 So hello friends, a very good morning to all of you. So today we are going to discuss exercise two of pair of straight lines. In the last video we have already covered the first exercise. So today we will discuss exercise two. Okay. So without wasting any time, let's start. So yes, pair of straight lines. Yes, here comes the first question of exercise two. So the question is saying find the angle between the pair of straight lines, given by this equation y square sine square theta. So let me write the equation. y square sine square theta minus x y sine square theta and plus x square cos square theta minus one right is equal to zero. So yes obviously this equation is giving us a pair of straight lines. It is a homogenous second, second degree equation. So it will, it will give us a pair of straight lines. Now we have to find the angle between this pair of straight lines. Okay, so let me discuss one theory, but not theory, but how do we find the angle between the pair of straight lines. We used to find it by this formula right tan theta is equal to two two times under root of h square minus AB upon mod of a plus b right. Now, this is the angle between the two lines. Okay, and for this equation. This is our standard second degree homogenous equation that is a x square plus two h x y plus b y square is equal to zero. So if, if our pair of straight lines is given by this equation, then the angle between these lines been angle between the pair of straight lines is given by this formula, tan theta is equal to two into under root h square minus AB upon a plus b right. Now, if you observe, if you observe, if a is equal to be or a equal to minus be right a equal to minus be, or you can say if a plus b is equals to zero, then our theta comes out to be 90 degree. Why because this if this will be zero degree or 10 theta will approach to infinity. That means our theta will be 90 degree. So what is a plus b what is a plus b a plus b is nothing but the sum of the coefficient of x square and y square coefficient of y square. So let's observe the coefficient of x square here. So if you compare it if you compare this equation with our standard equation. The value of a here is a is what coefficient of x square that is force is square theta minus one right and what is b b is the coefficient of y square that is sine square theta. So if you add this, if you add the coefficient of x square and y square what you are getting, we are getting cos square theta minus one, and plus sine square theta. Sine square theta plus sine square theta will be one minus one that is zero. So the value of a plus b is coming out to be zero. What does it signify it means the angle between the pair of straight lines will be 90 degree. The angle will be 90 degree angle will be 90 degree between the pair of straight lines which is given by this particular equation. So, if you see the options, we are getting pi by two so this option B is correct for this question. So hope this is clear to all of you. So the only thing what we use to know what we need to know is, if the coefficient of x square and the coefficient of y square, if their sum is coming out to be zero, it means the angle between the lines will be 90. Okay. Moving to the next question. Let's see this question number two, the angle between the lines given by the equation, a y square minus one plus lambda square into x y minus a x square equal to zero, each same as the angle between the lines. So let's analyze or let's observe the given equation, it is a y square minus one plus lambda square into x y and minus a x square is equal to zero. So if you observe here, if you observe here, here also the coefficient of x square and the coefficient of y square is minus a n plus a means the sum of that the sum of coefficient of x square and y square is coming out to be zero in this case also. Right. So the sum of coefficient of x square, or you can say coefficient of x square plus coefficient of y square is coming out to be zero. What does it mean, it mean the angle between them will be 90 degree. Okay, so the question is not asking the angle between these lines, but it is saying that we have to find the equation of lines whose angle is who's the like we have to find the equation of that lines where the angle between the lines is 90 degree. Okay. So we have to go option wise, there is no other option right. So, if you see means angle between this angle between required lines would be 90 degree or pi by two, whatever you can see. So if you observe the first equation, it's a five x square. So if you see x square plus two x one and minus three y square is equal to zero. So here if you see a plus B is not a plus B is not coming out to be zero. Okay, it means the angle between these lines will not be 90 degree. Okay, a for a in this case is five and be in this case is minus three so basically a plus B is coming out to be two, which is not equal to zero. So, let's check the second option. Here you see x square minus two x y minus three y square is equal to zero. So here if you see the sum of a plus B we are having one, and what is the coefficient of wise for minus three so here it is minus two and that is also equal to zero so this will be not our answer. The second option be option will also be incorrect. Let's see this third option x square minus y square minus hundred is equal to zero. So here if you see a plus B. So what is a here is one B is minus one yeah it is coming out to be zero and the angle will be 90 degree between these lines between this pair of lines. Okay, now let's see the next option. D option is saying x y equal to zero. Now what, what is the meaning of x y equal to zero. Basically, it is representing our coordinate axis that is x equal to zero and y equal to zero. So this x y equal to zero means x equal to zero and y equal to zero, and these are nothing but our coordinate axis. Okay, and what is the angle between the coordinate axis, it's 90 degree obviously. Basically, we are having this option C and option D as the correct option, right, because we need to, we had to identify the equation of lines where the angle between them is 90 degree. So we are getting this option C and this option D as the correct option. Okay. Now moving to the next question. Here it is saying, which of the following pair of straight lines intersect at a right angles. Okay, so same concept, we have to see the coefficient of x square and y square. So what the following pair of straight lines intersect at right angles. Okay, so let's check this option a it is to x square is equal to x y and plus 2 y square. Okay. So here we can write it as 2 x square minus x y minus 2 y square is equal to zero. So clearly here a plus B is coming out to be zero. A is to B is minus to obviously a plus B will be equal to zero. That means the angle will be 90 degree between these pair of straight lines. Now let's see the option B. So which see the option B it is x square plus y square plus 2 x y is equal to y x or x y. Let me write it as x y money x y and plus 3 x square. So if you take this to the left hand side, we will have plus x square minus 3 x square minus minus 2 x square. Plus 2 x y minus x y will be plus x y and plus y square is equal to zero. So here if you observe a plus B is how much is minus two and B is one so it is minus one and that is not equal to zero. Hence this option B is incorrect. Now let's see this option C. It will be 2 x y and plus 2 y square minus x y equal to zero. So basically there is no x square term at all. So no 2 y square. Okay, 2 y square and what minus x is equal to zero. So this will be also not our option. And let's see this option D y is equal to plus minus 2 x. So we can write it as y equal to 2 x and y is equal to minus 2 x. Okay, so basically, the pair of equation pair of a straight line we can represent it as y minus 2 x and into y plus 2 x is equal to zero. Now let's open it. So it will be y square plus 2 x y minus 2 x y. Okay. And minus 4 x square is equal to zero. So this plus 2 x y minus 2 x y will get cancelled out. And we are having minus 4 x square and plus y square is equal to zero. So here also a plus B is not coming out to be zero. So hence this option D is also incorrect. So only option where we are getting the angle between the lines to be 90 degrees this option A. So let's take this option. This will be our correct answer to this question. Now, moving to the next question, the question for question number four is saying, if H square is equal to AB, then the lines represented by this homogeneous equations homogeneous second degree equation are parallel coincident or perpendicular. So this is our standard second degree equation to H x y plus B y square is equal to zero. And we know it will represent a pair of straight lines passing through origin. The angle between those two lines will be two times under root of H square minus AB upon mod A plus B. Okay, now the question is saying if H square is equal to AB. What does it mean? What does it mean this complete thing in the numerator will become zero. Right. This complete thing in the numerator will become zero. That means our tan theta will become zero means the angle between the lines will be zero. Okay, so basically, what does it mean angle between the lines will be zero means what the both the lines will be coincident. No, they will lie one. Whatever one of the right. So they both the lines will represent will coincide. Okay, that's why our time theta is coming out to be zero. So if that time theta is zero, it means if the angle between the lines will be zero. It means it means both the lines will be coincident. Right. Both the lines will be coincident. So from here, we have derived, we have seen two things. First of all, if a plus B is equals to zero. That means the both the lines will be at 90 degree and or both the lines will be perpendicular and if H square is equal to AB that will represent that the lines are actually coincident. Okay. So I hope everyone is clear on this. So, moving to the next question. It is saying the equation a x cube minus nine x square y minus x y square plus four y cube equal to zero represents three straight lines. Okay. So the given equation is representing three straight lines. So the equation is saying if the two of the lines are perpendicular, then we have to find the value of a. Okay. So the given equation is representing three straight lines. Okay, so let's first write the equation. So that is a x cube a x cube minus nine x square y minus x y square. Okay, plus four y cube is equals to zero. So, what can we do. So let's first divide this equation by four. Okay, so, and I'm taking this last term means why Q y term in the making it as first term. So, I will write it as why Q okay and I'm dividing both sides by four. Okay, then minus one by four. Then minus nine by four x square y and plus a x cube is equals to zero. Okay, now I will divide the equation by x cube. So I will have y by x cube, then minus one by four. Y by x square, right, then minus nine by four. This is same thing what we like for deriving the equation of straight lines. For homogeneous equation, if you remember how we used to derive the equation of lines. Okay, so the same way I am doing same process I am doing here. So it will be minus nine by four, then y by x. Okay, and what we are left with, we will be left with a. Okay, okay, so here we have to divide by four also. So, one by, sorry, a by four is equals to zero. So I will assume this y by x to be, I will assume this y by x to be any constant or sorry, I will assume this to be any variable x. Okay, so this will be basically x cube minus one by four. So here minus nine by four x and plus a by four is equal to zero. So basically it is a cubic in X. It is basically cubic in X. And for a cubic equation, we know the product of roots, right. So let's say it's a roots are it's roots are M1 M2 and M3 M1 M2 M3. Okay, okay. So what will be the product of the roots M1 into M2 into M3 that will be equal to minus a by four means that used to be minus D upon a. So D, the role of D is being played by a by four. So minus a by four and what is a is the coefficient of this x cube so it will be one only. So the information is given here that out of these three lines known two lines are perpendicular to lines are perpendicular. What does it mean, it means this the product of M1 into M2 or anything M1 into M3 or M2 into M3 whatever you need. But the information is like two lines are perpendicular so the slopes of these two will be minus one, right. So from here, this product of M1 into M2 will be minus one. So from here we get the value of M3 is that as a by four. Is it okay. The value of M3 as a by four. Now, this M3 is the root of this equation. M3 is the root, right M3 is the root of equation. So what does it mean, it will satisfy this equation. Right now it will satisfy it will satisfy the equation. Sorry. So since M3 is the root so it will satisfy the, it will satisfy the given equation. It will satisfy the given cubic equation. So I will put a by four here. So it will be a by four car cube, then minus one by four, a by four car is square, then minus nine by four, a by four and plus a by four will be able to zero. So we can take a by four common from all the terms. So we will be having a square upon 16. And here we will be have a by four into one by four that is a by 16 and minus nine by four plus one equal to zero. Okay, so a by four and what is there. So if you simplify it, these two terms. So one nine. So here it is written as one minus nine by four. Okay, one minus nine by four so nine minus nine minus nine. Here it is what one plus. I'm not making any mistake. I don't think so this will be basically for right for minus nine so minus five by four. So in place of this I'm writing minus five by four. So minus five by four equal to zero. Now I will take this LCM. So 16 and a square minus a minus 20 equal to zero. So from here, what we can have, we can have this either a by four equal to zero, or a is square minus a minus 20 equal to zero. Now a by four cannot be equal to zero because from here we are getting a equal to zero and if that is the case, this will be like this. This one will be not there, and it, it will not represent a three straight lines right. So this value and rejecting, and further we will simplify this quadratic. So we will have a square minus five a plus four is it okay minus 20 equal to zero. So a into a minus five plus four into a minus five equal to zero. So we get the value of a five and minus. So we got the value of a five or minus four. So I am able to see both options. So a equal to five is this one and a is equal to minus four. So I think this is a multiple choice question. So both this option B and C will be both this option B and C will be correct. So this was our question number five. Let's move to the next one. Get these questions are without options. So let's find find the angle between the lines whose joint equation is given by this. So it's very simple. So, our equation is to x square minus three x one plus y square is equal to zero. Okay, and we have to find the angle between the lines given by this equation. So, this equation, this equation will represent a pair of straight lines passing through origin. Right. Why, because this is the homogeneity secondary equation. Now we know the angle. We know the formula to find the angle between these lines. So this will be two under root of h square minus AB upon mod A plus B. Now, if you compare it with with our standard equation, a x square plus two h x y plus b y square is equal to zero. So what is the value of a is to and the value of two h is minus three. So we get h, minus three upon two and age to h is minus three upon two and bees how much bees one support that put that substitute this value of AB and H in this equation. So we will have two into H square means what nine by four nine by four minus AB will be two and upon a plus B will be three. So this will be two by three. And within under root we will have four nine minus eight means one by four. So, what we are having two by three into one by two. So that will be one by three. So theta is coming out to be 10 inverse one by three. Okay. So this will be the angle between the lines given by this equation. So 10 inverse one by three will be answer to this question. So it's just we have applied the formula nothing else. And we have compared the equation with our standard equation from there we got the value of AB and H and we have just simply put that value in this equation. Now here comes the next one so that the lines this into y square minus this into x y plus this into x square equal to zero include an angle alpha between them. So here if you see. You have to find the angle between them right. So if we are going to apply this formula what we were what we have done in the last example so it will be I think very tedious to wonder what H is where minus AB upon a plus B. So here the H will value of a is this thing the value of sorry the value of a is this long term cost data into cost data plus 10 alpha, then B is this thing one minus cost data into 10 alpha. So what can we do. So, I don't think that applying this formula is going to help us. Let me first write the equation. So one minus cost data into 10 alpha. Okay, into y square. And minus this thing to cost data. Plus sign square theta. Into 10 alpha into x y plus cost data cost data into cost data plus 10 alpha into x square is equal to zero. Okay, let's approach this question without a standard process what I'm going to do. I will divide this complete equation by coefficient of this y square. So we will have this y square minus to cost data as it is plus sign square theta 10 alpha upon this one minus cost data into 10 alpha. Okay, this whole complete thing multiplied by x y, then plus cost data and this thing cost data plus 10 alpha upon one minus cost data 10 alpha right into x square is equal to zero. So, from here is something better, we are getting from this to cost data plus sign square theta 10 alpha upon. Make this sign square theta into, we can express the sign square theta as one minus cost data right so I'm writing here only one minus cost square theta. And that will be multiplied by further with 10 alpha. So, 10 alpha minus. Okay, first let me write that only one minus sign square theta was there so it will be one minus cost square theta. Okay. Further it will get multiplied with this so this complete expression will become 10 alpha minus cost square theta into 10 alpha. So this will become 10 alpha 10 alpha and minus cost is squared theta into 10 alpha right into 10 alpha and this whole multiplied by x one. Okay, so further we can we can do one thing we can take cost theta common from this, this term and this term, and we will leave this cost theta 10 alpha as it is okay I'm doing it in next step. I'm writing this to cost data is cost data plus cost data, and from, and I'm taking cost data common here. So it will be cost data, then one minus cost data. 10 alpha, right. So here, this is the benefit. Here we can cancel this term one minus cost data into 10 alpha. And this is a complete thing is within the bracket and one cost data is left here so that will be plus cost data. Okay, plus cost data plus 10 alpha upon one minus cost data 10 alpha. Is it okay. And it will be as it is cost data into cost data plus 10 alpha upon one minus cost data into 10 into x square is equal to 0. Okay. So this thing will get cancelled out. So this will be cancel each other. And what can we do. If cost theta becomes our 10 any angle 10 beta so we can apply the formula 10 alpha plus beta right. So, what we can do, let me assume late for any late for late for any beta, we are assuming our cost theta to be equal to 10 beta. Okay. Because the cost data, the range of cost data is between minus one to minus one. Sorry, minus one to plus one and 10 beta is defined everywhere. Right. The range of 10 is minus infinity to plus infinity. So we can take for any beta for any angle beta, we can take that cost data is equal to 10 beta. Okay. So what we can do, we can write it as we can write it as y square minus cost theta we can write 10 beta. Okay, 10 beta. And this will be basically 10 beta plus 10 alpha upon one minus 10 alpha 10 beta. So this will be formula of 10 alpha plus beta. Is it okay. 10 alpha plus beta and here it will be what 10 beta 10 beta into same thing here also 10 alpha plus beta 10 alpha plus beta. Okay, this thing was multiplied by XY. Yes. So this complete thing multiplied by X square and this thing was multiplied by our XY. So this is going to be zero. Now we can further divide the complete thing by X square. And so this will be Y by X the whole square and 10 beta plus 10 alpha plus beta. Okay. So this will be divided by X square. No, so it will become Y by X. And this will be 10 beta into 10 alpha plus beta is equal to zero. Right. So we can assume this Y by X to be M. Okay, so this will be M square into M. And this. So basically this is a quadratic in M. Okay, and whose roots are basically our roots are M one will be 10 alpha plus beta right and our M two will be what 10 beta. So the slope of one line is 10 alpha plus beta right slope of suppose suppose for any line. This is our angle beta. This is our X axis. Okay, and this is angle or this is our angle beta. So this is the slope of one of the line and the slope of another line is this thing alpha plus beta. Right. So the angle between the them will be this angle will be alpha only right this angle will be alpha only. So this is what we need to prove in this question. Right. The question was asking that to show the lines that the angle between these lines is alpha. So basically, if you see this as L one, if you tell this as L two. So the angle between angle between L one and L two will be what angle between L one and L two will be because the first angle is 10 beta means slope is this 10 beta and the slope of other is 10 alpha plus beta obviously the angle between both lines will be right. So this is what we were asked to prove. And I think we have done it. So, this is our question number seven moving to the next one. Find the angle between the lines represented by this equation. Okay, so we have to find the angle. So basically, this will be x square. I'm just writing the equation first. So x square minus two p x y. Okay. And plus y square is equal to zero. So, what is the angle angle is nothing but we used to give the angle by this formula two times under root h square minus AB upon mod of a plus. Now if you compare it with our standard equation, the value of a is one, the value of B is also one, and two h is how much minus two p. Okay, so two h x y, so two h is equals to minus two p. So the h is coming out to be minus be support that. So we are having 10 theta is equal to h square means what p square and AB is one. And what will be our a plus B, it will be two. So this is our 10 theta, right. 10 theta is coming out to be under root of p square minus one, or we can say theta is nothing but tan inverse under root of p square minus. So, this is easy one day. So this will be our answer to this question. Okay, so now moving to the next one, question number nine. So in this question, it is asking that, so that the lines this, this pair of line x square minus four x five plus y square. So obviously this will represent a pair of lines, and the third equation is given as x plus y equal to one form an equilateral triangle. So we have to show that three, these three lines have the three sides of the equilateral triangle. Okay. And once we are done with that we have to find its area also. Okay, so this given equation basically x square minus four x y plus y square is equal to zero. So this will represent a pair of lines passing through or isn't so let me just roughly draw. So, this will be our first line. And just drawing the rough sketch. Okay, so this is our L one. Okay, and this is our L two, and the third equation is x plus y equal to one. So this will be this line, right. If you put x equal to zero wise one. Y equal to zero x was one. Okay, now the triangle formed by these three lines basically this, this triangle, the question is asking for this time. So this is our origin. Let me call the point of intersection as a and b. So we have to prove that triangle a OB is a equilateral triangle, right. So, how can we prove that. We can write this equation of L one s. Why is equal to m one x right. And we can write the equation of a two as y is equal to m two x, where m one plus m two, where m one plus m two will be how much minus two h upon the minus two h upon the. So, if you see here, two h is how much minus four right. So minus two h will be four. So this will be four upon B. Okay, B is one is also so is one B is one. So m one plus into is four. And what is the product m one into m two is a by B, that is one. Okay. So, once we are done with that, our target should be like to find the equation. Okay, so from these two equations, I think we can have the value of our m one and m two. So let's first find out that. So m one minus m two guys square that that will be equal to m one plus m two guys square and minus four times m one into right. That is the value. So m one plus m two is for that is 16 minus four times one that is 12. So we are having m one minus m two. As how much. Plus minus two root two. No, four into three, no, so two root three, two root three. Okay, so, and we are having this equation m one plus m two m one plus m two is equal to how much four. So we are going to solve it. So if you solve it. If you solve it two times m one will be two times m one, two times m one will be coming out to be. That positive sign. Okay, I'm taking the positive sign only. So for four plus two root three, that is, if you take two common. So two plus root three. So our m one is coming out to be two plus root three, and obviously our m two will be then two minus three. So once we get the value of this m one and m two, we know the equation of these lines L one and L two. Now what I will do. What I will do, if any how we can prove that this angle is 60 degree right. So if we can prove that this angle is 60 degree. And this angle is 60 degree, we can say that that's a triangle, the given triangle will be equilateral triangle right. So let's try to prove that. So our first equation this EO or OB you can say the equation of OB will be, or means, don't name it in this way. And what we can do our, the equation of this we can tell, say it as two plus root three, no, so it will be having higher slope to minus three. Okay, anyhow. So let me write it as a Y is equal to M two X. And let me write it as a mind equal to M one X. Same thing, not an issue. So what we can do, we can write the equation of OB, OB is what Y is equal to M two, that is two minus root three X, and our equation of way is why is equal to. Two plus root three into X. Okay. Now we have to find between angle between these two lines right away and AB. Okay, so we can find it, we can find the 10 theta between these lines is M one minus M two. So two minus root three minus two minus root three upon one plus M one M two. That is one plus. This is our M one and this is our M two. So M one into M two will be four minus one hundred. Right. And this complete come on. So this will be two two chala jaya then two root three with negative sign but I'm. Since this is in more. I can write it as two root three and this will be how much. Four minus one will be three and one four. So this is coming out to be root three upon two 10 theta but we have to prove that 10 Twitter should be 60 degree. So please check whether I'm doing some mistake or what. So angle between these two lines will be M one minus M two M one minus M two upon. The angle between two lines we used to write in this way and one minus M two upon one plus M one M two. So M one in one plus M one M two will be how much. Four minus a square minus B square. Okay, okay. So this will be three more why I'm writing one. So one plus one that will be two. So yeah, we are getting root three. That means the theta will be 60 degree. Similarly, we can in the same way we can find the angle between this. What do you say AB and angle between AB and what. Okay, that will also coming out to be 60 degree. You can try it by your by your what what I was saying yeah the angle between this AB and OB will also be coming out to be 60 degree. So yes, we hence we can say this triangle a OB triangle a OB will be a equilateral trend will be equilateral trend. Now, having proved this we have to find its area also, we have to find its area also. So basically, this is our original. So I will drop one perpendicular from origin to this AB. Okay, so and I'm naming it as OD. So OD will be what OD will be the height of equilateral trend, height of equilateral triangle, equilateral triangle. Okay. And how to find this OD, we know the coordinates of O and we know the equation of AB. Okay, and that will be equal to this OD will be equal to how much the equation of AB is given as X plus Y. This plus Y equal to one. So that will be more of minus one upon under root of one is where plus one is where that is nothing but one upon one plus one that is. Okay, so this is our height. And what is the area of equilateral triangle in terms of height, it is. H is square by root three. Right. H is square by root three. So what is H is square one by two root three, one by two root three. H is square is one by two and root three as it is so this will be our area means this will be a square units obviously. So this will be the area of our equilateral triangle. So hope this is clear to all. So area of equilateral triangle in terms of height if we can write any how the area of triangle is how much area of triangle is for equilateral triangle it's root three by four a square. Okay, and this is where we just for writing it in terms of age, we have to change this a into H. So that relation you already know. So this is our equilateral triangle this is H. This sites are a and a. So obviously this H is square plus a square by four. This length will be a by two note that will be equal to a square or our H is square will be equal to how much a square. Right. So from here we can say a square is nothing but four H is square upon four H is square upon three. So same thing I have done here. Same thing I have put in this equation. So if you put here it will be a root three by four in place of H is where I'm writing for H is square upon three. So this 44 will get cancelled out and it will be okay this root three and this three here it will be. So it is a square upon three will be our area. Okay. So this was our question number nine. Is it done or what I missed something. One more question is there this question number. It is saying question 10. So, here we have to prove that triangle formed by the lines this and this is isosceles if. If this condition holds, then the triangle formed by these lines will be isosceles. Okay. So the second equation is a x square plus two H x y plus b y square is equal to zero. So from here we will have a pair of lines and the equation of third line is LX plus M y is equal to one. Okay. So, now we have to prove that the triangle formed by these lines will be isosceles triangle. Okay. Okay, so what can we do here. This is basically this will be our. This will be the lines given by the homogenous second equation L1 L2. Okay. And this is our third site. So basically we have to show that this triangle is this triangle is isosceles triangle. And the equation of this line is LX plus M y is equals to one. Is it okay. Now how can we prove that. How can we prove similar same concept what we have done in earlier question we can do one thing. We can find the angle, right, we can find the angle between this and this. And we have to show that these two triangles, these two angles will be equal. Okay. So, let me write the equation of L1 is Y is equal to M1 X and equation of L2 is Y is equal to M2 X since these lines are passing through origin so this equation we can assume. And one more thing, we can have this angle between these two lines. Right now. So, let me say this angle between, angle between AO and OB. Or you can say OA and OB. So let me write in this way, angle between OA and OB. So, that is, let me call it as theta one. Okay. So, our 10 theta one, what will be our 10 theta one, it will be M1 minus M2. Right, M1 minus M2 upon one plus M1 M2. That's okay. And let me call the angle between, angle between this line, AB, angle between AB and OB as theta two. I'm assuming the angle between AB and OB to be theta two. So basically this thing I'm assuming. I've assumed this angle to be theta one and I've assumed this angle to be theta two. Okay. And this angle to be theta one. So, what we can do now, our 10 theta two will be M1 minus M2. So, what is the slope of this AB line? If you see here, the slope of this line will be minus L by M, right. And the slope of this will be M1. So, M1 minus minus L by M, minus L by M, right. This will be the slope for this line. Minus L, yes, M1 minus M2 upon one plus M1 into M2 is minus L upon M. Now, for showing this to be an isosceles triangle, I'm assuming this to be a isosceles triangle. Okay. After assuming this to be isosceles triangle, let's see whether we are having this condition or not. Okay, so, assume this to be assuming this to be isosceles triangle, I am assuming this to be isosceles triangle. So, since we can write this theta one will be equal to theta two or we can say 10 theta one will be equal to 10 theta two. Assume the triangle to be isosceles. Okay. Let's see whether we are arriving at this condition or not, or arriving at this equation or not. If I write then obviously that means the triangle will be isosceles triangle. Now, so this 10 theta one will be how much. Okay, we normally write the angle between the lines within the model right so this will be more of M1 minus M2 upon one plus M1 M2 is the more will be equal to more of M1. It will be plus L by M, right. And here it will be one minus L by M into M1. Is it okay. Minus L by M into M1. Okay, and but whether it is going to help us or not. M1 minus M2. And this angle theta one. And we have taken this. Okay, let's figure it. So, a mod of a is equal to mod of B, what does it mean we can say a equals to plus minus B. Okay. So I think taking the plus sign know so if this thing will be equal to this thing with positive sign I think we will be having M1 is equal to M2 that will be a not valid. So let's first open it with negative sign. So I will have M1 minus M2 upon one plus M1 M2 is equals to minus of this. M1 plus L by M upon one minus L by M into M. Let's cross multiply it. So we will have M1. Okay, and minus L by M M1 is square. Then minus of M2 and plus L by M M1 M2, and that will be equal to. So I'm just taking negative sign inside. Okay. So, now I can multiply it so minus M1 and what minus of M1 is square M2 is it okay minus of M1 is square M2 and minus of L by M and minus of L by M into M1 M2 M1 M2. Okay. So, if you see here what we are having is three four one two three four yes so M1 plus M1 will be that will be two times M1. Okay, and minus M2 here M1 minus M2. So this will be M1 is square. Let's see the calculation whether we have done correct or not. So this will be M1. Okay, M1 minus L by M M1 is where then minus M2 then plus by L by M M1 M2. Okay, here we are having minus of M1 and minus of M1 and minus of M1 is square M2 then minus of L by M and minus of L by M M1 M2. So I think I have done correct only. So let's see to M1 we are having this to M1 minus M2. Okay, minus M2 and what else we are having we can have this thing plus M1 plus M1 is square M2. So we have covered this this these terms are taken these four terms are taken right plus M1 is square and M2 and here we can take L by M common right. So if we take L by M common what we are having M1 M2 then here we will have a minus of M1 square. Okay, plus of 1 and plus of M1 M2 plus of M1 M2. Is it okay? This will be equal to our 0. Now what we can do to M1 if we take M1 common from here or M2 common. So I don't think M1 if you see here. So basically we are having this relation. We are having M1 plus M2 M1 plus M2 will be equal to what minus 2H upon B. Okay, and M1 M2 will be equal to A upon B. Right. So this, this is the relation between M1 M2 and our coefficient of X square Y square and X square. So how to utilize this given information we have to find it out. Right now. So M1 minus M2. How can we find it? So L by M if you see here M1 minus M2. M1 M2 is fine. Okay, M1 M2 is okay. So if we take M1 common here. If we take M1 common here what we will get. I will get 2 minus M2. Is it okay? 2 minus M1 M2 and minus M2 will be there. Is it okay? If I take 2M1 then minus M1 square. Okay, this will be plus sign. So M1 square M2 plus L by M what we can take here. Here M1 M2 is okay. We are having the value for that. And if we see we can take M1 common here. Right. So M2 minus M1 will be equal to 0. But no we are not having this information. So just here what we can do. Prove that the triangle formed by lines this and this is isosceles. Okay, let's let's try it one more time because I think we are going nowhere from this equations. So just let me copy this. Copy this equation. Okay. So prove that the triangle formed by lines this and this and this is isosceles. Okay. So let me take these lines. So this line is LX plus MY is equals to 1 and let me name it ABC. Okay. And I am taking this line to be Y equal to M1X, Y equal to M2X, which is given by this homogenous second-degree equation. So anyhow, if we can prove that this theta 1 is equal to theta 2, right, we can say that the triangles are isosceles triangle. Okay, so I think this mistake we have done we have taken the angle between these two lines. We have taken this angle and this angle in the in our last attempt. So basically that is wrong. We have to take these two angles right this theta 1 and theta 2. Okay, so our 10 theta 1 is equal to 10 theta 2. I am proceeding assuming that the triangle is isosceles. Okay, and let's see whether we are achieving this condition or not. So our 10 theta 1 will be how much M1 minus M2. So the slope for this will be how much the slope for this will be. Slope will be minus L upon M. Okay, so M1 minus M2. So minus minus it will be plus L by M upon 1 minus M1 M2 M1 M2. So this will be again plus right. So mod is parallel. And mod of this thing. What is slope of this line is M2. So this will be minus M2. So this will be again plus L by M and one plus M1 into M2 M1 into M2 one plus M1 it means slope of this is M2 and this is minus L by M. So this will be negative sign right. And similarly here also it will be minus M. One minus M1 into L by N. So within the mod. So I am opening with it. This mod, suppose mod A is equal to mod B mod A is equal to mod B so we can say no A is equal to plus minus B. So same thing I'm doing here. So taking positive sign I think it will be giving M1 is equal to M2 right. So that is not valid here. So I will I'm going to open with with our negative sign that is M1 plus L upon M okay and I'm cross multiplying it simultaneously. So this will be one minus M2 into L by M and since I'm minus open current get so that will be our minus M2 minus L by M into this thing one minus M1 into L by M. Now multiply it, we will have M1. Then minus of L by M M1 M2. Okay, then plus L by M and minus L square by M square into M2 and that will be equal to minus M2. See the calculation part. So plus L by M and then a minus L square by M square M2. Here we are having minus M2, then plus L by M M1 M2. Is it okay. Minus minus will be plus L by M M1 M2 and then minus of L by M and plus L is square by M is square into M1. Okay, writing here. So it will be M1 M2 M1 plus M2. So I have considered this term I have considered this term. Then what we can say, here we are having less L by M minus L by M so two times L by M. So these two terms I have considered. And here we can write minus L by M if we take minus L by M M1 M2. Okay, so it will be minus two times L by M M1 M2 minus two L by M M1 M2. So these two terms are also covered. And here we can have minus two minus two L is square by M is square. No, M1 and M2 are there. So minus L is square by M is square. M2 here and minus L is square by M is square M1 here. So that will be equal to zero. Okay. So this M1 plus M2 we can write it as minus two H upon B. We can take this two L by MB as it is so two L by M then minus two L by M this M1 into M2 we can write it as A by B minus. Here what we can do we can take L is square by M is square common. So we will have this M2 plus M1 right. And again this M2 plus M1 we can write it as minus two H upon B. So minus two H upon B. Is it okay? Yes. So, if you see what we can take this BME square LCM right. And this two two we can cancel it out from all the terms. So BME square I'm taking LCM so we are left with M is squared. So minus M is square H then M or BML M is there. So BML B and M and here MB so one M is there minus ALM minus ALM. And here we are having plus of HLE square HLE square is equal to zero. Okay, so numerator if you see H we are taking common from here so L is square minus M is square and that will be equal to if you take LM common from here we can have A minus B. So this is what we get after assuming that the triangle is isosceles and I think this is the condition given in the question also H into L is square minus M is square into A minus B and no not M it will be A minus B into LM. Okay, so yes we are getting this. So hence the triangle will be isosceles hence proved we can say hence proved. So basically in this question we stuck in the last attempt but okay anyhow. So I'm deleting this slide. Yes, this one is our. So I think all the questions are clear to you guys. So, very soon we are going to take up the next exercise. For this pair of Australians so till then take care Tata goodbye.