicated picture Let's have another example. Let's have another question here. Let me pull out the question for my Okay This is a slightly complicated version of the the previous question that we had Hope everybody can read this question. If not, I'll just repeat Read this question out for you find the combined equation of the straight lines passing through the point 1 comma 1 right, so it is passing through 1 comma 1 and Parallel the lines represented by this and parallel to the lines represented by this equation Just want to see if you're able to extend that concept Whatever you you learned just now to solve this else. Don't worry. We'll talk more about it You may use x2 to represent x square in your final answer. Arthik, you need your mic Good morning. Okay. Let's see anybody So let me just get it clear for you all is the question clear So there are two straight lines whose combined equation is the one which is given to you on the screen. I want I Want the combined equation of another two straight lines Which are parallel to the lines mentioned here in the equation and passing through 1 comma 1 So let's say I'm just drawing a Rough scenario for this So let's say the two lines which were Represented by x square minus 5 x y plus 4 y square plus x plus 2 y minus 2 equal to 0 This line is represented by these two. Let's say L1 and L2 Okay, fine Now I want another two straight lines Let me draw it in blue Where these straight lines are parallel to the ones which have already been stated in our equation So as you can see this is parallel to this And this is parallel to this But these two sale lines would be passing through which point 1 comma 1 So I want Let me call this as L2 dash and L1 dash combined equation. That's what the question is asking you in order to solve this First of all, we need to see What are the slopes of the lines L1 and L2 from this equation? So from this given equation, I need to see the slopes of the line L1 and L2 Or somebody has given me the answer Absolutely correct It's Ruchel, right? That's brilliant. Absolutely correct wonderful Great Ruchel. Good. So in order to know what are the slopes of L1 and L2, I Need to first Ruchel, okay Sorry. Yeah, in order to know what are the slopes of L1 and L2 We need to work on the second degree terms. So the second degree terms in this equation where These three terms Correct Yes or no Now people ask me, so why are you only working with second degree terms? Why not this x2y and minus 2? Let me complete this first and I'll tell you why I'm only looking at the second degree terms By the way, this is factorizable. We all can factorize this very easy to factorize split the middle terms like this Take x common Take a minus 4y common and This becomes x minus 4y and x minus y see that I have purposely left some gap over here Now why I left this gap over here because see These second degree terms they decide about the slopes The other terms have come because of the presence of some constants over here. So I'm writing it as L and m So what I'm trying to say here is If you multiply these two equations, you will end up getting something like a second degree equation like this Are you getting my point? So when I am trying to solve this equation I will get something like x minus 4y plus L equal to 0 and x minus y plus m equal to 0 What is L and m? I don't care because I don't need them Only thing that I need are the slopes of these two lines so the slope of this fellow is One by four the slope of this is one Correct. Now if I want to formulate a line Which is having the same slopes. So let's say I want to write down the equation of L1 dash So let's say this is your L1 This is your L2 and I'm writing the equation of L1 dash So for L1 dash, can I just do this thing? I can just write it as x minus 4y plus p equal to 0 and L2 dash I can write it as x minus y plus q equal to 0 I just have to replace these constants because we have all learned in our class 11 straight lines chapter That when two lines are parallel They only differ in their constant terms if we can easily take the coefficients of x and y to be the same on both the sides Correct. Is the rationale clear? Why did I choose L1 dash and L2 dash to be these two equations? Any doubt regarding this? Okay. Now, how do I get p and q because unless until I get p and q my equation will not be completed So for that I would use the fact that these lines must pass through 1 comma 1 So when I put a 1 comma 1 here in the L1 dash equation I get 1 minus 4 plus p is equal to 0. That means p is going to be 3 and In the L2 dash equation when I do that I get 1 minus 1 plus q equal to 0. That means q is equal to 0 That means the combined equation of these blue lines Or you can say the pair of equations of pair of straight lines given by these blue lines will hold the equation like this x minus 4 y plus 3 times x minus y equal to 0 Right, you may expand it. You should expand it in fact in case they're asking you so I'll give you x square minus 5 x y plus 4 y square plus 3 x Minus 3 y equal to 0. I know this is a slightly challenging problem to take up in the starting itself But this is what I wanted to set the agenda that when you're looking at pair of straight lines You must think them as if there are two separate lines involved and Then how to work around with that problem according to the situation given must be very clear in light of that visualization Is this clear everybody? Any doubt anyone By the way, good morning to everybody who has joined in late. Please avoid joining late guys because a new chapter Initial agenda is missed. You will not understand anything Okay, better leave early for the class but never come late for the class Okay So with this Our good morning. Shea. No problem. I just anyway these videos are going to be recorded So please pay attention from now on so that you don't miss out anything further Now something very interesting that has come come up from the previous discussions is Remember the first example that I took x square minus 6 x y plus 8 y square it represented two straight lines Which was like this isn't it? This was our first example It tells you that I don't know how many of you observed if you have a second degree homogeneous equation like this Okay, I use the word called second degree homogeneous Equation what do you understand by this? Okay. First. Let me speak something about homogeneous equations before homogeneous equations Let us understand. What is the homogeneous function? What's the homogeneous function? Anybody who has an idea about a homogeneous function? Let's say homogeneous function in two variables Let me make it More specific. What's the homogeneous function in two variables? So this is something which you are going to use again in a chapter called differential equations, which is going to be taken up Towards the month of August or September Where we will be talking about homogeneous differential equations. So it's very important that you understand this right away when I say homogeneous function X f x y basically why two very because I have taken two variables So I chose the function name to be f x comma y. So it's just a way of saying that there are two unknowns in this in this function so if a function F x comma y is a homogeneous equation Oh, sorry homogeneous function equation. I'll come to it later on homogeneous function of degree n of degree n Then it will satisfy this particular property It'll satisfy this particular property that if you replace your x with lambda x and y with lambda y It will end up giving you something like this Where n is the degree of this homogeneous function But this is just a mathematical way of saying something which is very very easy to understand It just means that the degree of every term in a homogeneous function will be the same example X square plus y square plus x y this is a example of a degree to homogeneous function Okay, so if I call this as f x y you can see that it will satisfy this property So change your x with lambda x change your y with lambda y That means you are going to change it in the entire function and you will see that You can pull out a lambda square throughout common Okay, that is your function becomes lambda square times f x y Okay, this is just a mathematical notation dear friends. Don't like you know take it very very Like it's a rocket science concept something like that. No, it's a simple thing So now talking about homogeneous equations So if a function is related if a homogeneous function is equated to some, you know, let's say zero It becomes a homogeneous function So here if you see this is a homogeneous equation of degree 2 Now my claim is a second-degree equation My claim here is that a second-degree Please read these these statements very very carefully a second-degree homogeneous Homogeneous equation will always represent Will always represent a pair of straight lines Passing through the origin Passing through origin means what intersecting each other at origin Okay, passing means intersecting Each other at origin Now, why do I give why why do I make such a claim? so let me take See guys, I'm not claiming that My both the lines will always be real, right? But I'm claiming that they may be Non-real they may be the same line squared up. They may be two distinct lines But they will always represent a pair of straight lines passing through origin so let us say I take a General second degree. Sorry. I take a homogeneous second-degree equation like this Okay, so what I can do is I can first take a B common something like this Then I'll divide throughout with x square Very similar to the second approach which I had taken while solving the first problem Okay, and then what I'll do is I will treat this as if it's a quadratic in y by x Okay, so if it's a quadratic in y by x, let's say it has roots M1 and M2 So this entire thing could be Factorized something like this Are you getting this point? So this is a quadratic the part which I have shown with the curly brackets that is a quadratic in y by x and I'm assuming that M1 and M2 are the roots of that quadratic So this can be factorized just like we factorize a quadratic equation like this Okay, giving us two situations one is y by x could be M1 or y by x could be M2 Which means y is equal to M1x and y equal to M2x are the two lines Which actually make this pair of straight lines equation and you can see that they pass through origin or you can say they Intersect through origin they pass through the origin Now I'm not claiming M1 M2 will always be real M1 M2 may be imaginary also M1 M2 may be equal also M1 M2 may be real and distinct also Okay, in those cases we'll get different types of lines. So if M1 and M2 if M1 and M2 are real and distinct Real and distinct we will get two real lines We'll get two real lines Two real and distinct lines. Let me say That means something like this you will see Okay, so you'll see them Intersecting each other and this intersection will happen at origin if your M1 and M2 are real and equal Guys don't waste too much time on writing these serves. I'll be sending the PDF to you on the group okay, focus on understanding and Feel free to unmute yourself and ask me questions If M1 and M2 are real and equal that means it represents two Same lines or you can say it's the same line two real and Same lines Or you can say coincident lines Coincident lines Okay, so the two lines will be one above the other so one is let's say this white one other will be right on top of it Okay So there's no difference between these two lines and if M1 and M2 happens to be Non real or imaginary Then we'll get two imaginary lines. So the last case we are not going to talk about it We are only going to talk about the first two cases. These are important for us not the last one any question with respect to this See the roots of it will be imaginary just like we can have imaginary circles we can have imaginary lines also But the idea is you cannot represent them on the Cartesian coordinate plane getting the point, which is Yes, the representation is not possible In order to represent something which is imaginary We cannot represent it on a real plane or you can say Cartesian pain Cartesian pain is a real thing Okay You can represent it on a plane but now then you have to write things in terms of you know Z and all yes You can represent it in the organ thing Now one more thing which is very obvious from this exercise is that We learned that if this represents two straight lines Y is equal to M1 X and Y is equal to M2 X M1 plus M2 Can be written as minus 2 H by B now, how did I get this? Let me go back to the previous slide Okay, so here you can see that this equation has roots M1 and M2, right? So we know some of the roots some of the roots is minus B by a correct. Remember Vitas relation Let me write it Vitas Relation, I mean it's just a name of the person who came up with this We have been doing this since class 10th only Some of the roots of a quadratic equation minus B by a product of the roots of a quadratic equation C by a The same thing I'm going to use over here So if let's say this is my equation, I can say M1 plus M2 here would be this is going to be your B A is one over here. So minus B by a Okay, and M1 and M2 would be equal to C. This is your C and a is one. So that's going to be a by B So that's what I'm going to write in the next slide. So follow this up in the next slide. This result is to be kept in mind Very very important result. Please keep this result in mind because it will be required in a lot of problem-solving Remember this So if so, H is just any constant. Does it have any particular value? See in a variable, what does what role does X and Y play? It pays the role of all the coordinates that are present on that line. Yeah, correct So in the same way, this represents two such lines Right, one of the lines is given by this another line is given by this So the role of X and Y is the same as rule of any coordinate present on that particular line in this case Present on the same pair of straight lines. No, I was asking about H. Oh H? Yeah, what is the role of H? Like is it is it another just variable or like a constant? It is a constant. See, okay, these terms would be mentioned to you in the question. Don't worry. Okay. Okay. This is not a variable A, H and B will be some constants But would be given to you in the question itself. Okay Just like in the previous example, when I gave you X square minus 6 X Y plus Y square equal to 0 Sorry, 8 Y square, 8 Y square equal to 0 Such as 3. Yeah, A was 1, H was minus 3 and B was 8 So by using these relations, you can know the sum of the slopes of the two lines and product of the slopes of the two lines Which constitute that pair Are you clear? Yes, sir. Okay Now another important thing that we'll be taking up here is when do we get Real and distinct lines, when do we get Coincident lines and when do we get imaginary lines? So from here in this entire exercise, we can use our quadratic equation formula Okay, so we know that This line could be written as this pair of straight lines could be written as Something like this. This is going to be just Y by X correct Now the quadratic equation formula says that Y by X that is your slope That is M1 and M2 could be given as minus B plus minus B square That is 4 X square by B square minus 4ac, correct? So 4 A is 1, so C is going to be this. Okay divided by 2A Now if you want your M1 and M2 to be both real and distinct, we can say the term within the under roots That is this term must be greater than 0 Isn't it? Which actually gives us X square minus 4 AB greater than 0 It can be greater than or equal to AB No, if you want a real and distinct Okay, so this condition when you're at square minus AB You already know your H. You already know your A. You already know your B from the lines equation If you find that S square minus AB is equal is greater than 0 That means the lines which constitute this for example in this case if you check 3 square minus 1 into 8. Isn't it? Isn't it greater than 0? That means this will give me two real lines L1 and L2 which are distinct. So real and distinct lines in the same way if your H square minus AB is equal to 0. It will give you co-incident lines Okay, and if your H square is less than AB that is H square minus AB is less than 0. You will end up getting imaginary lines Okay, this H square minus AB term is very useful will come across this term later on also when we are talking about angle between the lines Meanwhile if this concept is clear, I have a question for all of you ready Anybody who is copying something Should we move on to the next question? Okay Here it goes. If the slope of one of the lines represented by represented by AX square plus 2 HXY plus BY square equal to 0 BN times BN times the slope of the other The slope of the other then which of the condition here is true? So it's a single option The first is for N square H square is equal to AB 1 plus N the whole square option B says for N H square is equal to AB 1 plus N option C for N H square is AB 1 plus N the whole square an option D for N square H square is AB 1 plus N Okay, either you can reply to me privately or I will run a poll over here. Okay. Those who are answering Please do it privately at the same time. Please hit on the poll button also So right now how many people have given the right answer? So just two of the people have responded so far Take your time guys. No need to rush I'm allowing one and a half minutes for this question. It's not more than worth more than that Richard, I'll disclose the answer after everybody is done or most of the people are done as 10 seconds more Jala, I'll give you Two minutes only seven of you have voted so far. Okay, so I can see responses coming from Richard both the Ruchir Svikas, Sai Vedya Okay guys Time up end of poll. I Can see the result on your screen Most of the people have replied with option C. Okay Let's discuss this So you're claiming that One of the lines has a slope and times the slope of the other so we can do one thing We can say let the slope be M1 and let this other slope be N M1 Okay, so instead of assuming M1 and M2 We know only that the sum of the slopes that is M1 plus M2 is minus 2h by B So that's our first equation the product of the slopes that is N1 into N times N1 is a by B. That's my second equation Correct. So now what we can say is from the first equation From the first equation M1 could be written as minus 2h by B 1 plus n If you take a M1 common 1 plus n will factor out over here, which will come down Okay, and this substitute it into If you put that into N M1 square M1 square will be minus 2h By B 1 plus n the whole square. That's going to be a by B Okay, so let's try to simplify this further So n times 4h square by B square 1 plus n the whole square is equal to a by B B and B will get cancelled take the entire denominator to the right hand side You get 4n head square is equal to a B 1 plus n the whole square Which option matches? Yes, option C matches with this. Option C is the right answer. Well done Any question regarding this? So knowing these formulas really help it help your case We'll move on to the next question now this question I'm going to Project on the screen. I'm not going to write it down. Okay. Hope you can see the question If the slope of one of the lines represented by this be the n-th power of the other Then prove this so since this is the prove that question just type done if you're done. Oh sure Krithika Guys, I just need to go back to the previous slide For a second to help Krithika out Let me know once you're done seeing this Okay, thank you. Oh I'm sorry. It's Karthik my bad Sorry Karthik So Richard is done Param is done because is done Good guys It's just a two-minute question should not take more than that My this is done. Say is done done Just 30 seconds more for others to wrap up. I know she's also done. Hi Abiram Mehul Request you to mute your mics, please. Thank you. Okay guys. Let's discuss Kirtan Dhruv Krithika all are done. Okay So this again very simple case. Let's say one of the slopes is M Other is m to the power n. Okay, the sum of the slopes is given as minus 2h by b and Product of the slopes is given is known to be a by b. Okay, so from the last equation here I can say m to the power n plus 1 is a by b. So we can say m is n plus 1th root of a plus b Correct. Let's substitute this in the first one You'll end up getting a by b to the power of 1 by n plus 1 and a by b to the power of n by n plus 1 is equal to negative 2h by b Okay, let me multiply with b throughout and pull out a take the minus 2h to the left-hand side So it becomes b times it becomes b times a by b to the power of 1 by n plus 1 Plus b times a n plus 1 and plus 2h equal to 0 if you simplify this remember b is there in the denominator So the power of b will get subtracted from 1 so it will become 1 minus 1 by n plus 1 Into a to the power 1 by n plus 1 Similarly here b to the power 1 minus n by n plus 1 into a to the power n by n plus 1 plus 2h equal to 0 So just Simplify this this becomes b to the power n by n plus 1 a to the power 1 by n plus 1 B to the power again 1 by n plus 1 a to the power n by n plus 1 plus 2h equal to 0 take You can take in this case a b to the power n common and raise whole thing to the power of 1 by n plus 1 Here you can take a to the power n b common and raise the whole thing to the power of 1 by n plus 1 plus 2h equal to 0 Okay, and this is what you wanted to prove This is the result. Yeah, this is the result Is that fine very simple activity Nothing difficult about it. Okay, moving on to the next question. Let me pull that question out Okay This is a question which is again a subjective type, but I will give you some options so that you know You don't have to type too much So the question says Find the condition that one of the lines given by this pair of straight lines May be perpendicular to one of the lines given by this Okay What are the conditions? I will just write it down for you. You just have to choose which one is correct so option a for H dash a dash H dash b dash, okay, this is option a Option B is option C is option D is None of these I'll end up whole for this Let's have three minutes to solve this question Okay, so we have response coming from one of the students Again read the question very carefully one of the lines is perpendicular to one of the lines of this It's just one answer. I've got we have just 15 seconds left. I think I can extend this to four minutes Okay, one more response. I have got Okay, time up. I would now request everybody to press any one of the options that you feel I Mean, let's say you're taking a chance and you have to Press one of the buttons So press that button now so that we can start the discussion. Okay, so we'll discuss it Okay, so Cool, so we'll stop the poll here and I let me show you the poll result The poll result says almost 74% people have voted for option B. Okay, let's check whether B is correct or not Okay, so how to approach this problem many of you would be struggling with that So let us say one of the lines from this particular Let me just change my color. So let's say one of the lines one of the lines is Y is equal to MX Okay, so can I say a line perpendicular to it? Right and passing through origin now How do I know passing through origin because this is a homogeneous second-degree equation? So can I say Y is equal to minus 1 by MX would be a line which will satisfy this particular pair of straight lines So if this satisfies this Then this must satisfy the other one This is the Core principle that we will be using to solve this question So I'm going to write it over here. So if I say let Y is equal to MX be a pair of straight lines that satisfy Let's say this is one of the lines with satisfies AX square plus 2 H XY plus BY square equal to 0 that means I Can substitute my Y as MX into this and write it as something like this Which means a plus 2 H M Plus B M square equal to 0. Okay, let me write it in a Decreasing degree of M. So this is a quadratic Which will be satisfied by this M Okay so a line perpendicular to this That is Y is equal to minus 1 by MX should satisfy the other Homogeneous second-degree equation Correct. No, that is nothing but it must satisfy a dash X square To H dash XY and B dash Y square So here also I can do the same thing which I had done with the previous one I can substitute Y with minus X by M So this will become a dash X square to H dash X minus 1 by MX Plus B dash minus 1 by MX the whole square equal to 0 Okay, just drop an X squared term from everywhere and write it in Descending powers of M. So we'll have a dash M square minus 2 H dash M plus B dash equal to 0 so this is my Second quadratic that I have now. What do I claim here my dear? I claim that both these equations must have one common root M so one and two must have One common root at least so what are the condition that we can use when two quadratic Equation have one root in common We can do one thing We can treat our M square for the time being as X M to be something like Y and Get two linear equations from them. That means what I'm trying to say is that a right? Let me write it like this So one equation other equation and I will solve for X and Y So if you solve for X and Y you'll be able to solve for M also correct So what I'm going to do is I'm going to use My cross multiplication method that I had learned in my class 10th Everybody learned cross multiplication method Okay, it's a method that we use to solve simultaneous equations. Okay, we'll learn this in Determinant chapter once again. It is basically known as the Cramer's rule Okay, how does it work? So under X I have left some in a gap So in that gap what I'll do I will hide the Column in which X falls so I'll hide this column Let me show you this column is hidden and I'll do a cross multiplication so 2h multiplied with B dash and Minus 2h dash multiplied with a and subtract them so something like this 2h B dash plus 2h dash B. Sorry h dash a All right, you recall this method you had done this in class 10th In the same way for why what will I write? I'll do a cross multiplication of these two guys. So BB dash Minus a a dash and For one again, we'll do a cross multiplication of this with so minus 2h dash B Minus 2h a dash Okay, is this clear everybody are you happy so far with these tips? Let me know any doubt. Please highlight Anybody Okay, I'm not naming people here. So if in case you have any doubt, please stop me right over here Okay, so everybody is fine. So what I'll do is first I'll just change this to plus and reverse the order in which I have written this so change this to plus so I'll write a dash minus BB dash And moreover, I will also change my x x was m square And y was m. So I'll change this to m square and I'll change this to m Okay, so from this equation What I'm circling out over here I can say I can say m is equal to a a dash minus BB dash by negative 2h dash B plus 2h a dash And from this equation, let me show it in the white color this equation Sorry for making all those amoeba type of structures over here So from that equation, I can say m square Let me write it in white m square is 2h B dash Plus 2h dash a by negative of 2h dash B plus 2h a dash now clearly This expression that is the expression in the white Is square of the expression in the yellow Because one is the representation for m square and other is the representation for m So can I not write the square of this equal to this? So if I do that, let me move slightly to the right of the screen. So from here I'm continuing So if I write the square of it, we'll end up seeing something like this 2h B dash plus 2h dash a by negative 2h dash B plus 2h a dash Is square of this guy so a a dash minus BB dash square by 2h dash B plus 2h a dash square now instead of writing square I can just cancel off one of the terms, okay Cross multiply and bring it to one side. We'll end up seeing something like this 4 h B dash h dash a times h dash B Plus h a dash plus a a dash minus BB dash square equal to 0 I think this matches with this matches with What do you think this matches with option? Option C it matches with Option C yes, what was the poll result? The poll result was maximum people said B so most of you went wrong there Now this taught me one more thing other than pair of straight lines concept The condition for two quadratics to have common roots guys that chapter is also on our agenda In the summer break Before I move on what's going on in your schools so how many schools are there here in the session HSR is there in the other is there guys can you just type it out one person If any of the school. Yeah, I think Akancha is also there What is going on in your school HSR guys, sir right now we're doing our differentiation and continuity Differentiation and continuity. Acha indanagar is doing integrals Right, I know indanagar always starts with integration. I don't know why Okay, Karthik jay says didn't start the classes. Okay, badly affected by corona. It seems Matrixes and determinants in freedom international My god guys, don't worry about these chapters right now. We'll have enough time to you know cover them up So It's very important to pay attention to these concepts which we are going to miss out if you don't do them now Okay, well, I have another question for you all Mehul didn't start classes. Okay. No worries. Oh The summer vacation is already on Okay, I was under the impression that they can reopen Once this issue settles down Okay, never mind. That's a good thing actually Okay, guys next question for all of you If two of the three lines See, I'm just making the questions more challenging now 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 plus d y cube Now from the previous concepts, we can easily say that this would represent three lines all of them meeting at origin. So it's a cubic homogeneous equation now So what does the question say if two of these three straight lines represented by this are at right angles Are at right angles That means our perpendicular then Which of the following option is correct? a square plus ac plus bd plus d square 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 equal to zero an option d a square plus ac plus cd plus d square equal to zero I'll run the poll once again I guess in case the poll result is coming in front of your screen Do let me know Because it normally tends to block the screen. Okay So If it blocks your screen, let me know I'll remove it Let's have two minutes for this question Time has already started need to do that side. That is the first question We don't need to pick up these equations It can be done without going into those complexities Akansha why you're not able to see the poll? Are you uh Is everybody able to see the poll? Yes Try try joining again Akansha. I think you should be able to hope you're not seeing it on a mobile phone Because mobile phone will completely block the the screen Anyways, you can answer me on the on the chat also Poll is just just want to see how many people are giving right options right answers Guys, we'll take a break at 11 15. Okay Okay, two responses. I have got so far Two minutes is already over Don't worry Sai. We'll discuss it Okay, I'll give you a minute more everybody have a minute more. Okay, Richard Guys, come on 15 seconds more But can we get like 30 seconds more? Okay, fine Let's have 30 seconds more But after 30 seconds everybody should test one of the options not required Krithika Don't worry about it. Just give you a best shot. We'll discuss it Once I've got sufficient number of responses. We'll discuss it Correct Richard. That's a good idea actually Okay, we are almost touching four minutes guys everybody. Please press Press one of the options On your poll start pressing it Take a leap of faith. Let's see how lucky you you are Everybody just 13 people have responded guys everybody all 28 of you Never mind because Okay, let me end the poll now So you can see the poll results 47% of you have said option a Okay Then see and then d of course. Okay, so let's let's look into this See again Try to treat this as if you are dealing with A cubic equation in y by x Okay So treat this as a cubic equation in y by x because y by x will act as your slope Okay So when you place when you just take x cube common or divide throughout with x cube you see something like this d y by x cube You'll have c y by x square You'll have b y by x and you'll have a okay replace your y by x with m if you want So this is our cubic This is a cubic in m Which will have three roots m1 m2 and m3. Let's say so let's say these are the Roots of this cubic equation Okay Now read the question two of these are perpendicular That means let's say m1 and m2 are perpendicular. Can I say they'll be multiplying to give you minus one? Okay Now again use beta's relation Beta's relation. Let's use So in beta's relation, it says m1 plus m2. By the way, I'll write down that relation completely in fact So that everybody is clear the sum of the roots Just like we had it in quadratic minus b by a here. I will get minus c by d Okay product of roots two at a time that is m1 m2 m2 m3 m3 m1 Okay, any other possibility? No That would be equal to This term that is b by d. Okay, and finally the product of all of them will be equal to minus a by d So see how alternating sign it takes minus c by d plus b by d minus a by d Just like we have minus a by a c by a you are extending that to incorporate a cubic equation Okay people who are not aware of such relations. Please Uh read through this you will find this in any uh, uh Jay textbook talking about theory of equations Now from the last equation over here, can I say m1 m1 into m2 is minus one So minus one times m3 is minus a by d Which means m3 itself is a by d And since we know m3 is a root of this equation is a root of this equation m3 should satisfy that equation So I will substitute m3 as a by d in your let me call this as the first equation And whatever I get should be my result. So d times m cube m cube means m3 cube Because m is a root correct Isn't it so m3 will be replaced in place of m And plus c times a by d square plus b times a by d plus a equal to zero Okay, so let me just simplify this further So this becomes a cube By d square a square c by d square ab by d plus a equal to zero Let me drop a factor of a throughout So one of the a's I'll just drop off So this will become a square This will become an a This will go off. This will become a one Okay, and I'll multiply through with d square If I multiply through with d square, this is what I obtain And this is nothing but which of your options Option a option a absolutely Is that fine everybody? So it just involves the use of beta relation and just substituting one of the roots in that cubic equation So wow, that was enough number of problems that we have taken to understand the nitty gritties of the basic concepts So now we'll move on to the next sub topic under this, which is The angle between the pair of straight lines Angle between the pair of straight lines the pair of straight lines Given by a second homogeneous second degree homogeneous equation. So still we are dealing with homogeneous only Don't worry. We'll take up a general second degree equation in some time, but let us first Try to master this second degree equation only. That means I'm dealing with only those lines which are passing through Passing through origin Pair of straight lines passing through origin Now, so let's say these are the two lines and I'm talking about The acute angle theta between them So guys, let me tell you we do not know their equations separately If you know the equation separately So let's say this was y is equal to m1x And this was y equal to m2x I don't think so anyone would have any one of us would have taken Any longer to get the answer like this, isn't it? Remember your class 11th pair of straight lines angle between The two straight lines whose slopes are m1 m2 is given by mod m1 minus m2 by 1 plus m1 m2 Okay, so here I'm assuming my theta is an acute angle Okay, this formula is known to everybody anybody who is seeing this formula for the first time now Unfortunately, m1 and m2 are not known to me separately I know them as a combination. That means I know what is m1 plus m2 What is that? Minus 2h by v don't forget these results And I know their product What is that a by v? So from this expression Can I can I get my result of tan theta completely in terms of h a and b? Let's try that out So the numerator part By the way, somebody mumbled something any question? So the numerator part if you see mod of m1 minus m2 could be written as m1 plus m2 the whole square minus 4 m1 m2 correct Denominator I would not disturb Why will not disturb the denominator because m1 and m2 term is already known to us Yeah, is that fine So now I'm going to just make these subtle replacements. So m1 and m2 Any question anybody? Correct Richard correct So now I'm going to make these replacements over here. So I'll end up getting under root m1 plus m2 square will be the square of this term which is 4 at square by b square minus 4 m1 m2 is a by b. So 4 a by b by By the way, I don't need a mod here because under root outcome is always positive So 1 plus a by b. Okay. We can further simplify this We can further simplify this by taking 4 out of the under root sign. So that will come out as a 2 by a b square In the numerator term, so you'll end up getting x square minus a b And that b will eat away the b of the denominator. So that will become 1 plus Sorry a plus b Please note down this result because we need to remember this I understand the last step This step okay, so 4 I took out common After that the denominator Denominator Okay, I multiplied see what I did here was I multiplied with a b square in the denominator in the numerator and denominator So there was this was the lcm. Correct Okay, and this under root sign will come down as mod b Correct. So this will come out of the under root sign as mod b. So this is gone now Okay, and this mod b here from here will cancel off this mod b outside. So here you can write it as b whole divided by mod b. Okay, so this and this will get cancelled off So leaving you with this is okay clear, okay again, this expression was Known to us before also remember when we were talking about the lines being real and distinct the lines being coincident and the lines being imaginary so here you can see that if x square minus a b is positive You will have some finite you can say non-zero angle between the lines. Correct That means your lines will have something like this Okay, so this is your Angle between the lines if x square minus a b is zero Note that the lines will be parallel, but more specifically they will be coincident So they will lie on top of each other. So I'm just drawing them very close to each other So yellow and white lines will be exactly overlapping on each other. Why am I saying that it is coincident? Not parallel Because theta equal to zero can also mean parallel, right? But why I'm saying coincident Because even the intercept is the same because they're passing both passing through origin. So they're passing through Yeah, they are both passing through origin. So it cannot happen that they are parallel and they're both passing through the same point It cannot happen. So they have to be coincident. Okay And if x square minus a b is negative The angles themselves because the lines are imaginary. You will not have a real angle there Another important thing that comes out from this exercise is that if a plus b is zero It implies that the lines are perpendicular Okay, we say this as if coefficient of x square In the equation of the pair of straight lines Plus the coefficient of y square in the equation of the pair of straight lines add up to give you zero That means The lines comprising that pair of straight lines are perpendicular So the lines are perpendicular Is that clear? I think this concept we have used in past to solve Locust questions. So I'll write it down useful for solving Locust questions So about a plus b is in the denominator. So if it's zero won't it become like So tan theta would be infinity Let's have a question on this. Okay, hope you can read this question Find the angle Between the lines Whose combined equation or whose pair of lines equation is given by this Or let me reframe this question prove that The angle between the two lines Prove that the angle between the two lines this value is going to be Two alpha where alpha is this value so that you can know whether you have done it correctly So prove that the angle between the lines represented by this Second-degree equation By the way, it's a second-degree homogeneous equation Will be two alpha Let's have a two and a half minutes for this It's just trigonometry guys. Nothing else. Just do your basic identities Take care of your basic trigonometric identities properly and you'll be done Where could you repeat the question my connection? Uh, do you want me to say repeat anything for you? Yes, sir. I didn't I couldn't hear the question The question was you have to prove that the angle between the lines whose equation whose combined equation is given by this Is two alpha Instead of find I have given you to prove it so that you know that you're going correct Type done once you're done so that we can discuss it Done because okay, but I'm also done Nice last 30 seconds guys So you also done Okay First let us write this in our ax square Two h x y by square equal to zero form. Okay, so I'll collect coefficients of x square Uh Just let me know in case I miss out on any term the coefficient of x square. I'll not expand it I can directly pick out from here would be sine square alpha coming from this term minus cos square beta Okay And x y term let me write down the x y term x y term would be two x y sine beta cos beta and y square term would have sine square alpha And from here I'll get sine square beta also Okay, so this is playing the role of my a Okay, this is playing the role of my h So I'm not mistaken and this is playing the role of my b So I know the formula that tan of theta would be under root of Two under root of h square minus a b by a plus b guys from my experience. I've seen people tend to forget this too Don't do that Okay So let me write it down. So h square h square would be sine square beta cos square beta minus a b a b would be this It's all about you know simplifying your This thing So it'll be mod of a plus b. So this ugly expression you get. Okay. So what I'll do is I'll just uh Try to simplify this further So in the numerator you'll end up getting sine square beta cos square beta You'll end up getting uh negative sine to the power four alpha minus sine square alpha sine square beta minus sine square alpha cos square beta And you'll end up getting plus sine square beta cos square beta. Okay In the denominator, we'll end up seeing uh Two sine square alpha minus one. Let me know if I have missed out on anything. Okay done, Ruchir also If you open the brackets If you open the brackets, you realize that this term and this term would stand cancelled off. Okay In these two terms if I take sine square alpha common, I'll get a one So You will finally see This result coming up A sine square alpha minus sine to the power four alpha My mod of two sine square alpha minus one Okay So Let's do one thing. Let's introduce everything inside the under root sign So I'll have sine square alpha and you'll have one minus sine square alpha Guys, we know that if it is mod, we can switch the position of These two terms and nothing will happen. We can also write this as like this. Isn't it? Yes or no and isn't this actually cos of two alpha So inside it'll go as cos square of two alpha Let me introduce this two inside also so that The numerator could be written as under root four sine square alpha cos square alpha by cos square two alpha Which is under root of sine square of two alpha by cos square of two alpha Okay, that is nothing but tan of two alpha in fact Okay, and so theta has to be two alpha, which is what we were required to prove It's just a you know, trigonometric identity application nothing else. There was no merit in this problem Okay So now we are going to move on to the concept of equation of The angle by sectors angle by sectors of a pair of straight lines Given by this homogeneous second degree. So we are still stuck with the homogeneous one only So now my question is let's say these blue lines represent my pair of straight lines given by this equation. Okay Let's say L1 and L2 are those two pair of straight lines Now I want to get the equation of The bisectors of these angles. So let me use some different color here. Let me use orange So these orange lines that you see here They are the bisector lines Let me call them as b1 and b2 So what do I want here? I want the combined equation of the angle bisectors just like you There's a combined equation of L1 and L2 How will I get this? So for that I would again use my concept that We had discussed in class 11th in straight lines If you choose any point h comma k on the bisector Can I say that perpendicular distance Of that point from the two lines l1 and l2 would be equal And it doesn't matter even if you choose here also h comma k H comma k this distance Sorry for that crooked line Will be equal to this distance Isn't it? So this is the underlining principle on which we are going to derive The combined equation guys again. I'm repeating I don't want the equation separately I'm dealing with the pair of equations Okay So I'll use my basic distance of a point from a line formula So I'll assume first that let y is equal to m1x And y is equal to m2x be these two straight lines The presented by this equation We could be there's so many ways we can do it Ruchir, but I would like to take up the shortest and most convenient way Okay, so now Can I say the distance of h comma k from l1 will be k minus m1h By under root of 1 plus m1 square mod of it This distance should be equal to the distance of h comma k from the other line l2 which is mod of k minus m2h by under root of 1 plus m2 square Right This is the distance formula I'm using guys. So don't be surprised. So how did I get this expression? It is from the distance formula Anybody who doesn't know the distance formula. Let me know right now so that I can write it down for you Okay, now without loss of time without, you know wasting time can we replace my k with y And h with x because sooner or later we'll we'll have to do that So let me Replace it right over here So i'm Okay, now we have to square both the sides and bring it to a second degree general a second degree homogeneous equation So let me square and cross multiply. So i'm directly writing those steps over here. So 1 plus m2 square y minus m1 x square Okay minus 1 plus m1 square y minus m2 x square equal to 0 Okay Now i'll be simplifying this pay attention while i'm simplifying this i'm going to hand pick my terms and write them So first of all, let me write down My y square term Okay, so y square term Okay We'll have coefficients 1 plus m2 square Minus 1 plus m1 square. So can I say m2 square minus m1 square? You can also do this simplification at your end also. You don't have to follow me for that Next I will take minus x square common and for that I will have again m2 square Minus m1 square remember m1 square m2 square will get cancelled. So it'll be only left with uh m1 square multiplied to 1 and m2 square multiplied to 1 Okay, so this will be the expression and my other terms like x y I will have I will have 2 x y 2 x y m2 minus m1 Minus 2 m1 m2 m2 minus m1. Let me just shift it to the right equal to 0 Just check if I have missed out on any expression. This is a minus sign. I'll write it once again This is minus m1 So from this entire expression, I can drop off m2 minus m1 So m2 minus m1 could be dropped off because we know it's not going to be zero So I'm dealing with such lines where m1 and m2 are distinct So m1 m2 cannot m1 minus m2 cannot be zero So I'm dropping off that factor from everywhere. So that will give me uh, I'm just clubbing y square minus x square also together So I'll be left with m2 plus m1 Okay, here I will end up getting 2 x y 1 minus 2 m1 m2. Please remember I have dropped I'll write it here. I have dropped the factor of m2 minus m1 Cool enough any doubt here Now we already know the values of m1 plus m2 and that is minus 2h by b Okay, we also know by the way, let me take it to the other side so 2 x y 2 m1 m2 minus 1 By the way, I should not have reversed the sign. I should have dropped this guy Drop this negative sign and maintain the same order Okay, any doubt in the simplification so far and this m1 m2 I'm going to write as What is m1 m2? a by b Let's also drop some other factors Uh, just a second I think this there will not be a 2 here 2 is already taken common out. I'm sorry 2 was already taken common out here. That's why By mistake I need to use a 2 over here now this 2 and this 2 just drop them off so you have y square x square h by b is equal to x y b minus a by b b and b also could be dropped off You can write this as x square minus y square by b minus a is equal to x y by h And this further could be just for the purpose of no writing x before y we can write it like this Please remember this equation going forward. You will not have time to derive them in your exam condition Please remember this very very useful The equation of the pair of bisectors Of so this is the equation of pair of bisectors Of the given line given pair of lines a x square Plus 2 h x y plus b y square. So this this line's pair of bisectors. That means your orange line Equations would be these These would be your orange lines Note it down. More importantly, it should be there The picture of it should be there in your mind. Okay, one quick analysis over here What is the coefficient of x square in the equation of the pair of bisectors? 1 by a minus b Equation a coefficient of x square. What is that 1 by a minus b all you know? Yeah Okay, what is the coefficient of y square in this minus 1 by a minus b And if you add them, what will happen if you add them, what will happen? Zero what is They are perpendicular, which is very obvious. No, because we have also done this in class 11 that your bisectors will always be perpendicular to each other Yes, I know So this is how we can actually see whether our result or verify whether our result is correct or not Now answer this question And I'm going to ask you What will happen if a is equal to b? How would this equation change? Would you say undefined? Yeah, undefined No guys No, sir If a is equal to b this equation just converts to x square minus y square equal to zero That means y is equal to x and y equal to minus x would act as the pair of bisectors Okay, see later on in your 3d chapter you would you know, uh study about 3d equations There you would see such weird expressions I'm just giving you a dummy example So this zero is not an operational. This is not an operation. This is not an operation that you're performing. This is symbolic Okay, so you can have such equations in you know, 3d lines You'll you'll study it soon in the month of early October or September you'll study that so presence of a zero here doesn't mean this equation has become undefined it just another way of saying that x square minus y square will be zero In the same way if h is zero it just means your pair of bisectors would have x y equal to zero which is actually your x-axis and y-axis equations So your bisectors would be the coordinate axes themselves Getting this point Okay, so don't be like oh what will happen now my a is equal to b I don't know what to do in this formula everything becomes undefined nothing to panic You just a equal to b means x square minus y square equal to zero is your pair of bisectors h equal to zero means x y equal to zero is your pair of bisectors Yeah, sure Karthik see what do I mean is when you say a is equal to b this expression would have become like this, isn't it? Okay, just take the zero on the other side and write it as x square minus y square equal to zero So that's how you get this equation It doesn't it should not give you a feeling that you're dividing by zero which is not defined and all It is just a symbolic representation It just says it doesn't depend your your pair of bisectors doesn't depend on h It is just trying to say that Similarly, if your h is zero it says it doesn't depend upon your a and b it will definitely be your x and y-axis Getting my point Okay, so with this we are good to take some questions Let's have few questions Okay, let me start with the easiest one Hope you can read this find the equation of the bisector Find the equation of the bisector of the lines If it is hidden, let me rewrite it for you three x square minus five x y Plus four y square equal to zero Quickly Combined equation always see the equation of the bisectors combined equation reward Uh, let me know once you're done or you can type it out on the on the chat box also Okay Richard white 25 x square Can you check that out once please Correct, uh, Richard Mittle is correct. No, no Richard Mittle also some mistake. Check it out My this correct my this correct Param is correct. Yeah, guys. It's very simple formula based No, brainer. So you just have to write x square minus y square By a minus b a minus b is three minus four is equal to x y by h h is minus five by two my dear Don't do those mistakes. Okay. So you can write five by two x square minus y square is equal to x y So you can say five x square minus five y square minus two x y equal to zero. This is your final answer Okay, simple x y by h This is the formula. I'll write it down here also x square minus y square by a minus b is equal to x y by h Okay Yeah, those who have made a mistake now, you know where you went wrong. Let's have one more question. Okay. Here you go So the question reads like this if pairs of straight lines x square minus Two p x y minus y square equal to zero And x square minus two q x y minus y square equal to zero be such that each pair bisects the angle between the other pair In other words, one of them is the pair of bisectors for the other one. Okay Then prove that pq is equal to one. This can come as an objective question also Then they can ask then pq is which of the following So it basically means that Like the lines constituting the first equation of bisectors for the lines constituting the second one advice And both of you can see their perpendicular lines because the coefficient of x square and y square is zero here adding up to zero here And also in this case also coefficient of x square and y square are adding up to zero So it is a situation where something like this is happening. So these are the orange perpendicular lines And bisecting them are two blue perpendicular lines, which I'm showing here Okay, so even these two are perpendicular and these two are perpendicular Let me know once you're done done done done. Okay Akritika, uh, is the question clear to you? Yes Drew done He are done. Ruchir Singh done Akansha done Param also done Anush also done. Okay guys most of you are done here So guys nothing to panic. All you need to do is just pick up The first pair of straight lines. So I pick up this guy Okay, and I would write down the bisectors equation So the equation of the bisectors or the pair of bisectors would be x square minus y square by a minus b a minus b is one minus minus one Correct is equal to x y by h h is minus p if I'm not wrong correct In that case this equation will become negative p x square Minus two x y Plus p y square equal to zero. I'm just simplifying it on the go Now this equation that you have is same as The given equation x square minus two q x y minus y square equal to zero That means they're both the same equations as per the question So what we can do is we can do a comparison You can do a comparison of these two equations So we can say the coefficients of x square would be proportional Two coefficients of x y that's minus two by minus two q And that will also be proportional to coefficients of y square Okay Now these two doesn't bring anything new to the table because this is always true So we have to take this into our account By the way, I have Written p over here my bad that was supposed to be q Okay, so from the white circled expression I can say by the minus two and this will go off So you can say p or minus pq is equal to one cross multiply and hence pq becomes one minus one It's a very very Commonly asked questions in regional entrance exams. So in ct comet k bits at v it etc. You can see this question Okay, let's have another one Let's have another one Let's have another Yeah, let's have this one Yeah question says if the lines represented by x square minus two p x y minus y square equal to zero Is rotated about the origin? Through an angle theta one in clockwise direction Other in anti clockwise direction Find the equation of the bisector of the angle between the lines in the new position You got the question right So one of the two lines is rotated theta clockwise and the other one is rotated by theta anti clockwise Then what is the equation of the bisectors of the new line? Ah, you got it correct param correct Richard This was actually a googly question Yeah, you all are correct guys. Well done. There would be no change in the equation So it will be our regular equation that we have seen that is x square minus y square By one minus minus one is equal to x y by minus p Okay, we have already seen this I think a little while ago So p x square plus 2 x y minus p y square equal to zero Okay Now guys, we are going to Uh, move towards general equation of second degree general Equation of second degree So now homogeneous one we are done with now we are moving on to the general case so, uh Why not here? Will will there be any change in the position of the bisectors if one is moving clockwise theta other is moving anti clockwise theta No That's the equal angle No change will be there no here It's like the two lines are just coming closer together, but the Bisectors will still be at the same point. So now she has got It's uh, let's do one thing. Uh, this is going to be again our time consuming topic It's a good time for now to have a break. Can we meet at 11 o'clock now? So eat something so we'll resume at 11 o'clock Okay, so 10 minutes break have something so that we can uh sit for uh till one o'clock the next two hours Okay, so this is the starting of the new concept. I thought I would give you a break now Okay So let's have a break