 Hello friends, a very good morning to all of you. Hope you all are doing good. So today we are going to discuss exercise four of pair of straight lines. And I have already copied my questions in the PPT. So let me first share my screen. And yes, so here we go. This is exercise for question number one from the pair of straight lines. So the question is saying if lambda x square plus 10 x y plus 3 y square minus 15 x minus 21 y plus 18 equal to zero represents a pair of straight lines, then the value of lambda is so it's a general second equation. Right. And we know for a pair of straight lines this delta must be equal to zero. Now what is delta. Okay, we will discuss first let me write the equation first. So it is given as this lambda x square plus 10 x y plus 3 y square minus 15 x minus 21 y plus 18 equal to zero right lambda x square plus 10 x y plus 3 y square minus 15 x minus 21 y plus 18 equal to zero. So for this equation to represent a pair of straight lines to represent this equation as pair of straight lines pair of straight lines are lambda must be equal to zero. So lambda lambda is basically defined as this ABC plus to FGH right minus AF square minus BG square minus CH square equal to zero. Okay, and in determinant form in determinant form if you say we write you used to write this delta as this a BC. Then here we get F, F, G, G and H and H. So this is the representation of delta in determinant form and this delta must be equal to zero. So I think everyone is aware of this I'm just recalling this. So let's write, let's write the value of this lambda. Okay, I will I will be using this determinant form. So I'm writing it in determinant form as what is a here, a here is lambda. Okay, and like, this is in the, what do you say we have to compare this with our standard equation second general degree equation. Okay, so so what is a second standard second, second degree equation. Right, it is a square equation if you see equation of second general second equation it is a square plus two H x y plus b y square plus two g x plus two f y plus a point to zero. That's a general second degree equation. So if you compare this with this equation, if you compare this standard equation with this, we are having the value of a as lambda. What is B, B is coefficient of y square so coefficient of wise where here is three and what is C, C is the constant of that is 80. F is minus 21 by two. So here it will be minus 21 upon two minus 21 upon two. And what is G, G is minus 15 by two. So here it will be minus 15 by two. And what is H H is five. So this is our five. And this must be equal to zero. Now we have to just solve this determinant and we will get the value of lambda. So let's quickly solve it. So expanding this determinant what we get lambda into this will be our three into 18 that will be 54 and minus of this. So 21 guys where how much 21 is where is 441 upon four. Then minus of five. What we get 18 into five is 90, right 18 into 590 minus 21 into 15. So 21 into 10 to one zero and one not five. So this is 315. So this will be 90 minus 315 upon how much. So, okay, let me write it clearly it is minus 15 by 12. No, it is minus 15 by 12. So minus five into this 18 into 590 and minus. Okay, this will be anyhow plus only so 315 upon four. And then here we will have minus 15 by two minus 15 by two into this thing. So this will be 21 into five how much one not five so minus one minus one zero five upon two and minus minus and minus that will be plus so 45 upon two. Now this is the determinant value and that will be equal to zero. So what we can do here. So four four into four 16 and four five 20 to one six minus four 41, then minus five. Here also we can take LCM four. So, 360 minus three 15 then minus 15 upon two. Okay, here it will be two only. So minus one not five and minus plus 45. So it will be minus 16 and that will be equal to zero. So further what we can write from here if you see we can take LCM four from all the terms okay so it will be lambda into 216 minus 441. So it will be five to minus 225 then minus five into this will be how much 45 right so minus 45 then it will be 15 into 60 that will be plus 900 that will be equal to zero. So from here what we can say minus 225 lambda minus 5525 225 and plus 900 is equal to zero. So we can say minus 225 lambda is equal to how much this will be 900 minus 625 that will be 675. Okay, so it will be minus 675 minus minus will get cancelled out and we got the value of lambda is three or 675 upon 225 that will be nothing but three. So this will be our answer. This will be our answer to this question. So lambda equal to three is, yeah it's there in option B. So our correct answer will be option B. Okay, so moving to the next question. Let's see this, the point of intersection of the straight lines given by the equation this 3 y square minus 8 x y minus 3 x square minus 29 x plus 3 y minus 18 equal to zero. So we need to find the point of intersection. So basically this general second degree equation is representing straight lines. Okay, now for finding the pair of this point of intersection of these two lines, what we used to do we used to partial differentiate this equation. First with respect to x and then with respect to why, and then we solve both the equations and we get the point of intersection. So if you see our given equation is this 3 y square minus 8 x y minus 3 x square. Okay, minus 29 x plus 3 y minus 18 equal to zero. Now what I will do I will partial differentiate it with respect to x. So basically, this thing will be zero, and here it will be means while partial differentiating with respect to x we treat the term consisting of why means we treat y as constant. So these three ways where we'll be treated as constant and sets the differentiation with respect to x will be zero. So this will be zero, then here it will be what minus 8 y. Then minus of 6x then minus of 29, this will be zero, this will be zero and that will be equal to zero right and our partial differentiating with respect to why this will be 6 y. Okay, then. Here it will be minus 8x. Okay, he has 6 y minus 8x. This will be zero, this will be zero. And here it will be plus three and this will be zero. And this equal to zero. So we got the equation first equation is minus 8 y minus 6x minus 29 equal to zero. And the second equation is 6 y minus 8x plus three equal to zero. Now what we do, we solve these equations, we solve these two equations and we get the value of x and y, and that point will be the point of intersection of this area of straight lines. So what we can do we can multiply the first equation by six and the second equation by eight. So I'm multiplying by six we will have minus 48 y minus 36x right and minus 6954 five. So minus of 174, and that will be equal to zero. And here we will have 48 y minus 64x plus 24 equal to zero. Now add both these equations we will have minus of 100x and minus of 150 equal to zero. So from here we get the value of excess minus three upon minus three upon. So this is our exponent of the point of intersection. Put this value in any of the equations. So if I put here x equal to minus one by two six y minus eight, what is x, x is minus three by two, and plus three equal to zero. So six y and four, so 24 plus 24 plus three equal to zero. So we got the value of y is minus 27 upon six, that is nothing but minus nine or what do you say minus nine upon two. Is it okay. But whether I have done any mistake, because minus nine by two I'm not able to see in any of the option x is there minus three by two. Okay, so six y minus eight x. So six y minus eight value of x is minus three by two and plus three equal to zero. So six y four times we are getting. Okay, okay, so four into three it will be 12. So it will be plus 12. So, basically, we will have this value as minus 15 upon six. Okay, yes, we are now getting minus five by two. So the y coordinate of point of intersection is minus five by two is the option if you see the correct option is this option D minus three upon two and minus five upon two. So, this is the method for finding the point of intersection. Okay, means if the general second degree equation is given, and that equation is representing a pair of straight lines. So we have to find this. Del s by del x equal to zero, and this del s upon del y or partial differentiating with respect to x equal to zero and partial differentiating with respect to y equal to zero and we solve these equations. Okay, from here we get the point of intersection. And what is this basically this is our s is completely this complete expression we call it as s. Okay, this is this. Now, moving to the next question, we are having this question number three. Okay, so if this equation, if this given equation represents two perpendicular lines, then we have to find the value of P and q. So, okay. Let's first write the equation. Sorry. So let's first write the equation so it is given 12 x square plus seven x y minus P y square and minus 18 x plus q y and plus six equal to zero. Okay. So if this equation is representing two perpendicular lines, means this coefficient of x square. Okay, coefficient of x square, and plus coefficient of y square. This must be equal to zero. Now what is the coefficient of x square here it is 12. And what is the coefficient of y square it's 15 so sorry coefficient of y square is minus p. So 12 minus p this must be equal to zero. So straight away we got the value of PS 12 right straight away we got the value of PS 12. Let's see the option. Okay, in option in all the options the value of these even as 12. So, now, our target is to find the value of q. So how can we find the value of q. Basically, this equation is representing pair of straight lines. If this equation is representing pair of straight lines so as we discussed earlier, the value of delta should be equal to zero, and we know the value of what is that. This is our ABC, right ABC F F G G H. So the value of this determinant should be equal to zero. So now if you see, we are going to put the value of ABC F G HR. So what is a here a here is 12. What is be here be here is we got the value of be no so be is how much coefficient of wise where so coefficient of wise where is minus 12 right coefficient of wise where is minus 12. And what is C C is six. Now what is our F. So F is basically q upon two right F will be q upon two. And what is G G is minus nine. So minus nine minus nine and what is H H is seven by two. So we are going to put this determinant value equal to zero. Now, our next task is to solve this determinant. So what we can do, let's expand on the first row. So 12 into six that will be minus 72, then minus q square upon four. And then minus times minus seven by two, and minus seven by two years, and then what so 42 upon two or this will seven by two into six no, so it will be basically 21. So 21 minus how much plus nine q upon two, so plus nine q upon two, and then minus nine into seven q upon four, seven q upon four, right, seven q upon four and minus of 12 into nine will be how much one not eight. This is equal to zero. Now we have to solve this. So okay, let's do it. So, taking four and see him here. 428. And four into seven is 28 so minus 288 minus of q square, then minus seven by two. Here we will have 42 plus nine q. Right, 42 plus nine q then minus nine. If we take four and same here, it will be seven q minus 432 right 432 equal to zero. So what can we do, we can take four and same right from all the three terms. So this will be basically 12 into minus 288 minus q square, then minus seven into 42 plus nine q, and minus nine times the seven q minus 432 is equal to zero. So it will be basically a quadratic in Q right, and we have to solve that. But it seems to be. Or one. Okay. So 12 into. Can we take anything common from you. 288. So in two 288 if you see. Okay, let's expand it. So 288 into 12. So, so let's set up and 576 and 288 so six, three, four, five, six, so minus three, four, five, six, minus 12 q is where that is a 14 this 294, then minus 63 q, then minus 63 q once again, then plus nine twos are 18. So this are a cacti is though, and nine four 36 and 88. So 3888 that will be equal to zero. So what we get finally is coming out to be minus 12 q is where minus 63 minus 63 will be minus of 126 q. Okay. And if you add this 294 in it. So it will be 6410 157 and 3750 and we have to subtract it from here so 3750 so it will be basically eight. Okay. And three. So 138 and with positive sign so plus 138 equal to zero. And we can take a six common no, so it will be two q square plus 21 q right. That means 21 q and minus two. Three means 23 equal to see. So two q square plus 21 q minus 23 equal to zero now we have to solve this quadratic and we will get the answer. So, it will be basically how much 23 into 246. So 23 into q we can break it as two q square plus 23 q minus two q minus 23 equal to zero. So we can take q common, we will have two q plus 23 right, and minus 12 q minus plus 23 equal to zero. So from here we get to q plus 23 into q minus one equal to zero. So what we got we got the value of q s q as one or q s minus 23 upon 22, sorry, minus 23 upon minus 23 upon two. So this will be the value of q p we have already got the value of p is twin. Now, we were finding the value of q. So from this, we are getting two values of q that is one and minus 23 upon two, so minus 23 upon two is there. So one is also. Yes, one. So this will be multiple correct question. So this option is also correct because value of q is one. Yes, the value of q is one or minus 23 upon two. So this option a will be correct for this question. So let's take the next question, question number four. So if the angle between two lines represented by this equation is tan inverse m, then we have to find the value of m. Okay, so this is a two x square plus five x y plus three y square. Okay, plus seven y plus four equal to zero. Okay, so we have a formula. Okay, but we have a formula for but that is for homogeneous second degree equation right. So this is the second degree equation if the pair of straight lines is represented in this form a x square plus two h x y plus b y square is equal to zero. Then we have the value of 10 theta is what two into under root of h square minus AB upon mod of a plus right. So this equation is our a x square means this is the general second degree equation right so a h square plus two h x y plus b y square and plus two g x plus two f y plus equal to zero. So basically if you see no, we can write it in terms of this, in terms of our homogeneous equation with the shifted like suppose point of intersection point of intersection of both these lines is x one y one right. The point of intersection is x one y one and here point of intersection is our origin zero comma zero. The point of intersection is origin here the point of intersection is x one and one. So basically, this is shifted right. This is only shifted there will be no change in the angle like angle will be same whatever the angle is for this, the same angle will be for this one. So this is what I need to say. So we can apply the formula here also. And the other way is like, we can find the equation of two lines from this right. And after having the two equations we can easily find by applying the formula of 10 theta that is m one minus m two upon one plus m one m two. That is one process. But what I mean to emphasize is that what I mean to say is that, like, there will be no change in the angle like angle between the lines will be same for this and for this equation. So, I am applying the value on this I'm substituting the value on in this formula only. So, what I'm getting two times, what is h square h square is five by two guys square that is 25 upon four and minus AB two into three that will be six. Right, a into B, and further we will have a plus B so more of five. Right. So what we are getting here we are having 25 four and one right. So, two into under root one by four upon one of five that is one upon five. Okay. And it is given out to be this is our 10 theta. And from here we get theta is equal to how much 10 inverse one by five. Okay, and so what is the value of m m is nothing but one by five. So this will be the value of him. Okay. So, our homogeneous equation if you see this will be the lines passing through or isn't right. But if that is represented in this form, if it is represented in this form. In general second degree equation, so it will be somewhere here right somewhere here, whose point of intersection will be here it is origin but here it is x one comma. So this whole thing is getting sifted no see this whole thing is getting sifted only the point of intersection will change but the angle between the lines will not change. Right, so this is a little and this is a little. So hope this this is clear to all of you. So, the angle between them will be so as per the question the value is a means coming out to be one by. So option B is correct. Let's go into the next question that is question number five. Let's see. Question number five the equation of second degree equation second degree this represents a pair of straight lines, then the distance between them is okay. So, the equation given is x square, plus two root two x y okay, and plus two y square, plus four x, plus four root two y, plus one equal to zero. So this equation is representing a pair of straight lines. Right, and we have to find the distance between them. So, what we can do we can individually write the equation by equation of straight lines. So, like, what, what is my approach I'm going to write the equation of straight lines individually okay. So if you observe this x square plus two root two x y plus two y square. So this is nothing but x plus root to do why the whole square this complete thing. Right. So, x plus a root to come holy square that will be this thing. Now what, how can we write the equation of equation of the straight line individually. This will be written by x plus a root to do why I'm assuming a constant a okay, and here I'm writing x plus a root to do why. Okay, and plus of B, and that will be equal to this equation. This will represent this equation. So compare Korean if we are going to compare the coefficient of comparing the coefficient of x and y. What we will get, what is the coefficient of coefficient of x from this. These two factors, it will be a when it will be get multiplied with this term will get multiplied with X, so it will be a and when this B is getting multiplied with this X, then we will be having the coefficient of x, so the coefficient of x will be a plus B, and what is the coefficient of x here, the coefficient of x here is four. So this is our first equation, and the second thing we are going to compare the coefficient of y. So how do we get the coefficient of y here. See the coefficient of why we can have when this thing will get multiplied right this a into root to do why. Okay, so a into root to do a into root to do, and how again this this be when this be will get multiplied with this. So, plus beat into a root to do this will be the coefficient of y and what is the coefficient of y here it is for root to do. So this is our second equation. So if you take a root to do common here, so we will have a plus be equal to for a root to do. Okay, so a plus B is coming out to be four. And what else if you multiply this that will be the constant term right, so this AB will be equal to how much one AB will be equal to one. So how can we get this. If we put AB equal to one, I mean, this is one equation and this is one equation a plus be equal to four. So if you see we can find the value of a minus be right a minus be police where will be how much 16 and minus four, so that will be 12. So a minus be will be basically to root three. Is it okay. And a plus bees for so a plus be is for and a minus be is how much to root three. So we got the value of two as two into root three plus two. So the value of a is coming out to be root three plus two. And so basically this a minus be will be with plus minus and right, because this is a we are taking the under root. So this will be I'm taking the plus sign only here. So if the value of a is coming out to be root two plus root three, what will be the value of B. So two plus root three plus be is equal to four. So from here we get the value of BS to minus root three. Right. And we can solve by taking the minus sign also. So I think the same value will come. So now we got the equation as how what our equation of first line is x plus root to why and plus a is what two plus root three two plus root three equal to zero and the second equation of second line is x plus root to why and plus two minus root three equal to zero. So basically these are parallel lines right. These two lines are parallel lines. So, what will be the distance between them distance between them will be this thing. C1 minus C2 right. So two plus root three, two plus root three minus two plus root three. Okay, and all divided upon under root of one guy square and plus root two guys square root two guys square. So this will be plus two minus two will get cancelled out to root three upon. How much this will be two plus one that is three. So root three. So this will be equal to two units. So it's given in option a. So this will be the distance between these two lines. So option is correct. Okay. We are getting the distance between live as between the lines as two units. So this was our question number five. Let's see this question number six. So find the area of the parallelogram formed by the lines this. Okay, so the first equation is giving us the pair of straight lines the second equation is also giving us a pair of straight lines. So basically we are having the four lines. Okay, and we have to find the area of parallelogram formed by those four lines. So what we can do we can simply get the equations first okay the equation of four lines, which are which are forming this parallelogram. So this is two x square plus five x y plus three y square. Okay. This equal to zero. So we can split the middle term as two x square plus three x y plus two x y and plus three y square equal to zero. Taking x common we will have two x plus three y and taking my common, we will have two x plus three y equal to zero. Two x plus three y. Okay, and x plus y equal to zero. So this is basically two equations. Two equations of line this one is our L one, this one is our any two, and on solving this we will have the next two lines. We can solve this equation also. So the given equation is two x square plus five x y plus three y square and plus three x plus four y plus one equal to zero. Okay, so this is what we have already saw no. So, yes, we can now compare the coefficient of x and y we can find the value of straight lines. So this will be basically two x plus three y I'm assuming the constant to be a here and then x plus y plus B and now I am going to compare the coefficient. So if you see, if you see what will be the coefficient of x here it will be a and plus a plus to be right, and that must be equal to three, and the second equation we will get by comparing the coefficient of why. So it will be a. Okay, and here we will get the coefficient of why is how much be into three. So that is three. So a plus three be will be equal to four. Now subtract this equation one from equation two, we will have the value of B as one. Once we get the value of a once we get the value of B we can easily find the value of a is two into a will be also equal to one. Okay, so basically our L three will be L three line will be this two x plus three y plus one equal to zero and our L four line will be x plus why. Okay, x plus why and plus one equal to zero. So we are having the four lines, right, we are having the four lines, which are forming the parallelogram. So if the line is given in this form, if the line is given in this form y is equal to MX plus C. Okay. So we know the area of parallelogram, we know the area of parallelogram we used to give it by model C one minus C two. Okay, into T one minus T two upon one minus M one M two right one minus M one M two. That will be M one minus M two. Okay, now if you compare this first two lines. If you compare this first two lines. Let me change the, let me change the color of the pen. So it will be basically. Okay, so can we draw. Let me draw the parallelogram of diagram. Okay. So yes, the this equation is to x plus three y equal to zero and line parallel to it will be to x plus three y plus one equal to zero. So this equation is x plus y equal to zero and the line parallel to it is x plus y plus one equal to zero. Okay. So basically, if you see what will be the value of C one here, what will be the value of C one here C one is zero, right. And what will be the value of C two. So this will be the C two right. So this will be value of C two will be minus one upon three minus one upon three, and what will be the value of M one. M one will be minus three by two the slope of these two lines these two parallel lines. So M one will be minus two upon three right minus two upon three. And similarly if you see what will be the value of C two, sorry. What will be the value of D one D one here is zero and D two that is constant term so constant term here is minus one and the slope here is how much minus right now put the substitute this value. So area will be area of parallelogram will be C one is zero minus C two is minus one by three so this will be plus one by three. Okay, then hold multiplied by D one minus D two that will be one and upon M one minus M two that is minus two upon three and minus M two that will be plus one. So this is coming out to be one by three upon one minus two upon three this will be also one by three modulus. So this is coming out to be one square unit. So this will be the area of the parallelogram. Okay, so important to note that area of parallelogram is given by this, but the equation is given in this term, equation is equation of straight lines isn't why is equal to MX plus C four. You know this formula, so you can easily substitute the value of C one C two D one D two M one M two and you get the area of the parallel. So what we have done we have just first find out the equation of lines. Okay, which are representing this we are which are constituting or which are making this parallelogram and once we get this we can easily get the answer. So, the area is coming out to be one square unit. So this will be our answer. No options available here. So I'm writing here. This is our answer. Now, let's see this question number seven. It is saying find the locus of the in center. Okay, find the locus of the in center of the triangle formed by this equation and this. Basically, this is the first equation will give a pair of straight lines, and the second equation will give a straight line so basically we are having three straight lines which are forming a triangle, and we have to find the locus of its in center. Okay, so let me first write the equation. So it is XY. It is XY minus four X minus four Y plus 16 equal to zero. Okay. So if we take X common from here first two terms we will have Y minus four. And if we take four common we will have Y minus four equal to zero. So Y minus four into X minus four equal to zero. Okay. So the first equation is giving these two lines. So our first line is Y minus four equal to zero. Our second line is X minus four equal to zero. So why minus four equal to zero means what Y equal to four. So this is our first line. X equal to four is our second line. And our third line is given as what X plus Y equal to a X plus Y equal to a. So, like, let's first draw a meter. Sorry, rough sketch for this. So, basically, this will be our X equal to. Yeah. Okay, so this is our Y axis. This is our X axis. Okay. And somewhere here it will be our X equal to four. Okay. And somewhere here it will be our Y equal to four. So this line is Y equal to four. This line is X equal to four. So what will be its point of intersection. Let me call this point P. So its point of intersection will be four comma four obviously, right. And this point is lying on this line, right. This point is lying on. If you join this, this is our line Y equal to X. This is our Y equal to X line. So this point P is lying on this line Y equal to X. And both are these lines Y equal to four and X equal to four both are at right angles. Okay. Now, let me draw this is this, this third line. So that we can see this triangle. So let me say this is, this is our, oh, sorry. Yes. So yes, what I am saying this is our third line X plus Y equal to a X plus Y equal to pay where this point will be a comma zero. Okay. And this point will be zero comma zero comma. Okay, so the question is asking for this triangle a triangle guy in center to look us. Okay, so let me ask you one thing. What is in center in center is the point of intersection of the angle by sectors. Okay, internal angle by sectors. So internal angle by sectors, if you see the point of this point B, the angle, if you if I name it as a what do you say, a B. Name it if I say angle ABP. Okay, ABP is how much it is 90 degree. This, this line is variable this line white line is variable right this may move in this direction also for this direction also. But, but the angle by sector of angle a PP angle it's angle by sector will always line on this line Y equal to X angle by sector will always lie. Y on Y equal to X. How many of you agree on this. Right. So if you see this, this angle ABP, that is 90 degree and it's angle by sector will always line on Y equal to X. So, no matter where this, this white line moves, whether in this direction or that direction, or opposite direction. Whether this line move towards origin or away from origin. The internal angle by sector, right, the angle by sector of P will always lie on this line. And from here, we can conclude that we can conclude that the in center will always lie on the line my equal to X. So basically in center in center will always lie. We'll always lie on Y equal to X will always lie on Y equal to six. So that will be the locus that will be the locus of incentives, right, somewhere here it will be there. This will be your incentive, and it will always lie on Y equal to excellent. So hope this is clear to all of you. Okay. So, now let's take the next question, question number eight question is saying if the equation to H XY plus two gx plus two f y plus equal to zero represents two straight lines. Then so that they form a rectangle of area mod of FG upon H square with the coordinate axis. So the given equation is this thing, the given equation is to H XY to H XY plus two gx plus two f y plus equal to zero. And it is representing two straight lines right, it is representing two straight lines then so that the area of triangle rectangle. Okay, so for representing this equation means if this equation is representing two straight lines. That means our data must be equal to zero. If you say a delta is what ABC so a zero deal ABC plus two FGH. So, this will be two FGH minus AF square that is zero minus BG square will be zero, and minus CH square minus CH square equal to zero. Basically from here if we get two FGH equals to CH square, so one inch and one inch will get cancelled because it is not equal to zero here. Right, so we got one equation that is two FG is equal to CH. Okay. And what else information is provided here. And then so that they form a rectangle of area this with the coordinate axis. Okay. So coordinate axis. Okay, what we can do, we can substitute the value of F in this equation. So we will be two HXY. Okay, plus two GX plus two Y, what is F, F is the CH upon two G CH upon two G and plus C equal to zero. So, we can take two X common from here. So we will be having G plus H Y G plus H Y right and from here if we see we can take C by G common. So we will have G plus G plus this two and two will get cancelled out. G plus H Y right G plus H Y equal to zero. So we got the two equations G plus H Y equal to zero and two X plus C upon G, two X plus C upon G equal to zero. Okay, now let's draw the rough diagram for this. So this is our X axis, this is our Y axis. And from here we get, this is our L1 and this is our L2. So from here we get Y is equals to. If you see L1 is, Y is equals to minus G upon H. Okay, and if we see L2, L2 means what the value of X is coming out to be minus C upon G, or two G you can say. So this is our X equal to, X equal to minus C upon two G. And this is our Y equal to minus G upon H. So basically this area we need to calculate this area, area of rectangle right. So area of rectangle if you see area of rectangle that will be mod of, right, mod of C upon two G into G upon H. Okay, length into height or length into width. So this will be equal to, if we substitute the value of C here, we will have two FG upon H. Okay, so two FG upon H and already it is two G here already two Z was there. Into G upon H, mod yellow. So that will be this two, two will get cancelled out. This G and G will get cancelled out and we will have mod of FG upon H square. So this is what we were required to prove. And we got this result. So hence it is proved that this will be our area of rectangle formed by the pair of straight lines and our coordinate axis. Okay. So this was our question number eight. Now let's see this question number nine. So the area of triangle formed by the lines this AX square plus two HXY plus BY square plus two GX plus two FY plus equal to zero, and the axis of X. This is general second degree equation. AX square plus two HXY plus BY square. Okay, and plus two GX plus two FY plus equal to zero. We have to find the area of triangle formed by this thing. So let me say these are the two straight lines. And this is our axis of X means Y equal to zero. And this is our L1 and L2. L1 and L29. Okay, so if we put Y equal to zero in this equation, if we put Y equal to zero in this equation, so it will give me the X coordinate of these two points, right? A, B, the X coordinate of this A and B will be, we will have when we put Y equal to zero in this equation. So let's do it. So it will be AX square plus zero plus zero plus two GX, okay, and plus equal to zero. So this is a quadratic in X, okay. And we need to find the area of this triangle, area of this triangle. And so X1 and X2, we can easily get it. Okay, X1 and X2 are the roots of this equation. But for finding the area, for finding the area of triangle, what we need to do, we need to do half into base, okay, and this height. Okay, so half into base into this, if I name it C and it's D, so half into base into height that is CD. So for calculating this AB, what we need to do, we need to find the difference of the roots. So X1 minus X2. So this is nothing but under root of D upon 2A. So what is under root of D? It is P square means 4G square minus 4A and C. Under root is ka jayaega and upon 2 times A, so 2A. So that is nothing but this will be our 2 under root G square minus AC, right, G square minus AC upon 2A. So this 2 and 2 will get cancelled up. So mod of 2A, okay. So no, this is basically mod of root D by only A comes, right. So this will be mod of A. This 2 will be as it is. So 2 under root of G square minus AC upon mod of A. This will be the, this will give us the, what you say AB length, right, this is nothing but our AB length. And what else we need to find? We need to find this CD. So basically CD is nothing but the Y coordinate, Y coordinate of the point of intersection. So if you see this CD will be Y coordinate, right, the CD length will be Y coordinate of the point of intersection of L1 and L2, right. Now what will be the Y coordinate? What will be the Y coordinate for this point of intersection? So basically Y coordinate will be, if you know, this is our, what you say AB, okay, ABC, F, F. And here we used to write H, no G, G, G, H, H. So this was our delta, right. This is a memory like how we used to find the coordinate of point of intersection. So we normally write it as AH, HB, GF and we again repeatedly write it as AH. So the Y coordinate will be basically this AB H square then HF, BG, right, and then this will be our numerator. So the Y coordinate will be the CD will be how much? This is H square minus AB will be in the denominator, okay, H square minus AB will be in the denominator. What we are getting from this multiplication H square minus AB. And from here we get AF minus GH, AF minus GH. Basically this will be our Y coordinate. So if you see this area, area of triangle will be half into AB, AB we can write from here, two times under root of G square minus AC upon mod of A, okay. And what will be the CD length? CD length is this AF minus GH upon H square minus AB, okay. So if you are able to further simplify it, it's well and fine or you can leave this question as it is. So this will be our area, okay, this will be our area. So these two and these two will get cancelled out. So our final area will be under root of G square minus AC upon mod A into AF minus GH upon H square minus AB, okay. So in this particular question, this value of AB, C and FGH is not known, but if you are having that, you can simply put that value in this equation or in this result and we will have the area of the triangle. Formed by the sphere of the straight lines and our axis of X, that is X axis or Y equal to zero, okay. So this was the solution to this question number nine. This is our question number ten. Question number ten after this I don't think we are having any question. Okay, so basically this is the last question of this exercise. So question is asking to find the equations of the straight line passing through the point one comma one and parallel to the lines represented by the equation this. Okay, so X square minus five XY plus four Y square plus X plus two Y minus two equal to zero. Okay, this is the given equation of a pair of straight lines. We have to find the equation of straight lines which are parallel to these lines and passing through this point. First, let's find the equation of these lines individually. So we will have X square four, okay minus four XY minus XY plus four Y square. Is it okay? And what we can write it for the further we can take X common so X minus four Y and we take Y common we will have X minus four Y. Okay, so X minus four Y and X minus Y. These are the means I'm solving this only second second degree terms right so I'm solving second degree terms we are having this. So, if we see our first line will be X minus four Y plus a equal to zero. Okay, plus a equal to zero. And our second line will be X minus Y plus B equal to zero. Okay, and now we have to find the value of sorry the equation of lines which are parallel to these two lines and passing through point this. So if we see we don't actually need to find the value of a and B that is not required that is not our target our target is to find the equation of lines which are parallel to these lines. So let me say our required lines will be of this form our required lines will be of this form. So, let me write it as 11 dash. So it will be of this form X minus four Y plus M equal to zero. Okay, and our L two dash line will be X minus Y and plus n equal to zero. Now this will pass through point one comma one right. This will pass through one comma one. So let's put the value of X and Y one as one and one. So we will have one minus four plus M equal to zero. So from here we get the value of MS or how much three minus three plus one minus three. So we're coming out to be three. And if you put one one here we will have n equal to zero. So basically our required lines will be this, our required lines will be this X minus four Y plus M that is three and into the joint equation we can give it in this way X minus why X minus why plus zero that is nothing X minus y equal to zero. So the required line will be this. If we ever want to solve it we can solve. So X square minus X Y minus four X Y, then plus four Y square plus three X minus three Y equal to zero or X square. This is X square, then minus five X Y, okay, plus four Y square, and plus three X minus three Y equal to zero. So this will be the required equation of lights. Okay. So I think and I hope everyone is clear on this question or rather everyone is clear on this complete exercise. So there were total team questions in this, which we have discussed all of which. Any questions and we have discussed each and every question. Okay, in detail to it. So I hope everyone. Everyone got the solutions okay and if you are having any doubt you got clarified from this video. Yes, we are done with this exercise and we will meet once again with the next topic or with the next exercise if left in this chapter. So till then Tata take care goodbye.