 Good evening, friends. Hope you all are doing good towards the preparation of J mains in advance and also for your boards exam. So in continuation with our video series, we are going to discuss exercise three of pair of straight lines today. So just time setting my screen and yes. So yes, we were discussing pair of straight lines and today we are going to discuss this exercise three. So let's check the first question. It is saying if coordinate axis are the angle by sectors of the pair of lines a x square plus two h x y plus b y square equal to zero, then following conditions are given and we have to answer which, which is true. So basically, if you see for this given second degree equation a x square plus two h x y plus b y square equal to zero. Now, what is the equation of the angle by sector for this, the equation of angle by sectors for the above pair of lines. We used to give it by this x square minus y square upon a minus b equal to x y upon each right now. The question is saying our coordinate axis are the angle by sector means our x axis right x axis and y axis are the angle by sectors. Okay, for this pair of straight lines. So basically x y x axis and y axis. This we used to represent it by y equal to zero and x equal to zero and combined Lee we used to denote it as x y equal to zero. So this is the combined equation of our coordinate axis. In this, we have to take, we have to see when this x y is coming out to be zero. Okay, so if you put our age equal to zero, if we put our age equal to zero, and see the symbolic representation of this equation. Okay, if we put our age equal to zero, and analyze this symbolic representation of the ever equation what, what do we get this thing will be multiplied into zero. So that will give us x y equal to zero. This is what we need right this is what we need x y equal to zero are the coordinate axis and these coordinate axis are the angle by sectors for this. So in this combination, we must have our age equal to zero right. So, after putting it equal to zero in this in this equation we are getting x y equal to zero means symbolic representation we are putting here. So yes, this option B is correct, that is a check one two zero. So, I hope everyone is clear on this. Now moving to the next one. Please check this question. If the line y equal to mx is one of the bisectors of the lines x square plus four x y minus y equal to minus y square equal to zero, then the value of it. Okay, so combined equation of a pair of straight lines is given here, that is, let me write the equation first. So this is x square plus four x y minus y square is equal to zero right. And what we have to find we have to find them. This will be basically angle by sector. Okay, this will be angle by sector it is saying by sector only. So the given line y equal to mx is one of the bisectors of the lines. So first let's write the equation of bisectors for this pair of straight line. So equation of bisectors is given by this equation x square minus y square upon a minus b what is a is the coefficient of x square here it is one minus be what is be coefficient of y square in this case it is minus one. So x square minus y square upon a minus b is equal to x y upon H, what is the value of H here to which is equal to four so our H will be equal to. Right. So, from here what do we get x square minus y square is equal to this to, and this to okay, let it let me first try it. So basically this to and this to will get cancelled out. So, our equation of bisectors comes out to be this x square minus x y x square minus x y minus y square is equal to zero. Okay, am I doing any mistake. Here it will be two here it will be a minus be it seems to be correct only. So let's check x square minus x y minus y square equal to zero. So, this is the equation of bisectors. Okay, for this pair of lines. Now what we can do from here. If you see, we can make a quarter take note, we can make a quadratic from here. So, if we divide it by. We can make quadratic also or we can simply put this y equal to mx in this equation, because this is one of the bisectors. So, I will put y equal to mx in this equation. So what we will get, we will get this x square minus x into y will be what mx and minus m square x square is equal to zero. So, from here we get x square. If I take x square common then it will be one minus m, okay, one minus m minus m square is equal to zero. And if we take m. So basically, we get this quadratic in m m square, plus m minus one equal to zero. So, from here we can find the value of our m. So, what will be the value of m m will be minus b is minus one plus minus under B square base where means one minus four a is one C is minus one upon to a so to so value of m is minus one right plus minus what is this five right. Oh, plus one five upon two. So yes, the value two values of him will be there the first value will be a root of five minus one upon two. Okay, and the second value of him will be minus root five minus one upon two. So I think this is multiple choice question so root five minus one upon two, this is an option a and root five minus one root five minus one upon two. Okay, if you take minus common here, what do we get a root of five plus one upon two. So yes, a root five plus one upon two so C is also correct. So option a and C both are correct for this particular question. So I hope everyone is clear on this. Now moving to the next one. Question number three. So, question is saying if one of the lines of my square plus one minus m a square x y minus m x square equal to zero is a bisector of the angle between the lines x y equal to zero CEO x y equal to zero means this equation what does it represent x y equal to zero means x axis and y axis right x axis and y axis. So this x y equal to zero, this is the combined equation combined equation of our coordinate axis. Okay. So, and this equation is given second degree equation m y square plus one minus m square x y and minus mx square is equal to zero. So one of the bisectors, right. One of the bisector of this pair of straight lines is a bisector of the angle between the lines x y equal to zero. Okay. So basically, y equal to x will be bisector right. This is our x axis, this is our y axis. So this will be the one of the bisector right. One of the bisector for this coordinate axis, the equation of this line we can give it as a y equal to x, having the slope of one means this angle will be how much 45 degree and slope of this will be one. So we will use this information in this question. Okay, so if one of the lines of this. Okay, one of the lines of this is a bisector. Okay, so this is basically a pair of straight lines passing through what is it. Okay, since because this is a homogeneous second degree equation right. Everyone agrees. Okay, so this will represent two lines and we can use this information. The slope of the two lines, the slope of these two lines will be equal to minus two h upon a and the product of these two lines will be what the product of these two lines, representing by represented by these two lines the product of the slopes of those two lines will be equal to a upon B. you all are aware of this AB and H. So A is the constant of X square B is the constant coefficient not constant is the coefficient of X square B is the coefficient of Y square. And this to H is what to achieve the coefficient of XY. So basically this will be minus two H no. And minus two H means this complete thing will be two H so minus of one minus M square and upon a coefficient of X square is what minus M. So this minus minus will get cancelled out, it will be one minus M square upon it right and M one into M two is equal to a upon B. So what is a A means coefficient of X square that is M minus M sorry, and what is the coefficient of Y square it's M. So this is coming out to be minus one. Okay. So what can we do here. If you see M one into M two is coming out to be minus one. So the information is given here that one of the lines is a bisector of the angle between the lines. So one of the slope will be one, one of the slope will be one. Right. So we can say our M two will be how much M two will be minus one and M one will be what our M one will be one. This is given in the question. Okay, it's given in the question. So from here we get the value of M two s minus one. Now I will utilize this equation one. I will utilize this equation one. So from equation one what we can say M one plus M two will be how much one minus one means that will the sum of the slopes will be zero only. The sum of the slopes will be equal to one minus M one minus M square by M. So from here we get what M square is equal to one or M will be equal to plus minus. So this is what we are getting the value of M. Now the question is asking to find cause inverse in cause inverse M. Okay, so it will be cause inverse of one. The second one, cause inverse and cause inverse of what minus. Now when does this cost data gives us one value. It's on the zero degree right. So the value of this will be zero degree and when does cause function gives minus one value. It's at pi cost pi is equal to minus one cost pi is equal to minus one if you see the graph of cost data. It is like this right. So this value is zero. This is pi by two and this is pi at pi the cost function is minus one and at zero So this is what this is a cause inverse of one will be zero and cause inverse of minus one will be five. So this also seems to be multiple correct question. Okay, so this option is correct. So P is in option say so both A and C will be correct pi by two no pi at pi by two it will be zero and three pi by two at three pi by two also it will be zero cost data gives zero value at or multiples of pi by two. So B and A will be incorrect. So correct option for this will be A and option A and C. So yes, let's take the fourth question. What it is saying the bisectors of the angles between the lines given by this equation where C is greater than zero are respectively parallel and perpendicular to the line and four options are given. So bisectors of the angles now. So first let's don't apply much brain on this. What we can do we can just write the equation of the bisectors for this given pair of equation. Given pair of straight line so the equation for this is a exit plus by a square is equal to C times B X minus a Y square. Okay, where C is greater than zero. So now we can open the square and expand it so it will be a square X square plus B square Y square. So this is two ABXY right to ABXY and that will be equal to what C times B square X square plus a square Y square minus two ABXY. Is it okay. Yes, so what we can do further. Okay, our target should be to represent it in the standard form. Okay, so this is what I'm targeting. So let's make it similar to our standard form. So this will be a square minus CB square right. This will be the coefficient of X square, then what will be the coefficient of Y square, it will be B square from this side and from here what we get C times a square. So C into a square C into a square. Yes. And what will be the coefficient of our XY. So coefficient of XY will be our to AB from this side. Okay, to AB from this side and to AB to ABC from this side. To AB plus to AB C to ABC. Is it okay to AB plus to ABC into XY equal to zero. Yes. So we have changed this equation into our standard form. Now what I will do I will write the equation of bisectors. Okay. So bisectors, I will, I'm going to write the equation of bisectors for this. So it will be X square minus Y square upon a minus B, or I can represent it with capital. A minus B is equal to XY upon pitch. Is it okay. Now I will substitute the value of a and B so X square minus Y square upon what is a coefficient of X square so it is a square minus CB square and minus coefficient of Y square. So here it will be minus B square plus C a square C a square is it okay. That will be equal to XY upon H what is H. So this complete thing is if you see this complete thing to H is equals to two times. Okay, we can take here to AB column. So to AB if we take common what we are left with one plus C. So that value of H is what so value of H is coming out to be this AB. Okay, AB into one plus C. Check it, whether I've done any mistake or what. So the coefficient of XY is basically two H. So, from here we got the value of this H as AB into a plus C. Okay, what else we can do whether we can simplify this. Yeah, it seems to. I've enumerate in denominator if you see the here it is a square minus B square a square minus B square. And if you take common C common from these two terms what we were left with a square minus B square. Okay. A square minus B square and that will be equal to XY upon AB into one plus C. Okay, if you take a square minus B square common from these two terms. We will have one plus C. So I'm writing here only. So it will be one plus C. Now if you see this C is greater than zero. So obviously this one plus C will be greater than zero. So we're just cutting this. So, finally, we get the equation of bisector as this X square minus Y square upon a square minus B square. Okay, is equal to XY upon AB. Right. So what can we do. We have to find the bisectors. Okay, this is these are the bisectors. And the bisectors of the angle between the lines are respectively parallel and perpendicular to the line. So this line is parallel and perpendicular to which line we have to identify that. So this is not a single line basically this is the equation of bisectors. So this is combined equation of two lines. So, let me see whether we can simplify it more. So AB X square, then minus of a square minus B square XY, and what minus AB Y square equal to zero. Basically, these two lines are perpendicular because X square coefficient of X square is equal to minus of coefficient of Y square and obviously the equation of bisectors will be at 90 degree means bisectors will be at perpendicular to each other. So now what we can do. We can find the slope for this, like M one plus M two for this if you see any of these are representing two lines. So M one plus M two for this will be what minus two H upon minus two H upon B, and minus two what is H here. H here is a minus of a square minus B square upon two and hold by B B is what minus of AB. So this minus minus gets cancelled out this to two gets cancelled out and we are left with this thing. A square minus B square means M one plus M two, we are getting a square minus B square upon minus AB. Or we can say, or we can say what minus B square upon AB will be B by a and minus a upon B. So this is the value of some of the slopes for these two lines. And what will be the product M one into M two will be obviously minus one because these are perpendicular lines. So if you see M one M two here. And we can find the value of M one and M two right, we can find the value of M one and M two from you. So, what can I do I can take this square M one minus M two square, that will be equal to this guy square B upon a minus a upon B guy square, and what minus of four times of this that will be plus four. Okay. So what we will get, we will get this thing B square. This is correct. Well, you know, AB B square minus. Okay, we have to just simplify it I my target is to find this value of M one and M two, so that then we can see these lines will be perpendicular and parallel to which given lines. So, four times of M one M two minus minus it will be four, and M one plus M two will be what M one plus M two guy square, that is, AB, or we can say B square minus a square, the whole square upon this plus four. Okay, and further what we get. Okay, why I'm finding this. Okay, to find the value of M one and M two. So, a B square if I take LCM. So this will be B square minus a square the whole square, and plus four times of a square B square. Okay. So, if you see, what can we do we are no space. Let me just copy this slide. So from here what we get. This will be B square minus a square, the whole square and so that will be be powerful. Let's write it in separate. So, plus, here from here we will get minus of two a square B square. So, is B square minus a square the whole square. Okay, let me write in this space. M one minus M two couple is square Joe. That will be B square plus a square couple is square upon AB couple is square that will be M one minus M two will be your distinct B square plus a square upon AB with plus minus sign, obviously. And what is the value of M one plus M two M one plus M two is B by a minus a upon B. So here also we can write it as we can write it as what B upon B upon a plus a upon. Right. So this is M one minus M two. And what is M one plus M two B by a minus a by B. So B by a minus of a by B. And here we are getting B by a plus a by B. So, if you see, if we add this, we will have two times M one will be two times of B by a. So M one will be B by a and once we get the value of M one is B by a what will be M two M two will be minus a by B. So, the slope of lines are coming out to be this. Now we have to check through the options. So which line is having the slope. A minus B by a minus a by B. Here we are having slope minus a by B and B by a. So minus B by a. So if you check here, if you see this option a option a is a slope is how much why we can write it as. B B X plus mu right B X plus mu. So our Y will be B by a B by a X plus mu by a. So this B by a, we are having one of the slope as B by a. So yes this one of the line will be parallel to this and the other slope is how much minus a by B. So minus a by B means yes the product of slopes will be minus one. So it seems to be correct if you see this option well. Right. So what I'm saying I'm taking this option a the equation of this is B X minus a Y plus mu B X minus a Y plus mu equal to zero. So here we can write a Y is equal to B X plus mu. So the slope for this line is B upon a B upon a X plus mu upon a this is the complete equation of the line. But what I mean to say slope of this line is B upon a and in our equation one of the lines is having this slope B by a. So this will be parallel to this line and obviously this will be parallel perpendicular to this. So it seems like option A is correct. Okay. Now let's see this option be a X plus B Y plus lambda a X plus B Y plus lambda equal to zero. So from here what we get B Y is equal to minus of a X minus lambda. So our Y we can write it as minus a upon B X minus lambda upon B. So here slope is how much the slope of this line is minus a upon B. Now we are having one slope as minus a by a P. So this line will be parallel to this line. And obviously the other line will be perpendicular to this line. So option B is also correct. Now let's see this option C option C is a X minus B Y a X minus B Y plus mu or plus gamma is equal to zero. Now here B Y is how much a X plus gamma. So why is a upon B a upon B X plus gamma upon B. No, none of the lines having slope a by B. So this will be. Similarly, I think here slope is how much minus B upon a so no we don't have any slope of that minus B by a so option B is also wrong. So I think both these options and B will be correct. So it seems like this is also a multiple choice question. So, yes, this option A and B both will be correct. Option A and B both will be correct and erasing this because we have done on the slide itself. So, yes, let's find the other question. Okay, we will, this one was question number four. So let's take this question number five. So the pairs of the straight lines a x square plus two h x y minus a square equal to zero, and B x square plus two g x y minus b y square equal to zero, be such that each by six the angles between the other, then prove that HG plus AB equal to zero. So first, if you see this first pair of equation, sorry, first equation, it is representing a pair of straight lines. So, okay, I'm just writing the equation first, then maybe talk. So this is a x square plus two h x y and what minus a square is equal to zero. This is one pair of straight lines and the other pair of straight lines is given by this equation. So B x square plus two g x y minus b y square is equal to zero. Okay. It is given that if the pair of history line this and this be such that each bisects the angle between the other. So basically I will, I'm going to write the angle by sector of these pair of straight lines. I'm going to write the angle by sectors of these pair of straight lines, and it will be equal to this equation only. So let me first write the angle by sector for this. So it will be x square minus y square upon a minus b. So a is a here B is minus a so it will be two way, and that will be equal to x y upon h. Okay. So from here if you see what we can get what we are getting h x square minus two a x y. Okay, h x square minus two a x y minus h y square is equal to zero. So this is the angle by sector angle pair of lines. This is also a pair of lines and these are the angle by sectors of these bear of straight lines. So that's per question these two lines are same. These two lines are same. Right. So we can compare it to coefficient of x square y square and x way. So from there what we can write, we can write this be upon each coefficient of x square. It will be equal to coefficient of x y that is 2g here. And what is here, coefficient of x y is minus two and coefficient of y square here it's minus p and coefficient of y square here is minus of h. Okay, so if you see this two and two will get cancelled out. So h g will be. Now I am doing this cross multiplication right. So h g is equals to minus of AB. Is it okay. Or we can say h g plus AB equal to zero. This is what asked in the question. So hence, it's proved. So basically the angle by sector of these pair of straight lines. As per the formula or as per the derivation what we have learned, we have written the angle by sectors for these pair of straight lines. Now we have compared the same line with this pair of straight lines because in the question it's given that both are same. Right. So from there, I have got this condition that h g plus AB equal to zero. And this is what we were asked to prove. So I think everyone got the got the solution and right. So this one was question number five needs. See this next question question number six prove that the lines this are equally inclined to the lines this. Okay, so two equations are given here to second degree equations for motion is second degree equation the first one is to x square plus six x y plus y square equal to zero. This is one equation and the second equation is for x square plus 18 x y plus y square is equal to zero. Now we have to prove that both these pairs of straight lines are equally inclined. Okay, so if we can say, if we can say that the angle by sector for this and the angle by sector for this comes up to be same then we can say both these lines are equally inclined. So I will write the angle by sector for this line angle by sectors for these pair of lines, it will be x square minus y square upon a minus be two minus one, and that will be equal to x y upon h, what is h h here is three. So three times x square minus y square, that is equal to x one. So this is the equation of angle by sector for this pair of straight lines. Is it okay. So I will write the same angle by sectors for these pair of straight lines. So it will be x square minus y square upon a minus be that is four minus what is be here. So here is one, and that will be equal to x y upon h h is how much h here is nine, because to which is it so it will be nine. So here if you see four minus a one is how much it is three basically. So three means three means from here we are getting the equation of angle by sector says three into x square minus five square is equal to x way. Both the angle by sectors are same. We are getting the same angle by sectors for both of these pair of straight lines. It itself prove that both these lines are equally inclined, means this pair of straight lines is equally inclined to this pair of equally inclined to this pair of lines. So given since the equation of angle by sectors are coming out to be same, we can say, therefore the given two pair of straight lines, two pair of straight lines are equally inclined, are equally inclined hence proved. Even two pair of straight lines are equally inclined and hence proved whatever is asking the question. So, yes, this is the solution for this question. Now, moving ahead, we will take question number seven. Okay. So question number seven is saying so that the lines bisecting the angle between the bisectors of the angles between the lines this this are given by. Okay, let me read the question first. So that the lines, okay, bisecting the angle between the bisectors of the angles between the lines. Okay, so first let me write this equation. Let's say x is square plus two h x y plus b y square is equal to zero. Okay, now what we have to write. We have to write the angle bisector for this. Okay, angle bisector. If we write the angle bisectors for this, it will be x square minus y square upon a minus b is equal to x y upon h. Is it okay. And now, so that the lines bisecting the angle between the bisectors of the angles between the lines means what I have to write the angle bisectors of this equation also. So angle bisector for this also. So how can we write, how can we write. We can write it as, first we have to modify this equation. So we can write it as h x square. Okay, and minus of a minus b x y, and minus of h, y square is equal to zero. Now I have to write the angle bisector for this. So the angle bisector for this line will be, let me change the color of the pen, the angle bisector for this will be x square minus y square upon h and minus of b minus a minus b no so a is h here b is minus h so minus h and that will be equal to x y upon h, right. So what is h here. It is minus of a minus b upon two. Okay, so if we can further. modify it, we will have x square minus y square, and minus sign is there, let it be as it is minus into a minus b, and that will be equal to this will be two, two, so two into two for h x y that will be equal to for h x y. So this is what given a minus b. And this minus sign we can write it here, so I can save this minus sign to this side, and x square minus y square into a minus b, and when I will take this thing to the left hand side it will be plus four h it's excellent. Okay, so this is as simple as this. So this is what we have to do. So yes, we have proved it. Basically, this is the, this is the homogeneous second degree equation, we have written the angle bisector of this line, and again we have to write the angle pi sector of the angle bisector. So this is what the question is saying, so that the lines bisecting the angle between the bisectors of the angle between the lines, right. So obviously it will be given by this equation. So we have applied the formula contains nothing else. So yes, we are done with this question. Let's next, let's see the next one, question number eight prove that the bisectors of the angle between the lines this and this are always the same. So I think we have to just write the angle bisector equation for the given pair of planes. So it is a x square plus AC AC x y plus what C y square is equal to zero. This is one pair of straight lines and the next one is three plus one by C. So this is one pair plus x y is also plus x y plus three plus one by a and y square is equal to zero. So let's first write the equation of angle bisector. So I'm writing the angle bisector as this AB. Let's write the angle bisector of these lines. So it will be x square minus y square upon a minus C, because coefficient of x square is a coefficient of B square is C. So that will be equal to x y upon H, what is H to H is AC, so H will be AC upon two. So AC upon two. So this will be the equation of angle bisector. Now let's try to find the angle bisector for this. So it will be x square minus y square upon a minus B that is three plus one by C minus coefficient of y square so minus three minus one by a, and that will be equal to x y upon H, what is H here to H is one so H will be one by two. Now this three and three will get cancelled out, we will have x square minus y square, and we can take NCM here so AC so this will be a minus C, right, and that will be equal to two times of x y. Now, what can we write here we can write x square minus y square upon a minus C, and that will be equal to two x y. Okay, this AC will be also so it will come in the numerator. So obviously both these lines are same only know both these lines if you compare they are same only. So this is what it is asked both these lines are seen. So obviously it is proved. Yeah, the equation will be always in, because this x square minus y square this AC you can put it in the denominator of the right hand side. So two x y upon AC. So yes, it is as simple as that. So this is question number eight. Question number nine. So I think only one question is left in this exercise. So, okay, we will finish it off. Question number nine is there. What is this, the lines represented by x square plus two lambda x y plus two y square equal to zero, and the lines represented by this are equally inclined, then find the value of lambda. Now see these pair of straight lines are equally inclined, right. So, basically the equation of angle by sectors for both these pair of straight lines should be same. This is the concept, which we are going to use you. So this is the first to x square plus two lambda x y plus two y square equal to zero. So, if you see the equation of angle by sector for this, it will be x square minus y square upon what one minus two that is minus one, one minus two, so minus one upon is equal to x y upon H. What is H here to H is equals to two lambda so H is lambda. So, this is for first and for second if you see for second pair of, okay, no need to write one and two. So, I'm writing here only. So one plus lambda into x square minus eight x y plus y square is equal to zero. So, angle by sector equation for this pair of straight line will be x square minus y square upon a minus b that is one plus lambda minus one, and that will be equal to x y upon H. What is H? H will be equal to minus four in this case. So basically these two are same, these two lines, these two lines should be same, means dot line, these two pair of straight lines are same. So, we have to just compare it. So what we can do x square minus y square upon x y, okay, that will be equal to minus one upon lambda. This is our first and here also we will write x square minus y square upon x y and that will be equal to how much this plus one minus one will get cancelled out so basically this will be minus lambda by four. So, on comparing one and two, what we can say on comparing one and two, we can say this minus one upon lambda is equal to minus lambda upon four. So minus lambda my sorry minus minus we will get cancelled out, we will have lambda square is equal to four, or you can say the value of lambda will be plus minus two. So this will be the answer to this question. So we got the value of lambda is plus minus two. So I think this is the last question for this exercise. So, yes, we have completed this third exercise. So, we will meet once again with the next exercise. Until then, ta ta goodbye, take care.