 I'm going to begin my today's session with the equation of bisectors. Okay, so we'll be discussing the equation of bisectors of the angle of the angle between the lines between the lines represented by the homogeneous second degree equation. Okay, so represented by or you can say given by, given by homogeneous second degree equation. So right now I'm focusing more on homogeneous second degree. Okay, but later on I will generalize it even if the pair of straight lines is not represented by a homogeneous equation, but instead represented by a general second degree equation. Okay, yes, Gaurav, you have any questions? Any other day also? Yes, yes, yes. See, take JEE today, KVPY tomorrow, it's fine. Okay, so you have available, a lot of time available in between also, right? So you can take it tomorrow also. Yes, will be open. Yes, will be open. Don't worry about it. The same link you have to go and take it the normal way. Fine. Okay, so let us say we have been provided with a pair of straight lines whose combined equation is a general second degree equation. Okay, now we need to know the equation of the bisectors. I would say you need to know the combined equation of the bisectors. So that will also be a pair of straight lines only, right? So here please understand in the last class we discussed that a pair of straight lines whose equation is given by a homogeneous equation that will always be two lines passing through the origin. Okay, so let's say one line is like this, another line is like this. So that will always represent a pair of straight lines passing through the origin. Okay, so let this line be y equal to m1x, this line be y equal to m2x. And we had also learned in the last class, we had also learned in the last class, what is the relationship between m1, m2, h, a, and b? Can somebody tell me what is that relation? If you want, you can write it on your chat box. m1 plus m2 is, tell me, minus complete 2h by b, not a, b, b, b. And m1 into m2 is a by b. Okay, so please remember this. Okay. Yeah, it's very easy to swap the position of a and b also. Okay, so it is m1 plus m2 is minus 2h by b and m1 m2 is a by b. Pranav Dharmagadi, where were you, my dear friend? Long time, no c, not coming in the class also. Everything okay? I mean health wise everything is okay. I hope you don't have any issues. Okay, fine. So you might, you might tell me separately also not on this forum. So you were not there for the last at least three, four classes, I think. Anyways, so now I have to find the equation of the bisectors. So let me make the bisectors with blue color. So let's say this is one bisector, so it is bisecting the, let's say acute angle between it. And this is another bisector, which is bisecting. Oh, let me change the color, why to keep the same color. Let me make it in gray. So the gray line is bisecting the obtuse angle between them. And of course, you would already know from your straight line chapters that this angle will be a 90 degree. So angle between the bisectors will be a 90 degree. Okay. And let's say these two bisectors, I am naming it as b1 and b2, just for the reference of calling them. So b1 is the acute angle bisector, b2 is the obtuse angle bisector here. Anyways, so I need the combined equation of b1, b2. So how do I get the combined equation? Any suggestion? Anybody has any suggestion? Anybody has? Okay, so can we realize that if you take any point on the bisector? Okay, if you take any h comma k point on the bisector, what was the basic principle of deriving the equation of the bisectors in straight line chapter of class 11? We used to say that the perpendicular distance of this point from the two lines would be equal. So these two distances used to be equal, isn't it? So assume that your line is y equal to m1x and y equal to m2x. So can I use the fact that if I take a generic point, pm, pn should be equal. So that is the core principle or that is the locus definition of getting the equation of the bisectors. So pm should be equal to pn. Now what is p? Now this line, this line, you may write it as a general form like this and this line, you can write it as a general form like this. So what is the distance of h comma k from y minus m1x equal to zero? So we'll say a simple mod k minus m1h by under root of 1 plus m1 square. This should be equal to the distance of p from the second line that is y is equal to m2x. So that perpendicular distance will be k minus m2h mod by under root of 1 plus m2 square. Now these kind of things you should start realizing on your own because your locus conditions, that will be tested for your comparative exams. So these type of situations you should be able to derive. Don't treat this as if a theory has been told to you. You should be able to get to these kind of equations. Now of course my equation only has a, b, and h. And here you can see there's m1, m2 also appearing. So I cannot generalize it over here. So what I'll do is I'll do some calculations here. I will try to get m1 and m2 out of this game by using these two conditions, that is m1 plus m2 is minus 2h by b and m1, m2 is a by b. So I have to use these two equations and try to eliminate my m1 and m2 from this locus relation or locus condition. Let us see how do we do that. So first I will, if you want I can put my y and x directly over here and then we can go for the simplification part because ultimately we have to replace k with y and h with x. So why not do it now only so that, yeah. So let's cross multiply and square it. So it will be 1 plus m2 square y1 minus m1 x square is equal to, is equal to 1 plus m1 square y minus m2 x the whole square. Okay. Now let us try collecting your coefficients. So first I will collect the coefficients of y square. So y square coefficients would be, so m2 square will come from here and m1 square will come from here. So m2 square minus m1 square. Let's collect similarly the coefficients of x square. x square coefficients will be m1 square minus m2 square. Okay. And let's collect the coefficient of xy also. Okay. So xy coefficient will be, if I am not mistaken, 2 m2 minus m1, sorry 2 m1. And the constant terms would be, if I am not mistaken, it will be minus 2, oh there are more terms, sorry for xy terms there will be more terms once again. I missed out those terms. So there will be more terms, you will have 2 m2 minus m1 and will also have minus 2 m1 m2 m2 minus m1. Please have a look at it and let me know if I missed out anything important. Now let's try to, let's try to simplify this. So this is m2 square minus m1 square. This is also minus x square m2 square minus m1 square. Okay. And here also I can take m2 minus m1 common. So you can take 2 common xy, m2 minus m1 common and you will have 1 minus m1 m2. Okay. Now m1 and m2 are not equal. So you can drop the factor of m1 minus m2. So this becomes m1 plus m2. I can club these two actually together. So I can say m1 plus m2 times x square, oh sorry y square minus x square. And here I will get 2 xy 1 minus m1 m2. Okay. Now if you see here, if you see here, you already have achieved terms like m1 plus m2 and m1 into m2. So please replace, please replace m1 plus m2 with minus 2 h by b and m1 into m2 with a by b. So when you make these replacements, let us see what do we get over here. So here what I am going to do, I am going to just swap the positions of x square and y square so that negative sign is taken care. Okay. And this is 1 minus a by b. Okay. So b also you can just remove from everywhere and multiply here with a b. Okay. So from here we end up getting x square minus y square into 2 h is equal to 2 xy a minus b. 2 2 also you can get rid of. Okay. So normally in the books you realize that they write the final equation like this. Okay. Maybe it helps you to remember the equation also. So this becomes your equation of the pair of bisectors b1, b2. So this is the equation of the pair of bisectors b1, b2. Okay. Now a quick analysis about this equation, but before that please note this down. Many people ask, sir, we should remember this equation. I would say yes, my dear, you have to remember it because it would be required in some complicated problem. And again, but remembering doesn't mean you'll sit and finally mug up all the formula existing in coordinate geometry. You have to keep solving problems in order to keep it fresh in your mind. Simple as that. It's like you don't remember to do a cycling. You do cycling every day and hence you end up remembering it. So please keep this in your mind in the same way. Any questions, any concerns with respect to the derivation or with respect to the concept used? I think after this step you should directly understand that it's all about simplification and getting your m1 and m2 eliminated from here. Okay. So let's do a few quick analysis on this. So related to the bisector equation, one thing that you would observe here is that the coefficient of x square and y square, they add up to give you zero. So what is the coefficient of x square here? The coefficient of x square here is 1 by a minus b. And what is the coefficient of y square over here? Negative of that. Correct. So when the coefficient of x square and y square add up to give you zero, what does it tell you about the component lines? What does it tell you about the component lines? In a x square plus 2hxy plus by square equal to zero, if a plus b adds up to give you zero, what does it mean? Yes, it means that the lines are b1 and b2 are perpendicular to each other. And that's precisely what we already have discovered earlier also. Okay. So while making the diagram, I already told you that these two bisectors will be perpendicular, as we have learned in class 11th as well. Okay. So that thing is getting valid over here. So that is good. Okay. Second thing important here is many people ask what will happen if a and b are equal. Okay. So you may get a pair of straight lines whose a and b are equal. Then what will you do? Because zero will come here in the denominator. So in that case, what do we say? Use a symbolic representation. If you use a symbolic representation, you'll automatically get x square minus y square is equal to zero. In short, x equal to y and x equal to minus y will become the pair of bisectors for those lines. For those lines where a and b are equal to each other. Okay. Similarly, you may also have a pair of straight lines given to you where your h may be zero. Correct. So if your h is zero again, use a symbolic representation here that if your h is zero, then this zero goes on the other side and your x, y becomes zero. That means your coordinate axes. That means your coordinate axes, that is your x-axis and the y-axis themselves, they act as your pair of or you can say component bisectors. Okay. So their combined equation is x, y equal to zero. So x, y equal to zero is what? Your coordinate axes equation combined, isn't it? x-axis is y equal to zero, y-axis is x equal to zero. So the combined equation of the coordinate axes themselves is x, y equal to zero. So please make a note of this. Let's take a small problem based on the same. Could you show the diagram on the last equation once? Oh, this diagram. Yeah. Yeah. Why not? Sharda, you want me to show you this equation also? Let me know once you're done. Great. So let's take questions. Yeah. So you have a question here which asks you to find the equation of the bisectors of the angle between the lines represented by this. Just a hands-on on the concept. That's it. Done. Simple. Just use of the formula. So your a is a three here, b is a four here, h is a minus five by two. So what is and this is a homogeneous equation also. So for homogeneous equations, our pair of bisectors, one important thing I would like to tell you here, even though it is a very trivial reminder that you are basically given a homogeneous equation and what we derived was only for a homogeneous equation. So what if it is a general second-degree equation for a pair of straight lines? How would the bisectors change? We will talk about it in some time. Don't worry about it. Okay. So as per the formula that we have used, so your a will be three, sorry, a will be three, b will be four. Okay. And your xy will be minus five by two. Okay. So you can write x square minus y square upon minus one is equal to two xy by minus five. Just do a cross multiplication and simplify it. So that will give you minus five x square plus five y square plus two xy equal to zero. You can write it as five x square minus two xy minus five y square is equal to zero. Okay. So this is your desired result. Is it fine? Any questions just as a hands on on the formula? That's why I gave you this particular question. Is it fine? Any questions? Any concerns? Okay. Should I go to the next question? If x square minus two p xy minus y square equal to zero x square minus two q xy minus y square is equal to zero, be such pairs, be such pairs of straight line that each pair bisects the angle between the other pair, the angle between the other pair, the other pair, then find the value of, find the value of pq, then find the value of pq. So read the question. So there are two pairs here. One is this pair, another is this pair and each pair, each pair bisects the angle between the other pair. Very good, Kinship. Aditya, no, that is not right. Check it out. Okay. Should we discuss it? See, very simple. Focus on one of the lines. Okay. Let's say I focus on the first pair. Let me call it as pair one, pair two. Okay. So as per pair one is, a is one, b is minus one and h is minus b. Correct. So if I have to write a pair of bisectors for, so what are the bisector equations for pair one? x square minus y square by a minus b, a minus b is two is xy by minus b. Okay. So here if you see the combined equation will be minus px square plus py square and you will have minus two xy of the bisectors. Now as per the question, the question center is claiming that this actually is the p two pair. This actually should be the p two pair. Correct. So in, in other words, we need to do a comparison of minus px square minus two xy plus py square equal to zero, along with the given bisector pair p two, the given pair p two, and that is negative, sorry, x square minus two q xy minus y square equal to zero. Okay. So let's do a coefficient comparison. So as per the given problem, these two should represent the same pair. So minus p by one is equal to minus two by minus two q is same as p by minus one. Of course, the first and the last expression doesn't mean the same thing. So you just have to compare these two. So from here, you can say two pq is equal to minus two. So pq value is equal to minus one. Okay. Actually, this problem is a, is a very well known question also, and it is asked in several regional entrance exams also. J is one of them. I mean, many a times this question is actually asked. Okay. So last year, I think people told me said this question is seen almost everywhere. So we have ended up remembering the result pq is minus one. Yeah. So please note this down. Any question that you have to let me know. Show the right part, right part is this part. So Dan, can you please draw the lines for this same set of lines? No. See, they both will be perpendicular lines. One pair will be this. Why? Because x square and y square coefficients are adding up to zero, isn't it? Another will be like this. The yellow ones are one pair. The gray ones are another pair. Okay. So this, this angle and this angle is equal. This angle and this angle is equal. Okay. And of course, the yellow ones are at 90 degree. The gray ones are also at 90 degree. This is the situation. Got it? So done. So each bisects the angle between the others and both are pair of perpendicular lines passing through origin. Any questions? So can we comment on the angles? Angles as in angles between what? The angle between this two has to be 45 degrees, isn't it? Won't it be? If this 90 degrees bisected, this is, this is going to be 45 degrees. Okay. Now we are going to talk about the general equations, general equation of pair of straight lines. I think we had already spoken about it in the beginning part of the chapter. So now we have gone past the homogeneous equations and now we are going to talk about such lines which not only have x square, xy, y square term but may also have xy and a constant term in the equation. Okay. So basically it just tells you that such a line is a line which is not or which may not be passing through the origin. Right? So their point of intersection is something which is, you know, not the origin. Okay. If at all they intersect, so I don't mean to say that they will always intersect. They may also represent pair of parallel lines also. Okay. So in such cases, your point of intersection of the lines, if at all they intersect, need not be origin or will not be origin depending upon your GFNC value. Okay. So the first thing that I'm going to do is, if these lines intersect, so let's say I call this equation as E. So if E represents the pair of lines that intersect. Okay. Then let us find out the point of intersection. So let's find out the point of intersection. Okay. How would you find the point of intersection? So for point of intersection, I already told you the process is you consider this equation to be or you consider this term to be our expression to be S. So simultaneously solve dou S by dou X equal to 0 and dou S by dou Y equal to 0. Okay. So let us do that. Let us figure out what is the point of intersection. And is there any way that we can remember the point of intersection? So dou S by dou Y, sorry, dou S by dou X will give you 2AX, 2AX. This will be 2HY. This will be 2G equal to 0. Drop the factor of 2s. It is not going to be important. Okay. Similarly, dou S by dou Y. Dou S by dou Y will be 2BY. So let me write BY. This will give you 2HX. So let me write HX and this will give us 2F. So let me just write an F because 2, 2 factors. I am just removing it off. Now, let us use our Cramer's rule to get my value of X. In fact, I should write minus Y and 1. Cross multiplication method I am using. So HF, HF minus BG. This will be AF minus GH. Okay. And this will be AB minus H square. Okay. So we normally write X value as HF minus BG by AB minus H square and Y value as AF minus GH by H square minus AB. So remember, there is a minus sign sitting over here. So I just switched the position of H square and AB. Now, in order to make it consistent, okay, so we will write it as BG minus HF by H square minus AB comma AF minus GH by H square minus AB. Is this fine? Now, there is a way to remember this result. If you remember the delta expression that we had seen when we were talking about the condition for a pair or condition of a second degree equation to represent a pair of straight lines, the delta value or the delta expression used to be this, isn't it? Okay. So in this, what do you do? You take the first two rows elements like this, A, H, H, B, G, F and then repeat AH over here. So I am just telling you a way to remember. Memory 8. Now, see here, this term will be your denominators AB minus H square. Now, here you have to change the position. It doesn't matter even if you keep AB minus H square. So this term, the term which is coming from this operation will be your denominator for every term. Okay. This HF minus BG will be the denominator for the X coordinate. Okay. And similarly, GH minus AF will be the numerator for the Y coordinate. So you can remember the result like this. So the yellow one, first write HF minus BG divided by the white one, that will become your X coordinate. Similarly, GH minus AF, that is your second yellow brisk cross divided by the white one, again, that will become your Y coordinate of the point of intersection. So this is how you can remember the formula for the point of intersection. In case you want to use it, else if you have numbers, you can always use the numbers to get to solve this equation. Okay. So I have told you how to remember the point of intersection just to save time. If you are not able to remember this, you will not lose anything other than few seconds, maybe a half a minute or something. Okay. So this is the coordinate of the pair of, sorry, the coordinate of intersection of the pair of straight lines, if at all they intersect. Now there are certain things over here which will tell you when will they not intersect. Okay. When will they not intersect? Okay. So we will do a small analysis over here. Are they not completely correct? We will talk about it. We will talk about it. Our next discussion will be based on that only. So now we will be talking about what is the situation when this pair of straight lines, okay, may be representing two parallel lines or may be representing two coincident lines. So what are the conditions that it must satisfy? So here I'm assuming that they're intersecting at a point. But who knows that they may be intersecting at no point at all. That means they may be parallel. Right. So they could be three situations. They could be like this. They could be like this, or they could be one over the other. Okay. Who knows? Isn't it? So we'll be trying to analyze this situation. Right now what we have seen is a very sober case where they are intersecting at one point. But again, it all depends upon what is your h a and b. So this is just a formula, which needs to be analyzed into more detail. Okay. So if the pair of straight lines represent parallel lines, what will happen? If they represent coincident lines, what will happen to these coordinates? That also we need to look into our analysis. Okay. So let me go to the next page and start doing that analysis. If everybody has noted this, can we move on to the next page? But before we move on, I would like to take a simple question just to give you an idea about the use of do s by do x and do s by do y equal to zero. Given that, given that this represents a pair of straight lines, which intersect at a point. Okay. Find the point of intersection. Find the point of intersection. I would request not to use the formula, but directly come from your do s by do x equal to zero, do s by do y equal to zero and try to solve them simultaneously. Two comma one is absolutely right. Simple. So again, go from the basics. The basic was we have to simultaneously solve do s by do x and do s by do y equal to zero equation. So this is going to give you a four x. Okay. This is going to give you a minus seven y minus one equal to zero. Okay. And this is going to give you a minus seven x minus eight y plus 22 equal to zero. Let's solve them simultaneously. Okay. So let's multiply this with a okay, or we can use our cross multiplication method itself here. So this will be minus 154 minus 154 minus eight. This will be 88, 88 minus seven. And this is going to be minus 32 minus 49. Yeah. So this is, this is nothing but x by 162. This is nothing but y by minus 81. And this is one by minus 81. So your x value has to be two y value has to be one. Yes. So point of intersection is going to be two comma one. No issues. Okay. Now let's talk about conditions, condition for a second degree equation to represent, to represent coincident lines, to represent coincident lines. Coincident lines means both the lines are actually the equation of the same thing, but may differ from each other by a proportionality constant. So let's try to figure out what are the conditions that must be honored if you are basically talking about a pair of straight lines which are coincident to each other. Okay. So let's try to look into that situation. So here, let us say our pair of straight lines, our pair of straight lines is composed of, this is composed of lx, I'm so sorry, lx plus my plus let's say n and l dash x m dash y plus n dash. Okay. So let us say this pair of straight lines is composed of these two lines where the lines are coincident. So lines are coincident means l by l dash m by m dash and n by n dash should be equal. That's the only way you can basically claim that the two lines which are basically making this pair of straight line equation are coincident. Okay. So only coincident lines will show this particular type of characteristic, isn't it? So we already know that if these two lines are coincident lines means they're one and the same equations, then I can say that the ratio of the coefficients of x should be same as the ratio of coefficient of y should be same as the ratio of coefficient of your constant term because they're representing same lines just differing. They may differ in constant terms. That is why I'm taking the ratio. I'm not doing a direct comparison. I'm not saying l is equal to l dash. Okay. Because something like this could also be there. No plus one equal to zero and the other is three x plus three y plus three equal to zero. Both are coincident lines or you can say both are like equations, two equations of the very same line. Okay. So this could be a scenario. Now, first of all, what I'll do, I'll do a coefficient comparison. So let's do a coefficient comparison. So let's take our x square coefficient on both the sides and compare it. So l l dash will become a m m dash will become b. Correct. And n dash, that is a constant terms will become a C. Correct. Can I say l m dash l dash m will become a two H. Okay. So when this multiplies with this, this multiplies with this, I'll get a x y coefficient and that is equal to why am I y is looking like a G. Yeah. So that will become two H. Okay. Similarly, l n dash l dash n is equal to two G. m n dash m dash n is equal to two F. Okay. Now, since this condition is true, since this condition is true, can I say, oh, sorry, can I say from here, let me write it here, can I say, if I take the first pair, can I say l m dash minus l dash m is zero. Can I also say m n dash minus m dash n will also be zero from this. Okay. And finally, we can also claim that l n dash minus l dash n will also be zero. Okay. Now, if you use this, can I say it is as good as saying l m dash l dash m whole square is zero. And what is whole square? Whole square of this can also be written as l m dash plus l dash m whole square minus four l l dash m m dash. Basically, I've used a minus b whole square as a plus b whole square minus four a b formula. Okay. Now, why did I actually write it like this? Because you have information about l m dash plus l dash m, which is actually two H. Okay. So this will become two H the whole square. And you have information about l l dash, which is a and m m dash, which is b. Okay. So please note, h square is equal to a b is one of the necessary conditions for it to represent coincident lines. So this is just one of the conditions, Aditya. Okay. So please make a note of this. This is our first condition. Okay. The other conditions also must be met. That means the second one. So I'll write it similarly. Please do let me know if you feel I have jumped any steps which you wanted to know from me. Same thing. I'm not writing anything separate. It's the same approach. Okay. And what is this? What is this m n dash plus n dash m? That is equal to two F, isn't it? So I can say this is going to be two F the whole square minus four. What is m m dash m m dash is b and then dash is C. So this is b c. Okay. So this should be zero. In other words, f square should be equal to b c. This is your second condition. Okay. And similarly, from the last one, from the last one, l n dash, let me keep it in yellow only, l n dash minus l dash n is zero. Our square of this is zero. That means l n dash plus l dash n, the whole square minus four l l dash and n dash is equal to zero. That means g square to g square minus four ac is zero. So g square should be equal to ac. Okay. So now please note this down. Very, very important. So from one to n three, what we come to know, the conditions that are supposed to be met, the conditions for the pair to represent to represent coincident lines is number one. At square should be equal to a b, but that is not sufficient. Only this will not tell you because the square is equal to a b is for one more situation we will see in some time. Apart from that, you should have f square should be equal to b c and g square should be equal to ac. So many a times a direct multiple option correct question is framed in the J advance exam for this pair to represent a pair of straight lines, which of the following conditions must be true and you have to mark all the three. You have to mark all the three. Okay. Now from here, a lot of other conditions also come into picture because some derived conditions can also come into picture. For example, for example, if you multiply both the sides, you'll end up getting f square. So last two, if you multiply, you end up getting something like this and you already know a b is x square, isn't it? So you can write it like this also. Or you can say f h is equal to, this could be a condition. Right? Who knows? This could be a condition. Okay. So what I'm trying to say here is that the question that may be asked or given to you to solve may not be directly stating these three conditions, but something derived from it. Are you getting what I'm trying to say? So they may divide it also. They will say, okay, is this a condition to be true? Or is af square should be equal to bg square? Then yes, because from these two, this condition might come into picture. Are you getting my point? So these are the things which they may ask you in your options and you have to be careful about it. So these are the three basic conditions from here. A lot of conditions can be derived and that depends on the question. I cannot sit and give you all the conditions. So this is an example of a derived condition, something like that. Is it fine? Any questions? Any concerns that you have? Do let me know. Okay. Now we'll talk about when does it represent or what is the condition to be satisfied when they represent a pair of parallel lines? So primary conditions for ax square plus 2hxy plus by square plus 2gx plus 2fy plus c equal to 0 to represent parallel lines. To represent parallel lines. Okay. Actually, I just want to know if you are, Anurag is in HSR or is in YPR? Anurag BJ. He's an HSR student? Okay. If anybody is in touch with him, just do let him know that I was wondering why he's not coming for classes. This entire area has been very dicey. Okay. See, these are very, very critical moments for you. Right now, if you let hold, let lose the hold on your preparation, it is going to be a giant fall. Okay. So keep running. I know you are all very tired mentally, but keep running. If you stop here, you lose the race and that that repent will be more painful. Okay. So some of you have seen, I don't want to name the people here in this session. Some of you have given up badly in the coming in the last six or four or five months. Those who are consistent, that will definitely pay you off. Okay. So please don't let go your preparation hold at this juncture. There's life after class 12th also. Okay. Anyways, so let's discuss this condition. So for this, again, we will have a similar approach. In fact, I wanted to give this as a homework question, but I would just do it for you. Okay. So in this case, what do we say? If they represent parallel lines, let the parallel lines be, let the parallel lines be these lines. Okay. Where N and N dash are not equal to each other. Rest, I've kept the coefficient of x square y square. I have kept the coefficients to be the same. Okay. So here also if you compare, you get L square as A, you get M square as B and you get H as correct me if I'm wrong, that will give you LM. Okay. LM. This also LM. So 2LM is 2HXY. I'm just directly comparing it. And the X term would be G term would be, let me write it like this. LM terms to left and our given terms to the right. Yeah. Yes. So this is going to be your, this thing and X term will be, correct me if I'm wrong. So X term here would be N dash or LM dash plus LM or you can say L common N dash plus N that will be equal to 2G. And similarly ML, sorry, M N dash and M N. So M could be taken common. This is equal to 2F and constant is N N dash, which is equal to C. Now here, here, what condition do you think must be satisfied for them to represent, for them to represent parallel lines? Tell me. Same, I think if you see here, H square is equal to L square M square and that is also equal to product of AB. So this condition will still come out to be true. S square is equal to AB. So S square equal to AB condition will be true for both the scenarios, whether you have, whether you have the lines are coincident or whether they are parallel. So this condition is true even for, so true for coincident lines, coincident lines as well as parallel lines. What else can I say? What else can I say? I can see from here that L by M is equal to G by F, isn't it? A lot of things can be said here. If you square both the sites, it is as good as saying A by B is equal to G square by F square. That means A F square is equal to B G square also can be said. And if you remember this condition, I think I deleted it or something. So if I divide this, so this condition comes from these two equations also combined. So please be very, very careful about what are you basically planning to get from the options. So this condition is also true when you had coincident lines. But are the equations satisfying these two conditions also individually? That is also to be kept in mind. If F square is equal to BC and G square is equal to AC for this case, that will not happen. That is not going to happen. So please understand here that the conditions are very similar to each other. But we have to see in combination, they will satisfy a condition for you to represent coincident lines. And in combination, they will satisfy an equation for you to represent parallel lines. So if these two are simultaneously satisfied, then you could have any of the two cases. We have to check out which will be the differentiator for it to be coincident line and which will be the differentiator for it to be parallel lines. And now I will tell you the differentiator also. I will tell you the differentiator also. The differentiator is, in this case, F square minus BC and G square minus AC is the differentiator. So in the second case, when it is representing parallel lines, in this case, you would realize G square will not be equal to AC and F square will not be equal to BC. So these two conditions will not be satisfied. G square is not equal to AC and F square will not be equal to BC for this case. Now many people say, sir, how did you get this? How did you get this? Now see, if the two lines are representing parallel lines, so if let's say one is LX plus MY plus N is equal to 0 and LX plus MY plus N dash is equal to 0. Now what would be the distance between the two parallel lines? If I ask you, what would be the distance between the two parallel lines? If you have remembered, we had done distance between parallel lines in our class 11th straight line chapter, I'm sorry, it is N minus N dash by under root of coefficient of X square and Y square. Now if you see N minus N dash modulus could be written as N minus, sorry, N plus N dash the whole square minus 4 N N dash under root divided by under root L square plus M square. Now what is N plus N dash? Let's try to get it either from this equation or from this equation. So let us use this one first. I could use the second one as well. So if I use this from here, I would get N plus N dash as 2 G by L. So let me put it over here. 4 N N dash, N N dash can be obtained from the last equation, which is equal to C. Okay, whole divided by L square M square. Now let me just write it completely in terms of G, A, C, etc. So this will become 4 G square minus 4 L square C under root by under root of L will come down under root of L square plus M square. Now remember L square was A. Remember L square was A. So what I'm going to do here, I'm going to write this as 4. In fact, let me take it outside the under root. So 2 under root of G square minus A, C. This will be root A. This will be A plus B because M square is a B. So in short, you'll end up getting something like this. In fact, I would like you to note this result down because this result gives you the distance between the parallel lines which comprise the pair of straight lines. So please note this down. And if they really represent pair of parallel lines, remember the numerator term should not be 0 because then the distance between them will become a 0. So if they represent parallel lines, this will not be 0. While in the previous case, if you see G square was equal to A, C, look at the last condition. G square was equal to A, C. That means your distance will become 0. So this is the distinguishing factor. This will tell you that, oh no, this is not a coincident line. This is going to be parallel lines. So there will be a lot of common conditions and there will be a lot of distinguishes also. These are the distinguishes. Are you getting my point? But this individually will not help you because this may be not 0 for even lines which may be not in either of the two. So please do not read them as a self-sufficient conditions. You have to analyze several conditions in order to come to a conclusion. Now many people say, sir, if you had used the second condition that is n plus n dash, n plus n dash is 2f by m. Then what would you have got? If I had used n plus n dash is equal to 2f by m, if I had used this condition, then my distance between two parallel lines, I don't know why my voice is cracking, then the distance between two parallel lines here would have become 2 under root of f square minus bc by under root of b a plus b. Correct me if I am wrong. And both of these are equal in value. If they represent parallel lines, they both will give you equal result because both represent the distance between the two lines. And here also if you see, if they want, if you want them to represent parallel lines means the numerator should not be 0. That means this term should not be 0. Then only you will get a parallel line. Okay. So many people ask me, sir, this result and this result. Do we need to remember? Do we need to remember? Yes. So please remember this as well. Sir, how many things to remember? Don't worry. This is the last set of formulas that you will be remembering. Okay. Any questions? So mostly my experience tells me that the question that will be asked from pair of straight lines, if at all the examiner has to confuse you, they either put it from these concept of coincident and parallel lines or they ask you the concept of homogenization, which I am going to do in next few minutes with you. Okay. So just to summarize what I want to say, if you want to have parallel lines, X square should be equal to AB. Okay. And we also discovered that BG square should be equal to AF square. Right? But we also discovered that G square will not be equal to AC and F square will not be equal to BC. Okay. Just these two, if you take, it may represent coincident line also. Getting my point, what I am trying to say. Okay. What happened to my voice here? Okay. Good enough. Any questions? Any concerns? Do let me know. Now coming to the equation of bisectors. Coming to the equation of bisectors for a general second degree pair of straight lines. For a general equation of pair of straight lines, in this case, there is shift of origin, which basically helps you. Now see, if this was origin, okay, if this was origin, that means your pair of straight line was a homogeneous equation, homogeneous second degree equation, then your equation would have been this, which we had already seen in our derivation today in the morning. Okay. First thing that we did was we derived this equation. But now if you want X1, Y1 to come over here, then what are you doing? Where are you shifting your origin to? Can anybody tell me that? If you want X1, Y1 to come in place of the origin, right? Where are you shifting your origin to? If you want this to become X1, Y1, where should the origin go? So, minus X1, Y1. Yes. So, if you shift your origin to minus X1 minus Y1, how would this equation change? How would this equation change when you shift your origin to H comma k? How does your X become? It becomes this. Y becomes this. Am I not right? Yes. In a similar way, small X will become capital X plus H. H is my dear minus X1. Don't confuse. Okay. Now, it is not a good practice. It is not a good practice to write things in capital. So, we'll write things in small. So, please make a note of this. So, if at all you are given a pair of straight lines, which you realize is not a homogeneous second degree equation, that means they do not represent a pair of straight lines passing through the origin, then just quickly do a small activity. Find out their point of intersection because if the question has come to find out the bisector, they must be intersecting. So, what does it find a point of intersection? Just find it out. So, what is the algo here? The algo here is number one, get X1, Y1. How? By simultaneously solving these two equations. Once you get it, all you need to do is use your equation X square minus Y square by A minus B is equal to X Y minus H, where your X will now be replaced with X minus X1 that you obtain from step number 1 and Y will be replaced with Y minus Y1. Sorry, this is minus. And this X1, Y1, you would have already obtained from step number 1. Is it clear any questions? Any questions? Any concerns? Okay. Good. So, we are now at the other question. So, we are trying to get like over. Yes. Yes. So, what are you doing? See, you're using the concept of shifting of the origin, a general second degree equation may not pass through origin. I mean, may not intersect at origin. So, what you're doing is you're shifting the entire situation such that the pair of straight lines intersect at X1, Y1. So, in short, you are basically using the pair of equations at the origin bisectors and you're shifting it by X1, Y1. Origin has been shifted in such a way that they represent the bisectors of the general second degree equation. So, it's a shifting of origin concept that is utilized, nothing else. So, we'll just take, I'm just searching for a good question if at all. I have one. Okay. Maybe we'll take the same. Yeah. Find the equations of bisectors of, by the way, this equation which I'm giving you, I've already given you in one of the questions before to find out the point of intersection so that now you don't waste time finding the point of intersection. So, for this, the point of intersection was 2 comma 1. Okay. So, please write down the equation of the bisectors. I just want everybody to have a hands-on on this. This is not a difficult question. Just have a practice of how do you write it? Okay. No need to simplify. No need to simplify at all. Just write it and tell me done. It is just a one-minute activity. You just have to fill in the requisites here, x1, y1. So, x1 is already 2, y1 is already 1, a minus b. So, in this case, a minus b will be 6. Okay. And I'm not asking you to simplify it also. Let it be in this form itself. You can simplify it when actual thing comes. And h here is minus 7, so this two will go up. Okay. Simplified and that is your pair of bisectors for this general second-degree equation. Okay. Now, the last concept for the topic, the concept of homogenization. Oh, yeah. Homogenization. Okay. So, what is this homogenization concept? A very, you can say important concept, especially for people preparing for J advance and all. This concept comes from the concept of homogeneous second-degree equation. Right. So, if you know there's a homogeneous second-degree equation and your h square is greater than equal to a b, then it must represent a pair of straight lines passing through the origin or pair of straight lines intersecting at the origin. Okay. Now, when you have, remember that concept, which I just now told, now I'm giving you another situation. Let's say if you have a conic. Okay. Any conic, whether it is circle or whether it's a parabola, it's hyperbola ellipse. Okay. So, let's say this is a conic. Okay. It could be pair of straight lines also. Okay. So, let's say this is a conic given to you. Okay. This conic is basically cut by a line. Let's say blue line. Okay. And this blue line equation is lx plus my plus n equal to 0. Okay. And let's say the point of intersection here is p and q. Okay. Now, if you are connecting, if you are connecting the origin to these point of intersection, okay, you can see that here it is creating a pair of straight lines. In fact, if you can extend this further also. Okay. Just to make it realize that it is like looking like a straight line only. So, the combined equation of l1 and l2, which is basically a pair of lines connecting the origin to the point of intersection of the line with this conic, then combined equation is given by the combined pair of l1 and l2 equation is given by homogenizing the line with this conic. Now, what is the meaning of homogenizing the line with this conic? So, what it does, it converts the entire conic equation into a uniform second degree equation like this. All of you please pay attention. So, this 2 gx. Now, see, this is a first degree term. See, this is second degree term, second degree term, second degree term. So, I need to make a second degree equation, then only I will get l1 and l2 equation combined because we have learned that a pair of straight lines, which is intersecting at the origin is a homogeneous second degree equation. That means all the terms in that equation must be uniformly second degree, second degree, second degree, second degree. But this is first degree, 2 gx. So, what I do here, I will do it in the next step. Let me write a 1 for the time being there. Similarly, 2 fy is also a first degree. So, I have to make it second degree. So, I will put a 1 there as of now. Just note it down. As of now, I am putting 1. And c is again a 0 degree. So, I will put a 1 square there. Now, see what I am going to do. From this equation, I am going to do lx plus my is equal to minus n. So, lx plus my by minus n is equal to 1. And this 1, I am going to put over here, here, and here. Let's see how. So, this is how I am going to do it. So, second degree terms are okay because they are already second degree. Let's not bother them. And this 1 over here, I will write it as lx plus my by minus n. Now, you can see here, I have created a second degree here as well. See, 2 gx into lx, x square will come second degree term. 2 gx my xy will come again a second degree. Similarly, here also, I will write the same 2 fy lx plus my by minus n. Again, a second degree term is created. Second degree terms are created here. 2 fy x lx, second degree term will be coming up. And the last term, I will just use the square of this which is lx plus my by minus n the whole square. So, not this equation will be, I mean, I am not simplifying it. And you don't need to simplify it because you can directly use your actual figures given to you in the equation. So, this second degree equation will be representing a pair of straight lines connecting l1 and l2 from the origin. Now, many people say, sir, why it could represent any curve passing through PNQ? Any conic passing through PNQ? Agreed. Any conic passing through PNQ can be given by this equation. But can any conic satisfy that it is being a second degree equation where h square is greater than ab? Only a pair of straight lines can satisfy that condition, right? So, this actually will only represent a pair of straight lines connecting the origin to the points PNQ. There cannot be any other conic. So, here you are basically using the combined effect of family of curves along with condition of homogenization which basically will lead only to one family member and that is a pair of straight lines connecting origin to PNQ. Very important concept and people even don't know that, oh, here I could have used the concept of homogenization. It is so hidden. It is so obscure and discrete that many people are not able to figure it out that, yeah, I could have used concept of homogenization to solve it. So, please note down, make a star next to this in case you want to. We will take a question on this, don't worry. Yes, right. If you would have just taken this plus lambda the line, you could have got a conic passing through the intersection of the line and the given conic. The first equation is any conic which has been intersected by that given blue line where above, okay? And this what you have got is the equation of L1 and L2 combined. The first one represents the curve. See, I have written the first one represents this curve. So, curve will be known, line will be known and then L1, L2 equations combined you are trying to find out. Okay. All right. So, we'll take some question. I think I have taken one question just for your understanding. We'll do this question and of course, I'll be sending you DPPs for you to work on. So, you can consider this to be our last question for conic section. So, we completed circle, ellipse, parabola, hyperbola pair of straight lines successfully. Do you think we could have completed this in our first year itself? So much of volume was there. Not possible. Prove that the lines joining the origin to the point of intersection of the straight lines with this curve, the angle between them is this. So, they have, they're basically coming from a different perspective. Then they're not only asking you for the lines combined equation, but also to find out there the angles between them. So, if you get the combined equation, you can always get the angle between them. So, please try it out. I'm giving you three minutes to try this out. Just homogenize. Homogenize, simplify. See what you get. JADvance people, I'll be sending you the link just after the class for the test. And KVPY, see, it is your call. If you want to write KVPY today and JADvance tomorrow, that is fine. JADvance today, KVPY tomorrow, that is also fine. Your call. You can write it anytime. You can write it anytime. Before I forget, tomorrow, note down everybody. 1130 to 130, you will have chemistry class tomorrow. Okay. Fine. So, Rajaji Nagar, sorry, YPR people, please note this down. 1130 to 130, you will have chemistry class conducted by Prabhatsar. It is only for YPR. No, it is for everybody, all batches. Combined class. It is for everybody. Just like I took maths class yesterday. Same chemistry class will happen. Okay. Why 1130? Because Yeshwantpur has some, you know, school classes. That is why we are keeping it at 1130. As we would have kept it at 930. Okay. Reminder will be sent. Don't worry. Today itself, reminder you will receive. All right. So, let's complete this problem. I think many of you would have done it also. So, we need to homogenize this line equation with this equation. So, what I'm going to do I'm going to write 4x into 1. Now, see how this one I'm going to write. So, Y minus 3x by 2, I will call it as a 1. Similarly, 8y and 1 will be again y minus 3x by, oh, so sorry, y minus 3x by 2. Okay. Minus 11, y minus 3x by 2, the whole square. Okay. Now, I think you have to do some simplification over here. So, I'll be just simplifying this result. Multiply throughout with a force so that, you know, that last term denominator is taken care of. But in the interest of time, I will directly write it down. So, you should be getting something like this minus 20x square minus 32xy plus 28y square minus, oh, sorry, I'm so sorry. This should give you something like this, 119x square minus 34xy minus 17y square. Am I right? That simplification you can do here. I mean, that is not a rocket sense. Okay. Now, for this equation, you need to find out what is the angle between the lines. So, we have learned that tan theta is 2 under root of x square minus AB by mod A plus B. Okay. So, ask yourself, who is playing the role of H here? 17 is playing the role of H, correct? So, 17 square minus AB. So, minus AB plus will be 17 into 119, correct? Okay. Many people say, sir, you can simplify this more. You can write it as 7x square minus 2xy minus y square, isn't it? It can go by a factor of 7. So, why to use such a heavy figure? Yeah. Let's use a simpler figure. So, x square here will be 1 minus 7 AB will be plus 7 by mod of A plus B. Mod of A plus B will be mod of 7 minus 1. So, that's nothing but 2 into root 8 by 6. That's nothing but root 8 by 3. That's nothing but 2 root 2 by 3. So, your theta is tan inverse of 2 root 2 by 3. Is it fine? Any questions? Any concerns? Is it okay? So, thanks to the concept of homogenization that you will be able to find it pretty quickly. If somebody doesn't know the concept of homogenization, what will you do? He will unnecessarily sit and find the points of intersection, waste of time. Then he will connect. He will find the slopes of the two lines connecting origin to that point of intersection. Then he will use the formula m1 minus m2 by 1 plus m1 m2 mod is tan theta. So, this takes away that, you know, lengthy, you can say, process and does it in a simpler way. So, homogenization, wherever you feel it is going to help you out, please do so. And with this, we come to the end of our conic sections.