 Hello students, a very good evening to all of you. So today our point of discussion will be pair of straight lines. Okay, we will do the first exercise from Arihan's book on this topic. So, here comes the first question. It is saying the lines given by the equation 2 y square plus 3 x y minus 2 x square into x plus y minus one equal to zero format triangle. And we have to comment on the triangle whether it will be equilateral or right angle triangle or isosceles triangle or obtuse angle triangle. So let me write the given equation first. So it is given that 2 y square plus 3 x y minus 2 x square. Okay, this whole thing multiplied by x plus y minus one equal to zero. Now this equation is giving, giving us the sides of the triangle right in that also if you see this first bracket, this will be giving a pair of straight lines right. This will be giving the pair of straight lines. And since this is a homogeneous second degree equation. So basically it will give a pair of straight lines passing through origin. Right. These straight lines will pass through origin. Okay. And so these, these two lines will be the two sides of the triangle and one third side will be this side means third side of the triangle we can write the equation for that third side from this equation. So basically the third side of the triangle and from this first bracket, we will be having the two sides of the triangle. So if you see if we factorize it by splitting the middle term, what we can get this two into two minus four so basically we can break it into two y square. This middle term we can write as four x y minus x y minus two x square is equal to okay, or not equal to I'm living as it is, I'm keeping it as an expression. So what I can do I can take two y common from here, then y plus y plus two x right, and from here if you see we can take x common so y plus two x, so y plus two x, okay, and two y minus six. So basically, this will give us the two sides y plus two x equal to zero and this two y minus x equal to zero, this will be the two sides of the triangle. If you observe here the slope of this line is how much minus two and the slope of this line, the slope of this line is how much one by two, right, and if you see the product of m one and m two is coming out to be minus. So basically this line and this line are perpendicular to each other. So, regarding triangle, we can comment that the same triangle will be a right angle triangle. Now, basically we don't need to do these lot of things right, because if you observe this, if you observe here. This is a bit of theory part. So if you observe this equation to why square plus three x y minus two x square right. Here if you see the value of a means is minus two and be is two. So basically is what coefficient of x square. Okay, coefficient of x square. The coefficient of x square is minus two and coefficient of y square is two, and here coefficient of Y square is two. So if you see the sum of a plus B is coming out to be zero, right. So whenever, whenever this homogeneous equation in second degree. And whenever this product of sorry the sum of the coefficient of x square and y square comes out to be zero. It means the given straight lines the given pair of straight lines will be perpendicular to each other. The given pair of lines or you can say given pair of straight lines will be perpendicular to each other. So by observation also we can answer this question, but okay by our calculation. Also we observe here. Okay, so if you have will command on this pair of straight line you can simply take the answer as right angle by seeing that the coefficient of x square the sum of the coefficient of x square and y square is coming out to be zero. Hence the pair of straight lines will be right angle pair of straight lines will be perpendicular to each other. And if the two sides of a triangle is perpendicular definitely it will be a right angle trying. Okay, so moving to the next question. Here it is saying the area of the triangle formed by the lines, this and this. Okay, so basically this y square minus nine x y plus 18 x square is equal to zero. Obviously this will give us a pair of straight lines passing through or is it pair of straight lines passing through or is it. And our third equation is y equal to nine. So first, let's write the equation of this pair of straight lines individually means what I can do, we will be writing the equation of straight lines from this equation. So if you see nine and okay we can make we can split the middle term as y square minus six x y minus three x y plus 18 x square is equal to zero. So taking y common from the first two terms we will have y minus six x. And here what we can take, we can take three x common. So why minus six six, and this equal to zero. So basically why minus six six. Okay, and why minus three x equal to zero. So why minus six x equal to zero will be the one line and why minus three x equal to zero will be the other line, right. Now we have to find the area of triangle formed by these lights. So basically if we if we draw a rough diagram for this. Okay, so this is our x axis, this is our y axis somewhere here, we will have y equal to nine. Okay, this is our y equal to nine line and this line will be y equal to six x and y equal to three x right. So, basically, if you see this line will be your y equal to six x. Okay, so the slope of this line is higher. So y equal to three x and this line is basically what y is equal to three x. So basically the question is asking this thing, right, the area of this triangle, we have to find the area of this triangle O A and B, we need to find the area of this triangle. Now what we can do, if you see, we can have the coordinates right, we can have the coordinates of this point A and B. coordinates of A will be coordinates of A will be the Y coordinate is definitely it's nine. Okay, and when you put nine here, when you put nine here it will be three nine by six. Okay, so nine by six means how much three by two. So the x coordinate of A will be the x coordinate of A will be three by two and it's a Y coordinate will be nine. Similarly, what will be the coordinates of our B. So y equal to nine definitely the y coordinate will be nine and you have to solve this thing y equal to three x is equal to and y equal to nine. So keep put the value of nine here so x will be three. So the coordinates of B will be this three by nine and anyhow you know the coordinates of the origin. It's zero comma zero. Now the area of triangle, area of triangle A O B right, this is what we need to find. This will be equal to half times base base means this AB we can take the base and this height. Okay. This I suppose I'm naming it as D. So half AB into AD. Now half AB is how much AB will be three minus three by two. Right. So it will be three by two and what is the height this AD will be how much y equal to nine. Okay, the y coordinate of A and B is nine. No, so its height will be means AD distance will be nine. So this is coming out to be 27 by four is square unit. So see where is this option available in the question. Yeah, it's there in option A. So our area of triangle A O B will be 27 by four square unit. Okay. So moving to the next question. It is saying the equation three x square plus two h x y plus three y square equal to zero represents a pair of straight lines passing through the origin, the two lines are. Okay, so the given equation is three x square plus two h x y plus three y square is equal to zero. Definitely it will represent a pair of straight lines which will passing through the origin. Okay, now the question is asking the two lines will be real and distinct real and coincident or real and distinct if this and this real and coincident if this and this. So, if you compare it, if you compare the given equation with our standard second degree homogeneous equation. So, a x square plus two h x y plus B y square is equal to zero. This is our standard homogeneous second degree equation. So on comparing, you can see the value of a is three. Okay, the value of B is three, and definitely it is as it is. It is equal to two H. So, there is no change in this H, H is as it is. We have to find the relation between H and AB, so that what will be the nature of the lines, what will be the nature of the lines compare, which are, which are derived from this equation three x square plus two h x y plus three y square. So the standard equation, if H is square is greater than AB, okay, if H is square is greater than AB, then, then the pair of straight lines, then the pair of straight lines, we have, we have learned this right. I hope you all are remembering, hope you all are remember, remember that that theory part. So, if H is square is greater than AB, our pair of straight lines will be, or the pair of straight lines are real and distinct right, are real and distinct. So, if there is an equality sign, if H is square is greater, is equal to AB, if H is square is equals to AB, our pair of straight lines are coincident, pair of straight lines, okay, are coincident. So, hope you all are aware of this. So, coincident, coincident. Okay. Now, here if you see H is square is greater than AB. So, in our case, what is the AB? A is three and B is three. So, if H is square is greater than nine, if H is square is greater than nine, then the lines will be real and distinct, right. So, real and distinct if H is square is equal to nine, option B is correct anyhow, real and coincident, if H is square is greater than three, no, this is not the case, real and distinct. No, for real, for the pair of straight lines to be real and distinct, H is square should be greater than nine, from here we get, so this is not the option, a real and coincident if H is square is equal to three, no. If H is square is equal to nine, then the pair of straight lines will be coincident. So, this option is also wrong. So, the only correct option is this option B, that is the lines will be pair of straight lines will be real and distinct if H is square is greater than nine, okay. So, hope this question is clear to all. Now, moving to the next one, here comes the fourth question. If one of the lines of the pair AX is square plus two HXY plus BY is square equal to zero, by six the angle between the positive direction of the axis, okay. So, this is AX is square plus two HXY plus BY is square is equal to zero, right. So, this equation will give us a pair of straight lines passing through horizon. Now, the question is saying that one of the line, one of the line of these of this pair of straight lines bisects the angle between the positive direction of axis. So, one of the line is one of the line, okay, one of the line is what one of the line is Y is equal to X, right. Because this is the line this this Y equal to X line bisects, right bisects the angle between the positive direction of axis, okay. So, we came to know that the one line of one line of this pair will be Y equal to X, okay. So, if you see it's a slope is slope of this line is how much slope of this line is one right. So, for this, for this a second degree homogeneous equation, we know one relation, right. So, this will be M1 plus M2 is equals to minus two H upon B, and, and one more thing M1 into M2, where M1 and M2 are the slopes of the pair of straight lines M1 into M1 into M2 is how much A by B. So, here if you see one of the slope of the lines is one. So, I'm taking that as one and the other slope as M. So, one plus M is how much minus two H upon B. Okay, and here if you see one into M will be A upon B. Okay. So, here we get the value of MS A by B, right. And we can put the here, we can put the value of M here. So, let's check it whether I'm doing correct or is there any mistake. Okay, let's put the value of M here. So, it will be one plus A by B equals to minus two H upon B. Let's take B LCM here. So, it will be A plus B is equals to minus two H upon B. Okay, so this B and B gets cancelled. And we got this A plus B is equals to minus two H. So, I think this will be the answer. A plus B is equal to two mod of H or A plus B is equals to minus two H. So, yeah, it's there in the option B. So, I think this will be the answer. Okay. So, it's a simple question only, since we know the slope of one of the lines, right. So, we from there, we found out the slope of the other line. On comparing these two equations means on analyzing these two equations, we got the value of M. On solving these two equations, we got the value of M. Okay. So, this relation we are getting A plus B is equals to minus two H. So, hope this is clear to all. Now moving to the next question, this question number five. It is saying if the slope of one of the lines given by A square X square plus two H XY plus B square Y square is equal to zero, be three times of the other, then H is equals to. Okay, so same concept we will be using the given equation is A square X square plus two H XY plus B square Y square is equal to zero. Now we are already aware of the fact that this will represent a pair of straight lines. Okay, whose slopes are M1 and M2. So, we are having one relation between M1 and M2. That is the sum of this M1 plus M2 will be how much minus two H upon B and the product of this M1 and M2 will be A upon B. Okay. So, okay, don't write A and B here in. So, what can we do because A and B is already there. So, what can we do here we can write it as capital B okay capital B and we can write it as capital B A upon B. So, M1 plus M2 here will be minus two H. So, who is playing the role of this B, it is B square basically, and who is playing the role of a is, is role is being played by a square and B's role being played by B square. Okay. Now, M1 is equals to it is saying that the slope of one of the line is three times the other. So, let me take, let M1 is equal I'm taking the M1 as M. So our M2 will be how much it will be three times the other so basically it will be three. So, from the first equation, this is our first equation this is our second equation. So from one if you see what we get M1 plus M2 means four M is equals to minus two H upon B square. Okay. And from two if you see what we get three into M square, three times M square will be equal to a square upon B square. Okay. So, if you see from here we get the value of MS minus two H minus two H upon four B square. Okay. Now what I will do I will put this value here. Put the value of M here. So, I will get three times M square M square means this four H square four H square upon this 16 right 16 B square. So, this will be our M square. Okay, so three into M square and that will be equal to a square upon B square. Okay, so B square B square will get cancelled out. So, B square B square will get cancelled out but we should have the B also okay in the final if you see. Let's check whether we have done it correctly or not. So, this, this is our equation. Okay, a square X square plus two H XY plus B square Y square. So, minus two H upon B, the sum of the slopes will be minus two H upon B so minus two H upon B square and M1 M2 is a upon B that is a square upon B square. Okay. So, from here if you see four M will be equal to minus two H upon B square and our M will be minus two H upon four B square and this the product M into three M square. So, three M square will be equal to a square upon B square. Now we are putting the value of M here. Okay, okay. So, this will be four H square and 16 B power four. It will be B power four and three M square that will be equal to a square upon B square. So, I think what we need to find, we need to find the value of H okay. So, this will go four times and we are having the H square the value of H square is coming out to be four times B power four A square upon this is three upon three upon B square right. So, this is nothing but two B square A square upon three because this B square will cancel it. So, two times B square, not two times no four will remain as it is. So, four into B square A square upon three. Okay, so the value of H will be coming out to be under root of this thing for B square A square upon three. That is nothing but plus minus two by root three AB right. So, this will be our value of H two upon root three AB with plus minus sign. So, it's there in option C and option D. Okay, so both will be correct I think so option C and D is correct for this question. Now, what we have done here we have used the known information that the sum of roots of the equation some of the sorry some of the slopes of the straight lines will be equal to minus two H upon B. And the product of the slopes will be equal to a by B. Okay. Now, this is our question number fifth. Now, moving to the next question, this is our question number six. The question is asking to find the separate equation of the two straight lines whose joint equation is given by this. The joint equation of straight lines or the pair of straight lines equation is given as AB into X square minus Y square plus A square minus B square into XY. This is equal to zero. So, if you open the packet what we will have AB X square. Okay. And here if you say it will be a square XY minus B square XY and this will be minus AB Y square minus AB Y square. Okay. So, I have just opened the packet and expanded this equation. So, if you see what we can take common here we can take X common here from the first two terms. So we will be having BX plus a Y right BX BX, since we are taking a X common so BX will be there and then a Y will be here. So, if you see we can take B Y common from these two terms. So, what we will be having, we will be having BX plus a Y right that is equal to zero. Now BX plus a Y into a X minus B Y equal to zero. So basically this will be the equation. This plus a Y will be equal to zero will be the first equation and this a X minus B Y equal to zero will be the second equation of the line. If you observe here if you observe here what is the slope of this line, the slope of this line will be minus B upon a right and what will be the slope. Sorry, don't write it as why the time wanting to make the slope and what will be the slope of this line. It will be basically a upon a upon B right and M into M dash is coming out to be what this is coming out to be minus one. So this is the same case, same case what we have done in the first question if you observe here, if you observe here. The coefficient of X square is a B and the coefficient of Y square is minus a B right. So a here is a B and B here is minus a B so a plus B is coming out to be zero. So, obviously, the pair of straight lines will be perpendicular to each other. So that is not the asked information the information was like the question was like write the equation of the lines so this will be our equation. So we have to add like the slope since the slope of this is coming out to be the product of the slopes of both these lines are minus one that we can see here in the equation of the pair of straight lines joint equation right because in the joint equation the coefficient of X square and Y square, the sum of coefficient of X square and Y square is coming out to be zero. Okay. So that is the additional information that is not asked in the question. Now coming to the next question find the coordinates of the centroid of the triangle whose sites are basically this equation will give us the two sites and this equation will give us the third site. So we have to find the coordinates of the centroid of the triangle. So basically for finding the coordinates of the centroid we have to find the coordinates of the vertices of the triangle. Okay, so if you see this will be 12 x square minus 20 x y plus seven y square is equal to zero. So these equation this equation will is giving the pair of straight lines that pair of straight lines will be the two sides of the triangle and one third side is giving us two x minus three y plus four equal to zero. So basically the third side is this the third side is given two x minus three y plus four equal to zero and from this equation we will get the two sides. So what we can what we can do here. How can we write the equations individually so 12 into seven will be 84 84 can we split the middle term. 84 means 14 and six yeah we can do it. So this will be 12 x square minus 14 x y. Okay, minus six x y plus seven y square is equal to zero. So if you see we can take two x common here. So taking two x common we will be left with the six x minus seven y. Okay, and here we can take y common so six x minus seven y equal to zero. Yes, so our equation of lines is six x minus seven y equal to zero and two x minus y equal to zero. So for finding the coordinates of the centroid, we need to have the coordinates of the vertices. So what I will do I will solve these equations. Okay, I will solve this side, the six x minus seven y equal to zero and this third side, that is two x minus three y plus four equal to zero. So here I will get the coordinates of one of the vertices, and from here on solving this two x minus y equal to zero, and two x minus three y plus four equal to zero, we will get the second coordinate. And obviously the third, third vertices will be the our origin, since these two lines are passing through the origin, that is zero comma zero. So solve it, what we can do, we can multiply it by three and then subtract it. So six x minus nine y and plus 12 equal to zero. Okay, now I will subtract it. It will be minus seven y plus nine y that is two y minus 12 equal to zero. So from here we get the value of y as six. Once we get the value of why we get the value of excess seven into six will be 42. So x will be seven right. So our x will be seven. Now on solving this, what we will get. Okay, we can subtract it directly. So three y minus why that will be two y. Okay, so two y minus four equal to zero. So value of why is two. Once we get the value of why we get the value of excess how much one right. So this is one of the vertices, this is one of the vertices and the third vertices is so suppose this is our triangle ABC. Okay, so this is our triangle ABC. So the vertices will be this will be definitely our origin. The second vertices will be seven comma six, and the third vertices will be one comma two. Now the centroid, we need to find the coordinates of the centroid. So centroid will be the coordinates x one plus x two plus x three, zero plus one plus seven by three, and the y coordinate will be zero plus zero plus two plus six upon three. Okay, so this will be eight by three, the x coordinate will be eight by three, and our y coordinate is also coming out to be eight by three. So this will be the coordinates of the centroid. Okay. So is this is what asked in the question find the coordinates of the centroid of the triangle. Yes, so this will be the coordinates of the centroid. Now, coming to the next question. This is question number eight. Since this a x square plus two h x y plus b y square equal to zero be two sides of the parallelogram and the line lx plus m y equal to one be one of its diagonals. So, so that the equation of the other diagonal is given by this equation. Why into BL minus HM is equal to x into a minus HM. Okay. So, a x square plus two h x y plus b y square is equal to zero. If the lines be two sides of the parallelogram. So basically these are the pair of straight lines passing through or is it. So we can do one thing we can make a rough sketch for this. Okay, so this is our y axis this is our x axis. Now I want to draw this pair of straight lines. So, how can we draw. So, let me draw in this way. And let me, this is our parallelogram and this is our diagonals. Okay. So this is a point, oh, let me call it as a B and C. Okay. And let me call this as a. The diagonal where the diagonals are intersecting is point M. So since this O A B C is a parallelogram. So basically diagonals of a parallelogram bisect each other we all are well aware of this fact. Now, the equation of away. Okay, the equation of away and equation of OC, we can write from here. So if we need to write the equation of away, it will be of this form y equal to M one x. Okay, and equation of OC will be why is equal to M two x. Okay. And where we are aware that where we are aware that M one plus M two is equals to minus two H upon B and M one into M two is how much a upon B. Okay, we are aware of with this. And what we can do basically we have to write the equation of the other diagonal. Okay, so equation of AC is given basically. The equation of AC is given as one of the diagonals equation is given. No. So equation of AC is given as L X plus M Y minus one equal to zero. So this is the equation of one of the diagonals. Okay, so we can do one thing we can solve. No, we can solve this way with this AC and we can solve this OC with AC. So care so that we can have the coordinates of this A and C. So, basically, I want to know the coordinates of this A and C, because by knowing that I can have the coordinates of M, because M will be the midpoint of this A and C, and once we will have the coordinate of M. And we are readily available with the coordinate of Oh, so coordinates of O and M is known to us then we can easily write the equation of the second time. So what I need to do I need to solve this O A, O A and AC, I need to solve the O A and AC, O A and AC. So, how can we do, is there any relation given between this M1 and M2, like the two sides of the parallelogram and the line this being one of its diagonal, so that the equation of other diagonal is this. Okay, so let's try to figure it out. So, I need to, I'm going to solve this O A and OC, let's see what's, what's coming out from this. So I'm solving this O A and AC. So equation of O A is Y is equal to M1X. So I will put in the equation of AC. So it will be LX plus M, in place of Y I'm going to put M1X, okay, M1X, then minus one equal to zero. So from here, from here we can have the X coordinate, X L plus M M1, right. I'm taking X common, so L plus M M1 is equals to one. So X is coming out to be one upon L plus M M1. Okay, so once we get the value of X here, what will be the value of Y, Y will be nothing but M1 times X, right. So M1 upon, M1 upon L plus M M1. Okay, so this will be the coordinates of point A. This will be the coordinates of A. This will be the coordinates of A because I'm solving O A and AC, so this will be the coordinate of A. Now, in the same way what I'm going to do I will solve OC and AC, solving OC and AC. So here also what I will do I will do put the equation of means put the substitute the value of Y in the given equation of diagonal so LX plus M into Y, Y is M2X in this case. M2X is equals to one. So again, I will take X. What we can do we can take X common from these terms it will be L plus L plus M M2. Okay. M M2 upon. No, one upon L plus M M2. So this will be the X coordinate and what will be the Y coordinate Y coordinate is nothing but M2 into this thing. So M2 upon L plus M M2. So basically this is, this is the coordinate of, this is the coordinate of our B. Is it okay. This will be the coordinate of B. Now once we are aware with the fact that the coordinates of A and sorry, not B, it will be the coordinate of C, it will be the coordinate of C. Okay. So once we are having the coordinate of A and C, we can simply find the coordinates of our point M. Right, so coordinates of point M will be how much coordinates of M will be X coordinate X coordinate will be X coordinate will be half times. Okay. So half times this, this thing, or we can write X1 Y1, and I can write this as X2 Y2. Okay. So X1 plus X2 means one upon L plus M M1. Okay. And one upon L plus M M2. Right. This will be the X coordinate and what will be our Y coordinate. Our coordinate will be half of, half of this Y thing, means M1 upon L plus M M1 and plus M2 upon L plus M M2. Okay. Is it okay. Now having known this, let me, let me write the coordinates of M as A comma B. This is our A and this is our B. Right. This is our A and this is our B. So what we can do. Now our target is to write the equation of this OB. Okay. So definitely OB will be of this form. The equation of OB if you say equation of OB. OB will definitely pass through the origin so the equation of this will be Y is equals to MX. Okay, where M is what the M will be M will be Y is equal to M. What will be the M what will be the slope of this, it will be B upon a right B upon a into X. Now, this B upon a is known to us. Why? Because B is given by this equation and A is given by this equation. So putting the value of this B upon a you can simplify this thing. You can simplify this B upon a ratio and you will have the equation of the OB. This will be the equation of OB. Take this ratio, find this ratio B upon a and you know the value of B and A. So definitely you will be able to figure it out and put this value here in this equation Y is equal to MX. It will be our equation of this equation of OB. Okay, so hope this is clear to all. Now, moving to the next question. This is our question number nine. Question is asking to find the condition that one of the lines given by AX square plus two HXY plus B square equal to zero may coincide with one of the lines given by this equation A dash X square plus two H dash XY plus B dash Y square equal to zero. So both are basically giving us both are giving us the equation of pair of straight lines. So AX square plus two HXY plus BY square is equal to zero. Okay, this is one homogeneous second degree equation. And the second homogeneous second degree equation is A dash X square plus two H dash XY and plus B dash Y square is equal to zero. Now this also gives us a pair of straight lines passing through origin. This equation also gives us a pair of straight line passing through origin. Now the question is saying one of the lines of this of this pair of equation and one of the lines of this pair of straight lines coincide means they are basically same. So one of the lines of this equation and one of the lines of this pair of equations are same. Okay, find the condition. So if that is the case what will be the condition find the condition that one of the lines this may coincide with one of the livens, one of the, may coincide with one of the lines given by this. So what we can do, we can assume the line to be Y is equal to MX. No, we can do that because that line will be definitely passing through origin. So I am saying one of the line of this is Y is equal to MX and one of the line from this second equation is Y equal to MX. So, definitely, this equation should satisfy both equation one and two, right. So what I will do I will, I will put the value of Y as MX in both the equations. So it will be a x square plus two hx in place of why I will put x MX right, and here I will put y into m square x square is equal to zero. So this is from one. And similarly, from the second equation what I will have from the second equation I will have this thing. A dash x square plus two h dash x into y in place of why I will put MX and this be dash and m square x square is equal to zero. Okay, so on simplification means on some after simplification we will have this as a x square plus two h m x square plus b m square x square is equal to zero. So x square if you take common, you will have b m square plus two h m plus a equal to zero. And similarly from the second equation, from the second equation we will have the same thing is same. No, so it will be come. It will be in the same way. It will be b dash m square. Okay, plus two h dash m and plus a dash is equal to zero. Okay, so no need to write this by observation we can say if this is coming for the first this will be our for second equation. Okay, so this will be b m square from this from this third we can say from this third we can say b m square plus two h m plus a equal to zero. And from this fourth equation we can say b dash m square plus two h dash m plus a dash is equal to zero. Okay, now we have to solve this. So we can use the concept of our cross multiplication. So it will be m square upon this will be the two h into a dash. Okay, m square upon two h into a dash. Okay, minus two h dash into a that will be equal to this minus m upon what it will be there b into a dash b into a dash b into a dash minus a into b dash and that will be equal to that will be equal to one upon how much b into two h dash or two h dash b two h dash b minus two h b dash two h b dash. Okay, so from here we can say our, if you take a minus common from here. So a b dash. Let me write in this way. Let me write in this way this m square upon this to if we take common what we can get. We can have two h or h a dash. Okay, or let it be as it is so two h a dash minus two h dash a is equals to one upon one upon two h dash b minus two h b dash and this will be our m upon be a dash minus a b dash or let me take a negative sign common here so that it will be cancelled from the numerator and denominator so a b dash minus be a dash. Okay, that will be equal to one upon two h dash b minus two h b dash. So what we need to find we have to from here if you see the value of m we are getting no so after getting him. From here we get a m s a b dash minus b a dash upon two h dash b minus two h b dash. Is this what we require to prove. Okay, the question is asking to find the condition. Okay, so basically our slope will be this no. So we can put the value of him here. We have got the value of him, we can substitute the value of him here and we will get the, we will get the condition because our target is to eliminate him because in question nowhere. There is a we have assumed the line to be y equal to max. So we can eliminate this m by putting the value of him in this equation. So, whatever you get that will be the required condition. So, I think this is the last question for this exercise. So, anyhow, we are done with this exercise one. So, very soon, we will be back with the next exercise. So, till then, Tata goodbye take care.