 Hello, good afternoon, greeting from Centrum Academy. Welcome to the YouTube live session on straight lines and circles. So those who have joined in the session, I would request you to type in your names in the chat box. Hello, good afternoon, Purvik, Shruti, Sai. Good afternoon, Rohan. All right, so let's start, it's already four o'clock. So this is the very first question for the day. Okay, so this question is basically picked up from straight lines. If H denotes the arithmetic mean, K denotes the geometric mean of the intercepts made on axis by the lines passing through 1, 1, then H, K lies on. Okay, so there's a line which passes through 1, 1, and H is the arithmetic mean of the intercepts made by the line on the coordinate axis. And K is the geometric mean of the intercepts made by the, geometric mean of the intercepts made by the line on the coordinate axis. Then we have to find the locus of H, K. So again, the rule remains the same. I will wait for at least four people today to answer this question before I start solving it. So unless until I get four responses, I will not start the discussion for the problem. Good afternoon, Shweta, Sondarya. Guys, am I audible to everyone? All right, so I've already got the response from Atmesh. Let's see what others have to say. So Atmesh says option one. Okay, Purvek also goes for one. All right, so Rohan also goes for option number one. All right, guys, so let's discuss this problem. So let's say I assume the line to be X by A plus Y by B equal to one. As we are talking about the intercepts made on the coordinate axis, it would be useful if I consider the line equation to be of the form X by A plus Y by B equal to one, where we all know that A is the intercept of the line on the X axis and B is the intercept of the line on the Y axis. Now, what has been given to me is that K, H is the arithmetic mean of these intercepts, means two H is equal to A plus B. Let's say I call this as the first equation. And K is the geometric mean of A and B, so K square is equal to AB. Okay, so these are the two equations that we get from the given information. And not only that, we have been also been stated that one comma one passes through the line. So let me substitute X and Y both as one, right? So this condition will be true, which implies A plus B by AB is equal to one. That means A plus B is equal to AB, okay? And in order to find the locus of H comma K, I need to get rid of the assumed parameters over here. Here the assumed parameters are A and B, right? So I can replace my A plus B with two H and I can replace my AB with K square. And now generalizing it, we can say that the equation of the desired line will be two X equal to Y square or Y square is equal to two X, which clearly makes option number one as the correct option, right? So almost whoever has answered has answered it correctly. And the first one to answer this correctly was Atmesh. Well done, Atmesh, very good, okay? Guys, simple question to start with. Any question, any concern with respect to the understanding of locus equation over here? All right, great. So let's move on to the next one. So this is the next question for you. So read this carefully, it's slightly lengthy. So we have been given the equation, we have been given the condition that there's a point A, A comma zero and B small B comma zero. And these are fixed-distinct points on the X axis, none of which coincide with the origin, of course. So that means A and B are not equal to zero, okay? And perpendicular to the line AC, the locus of the point of intersection of L and BC, if C varies along the Y axis is, by the way, this was actually given in bracket, so this is not the answer, is means you have to find out which of the following. So let's quickly make a diagram for the same. So when you draw the line AC, okay? So C is basically a variable point. I can call this as zero comma C, okay? And the perpendicular to the line, the question actually doesn't make much sense, but I believe the question says, the line L is perpendicular to AC. So there's a line L, okay? Which has been sketched from the origin, yes, yes. I understand the question is slightly incomplete. So there's a line L drawn from the origin, okay? Which is perpendicular to AC. Now what they're trying to ask is the locus of the intersection of L with BC. That means they're trying to ask the locus of this point. This is H comma K point, okay? So refer to this diagram. So I think the question should be here, coincides with the origin and L passing through origin is perpendicular to the line AC. So as this C varies along the Y axis, what will happen? Of course this perpendicular line will also change its slope and as it changes its slope, it will cut BC at different points, okay? Let me call this point as P. So we have to find the locus of the point P. Hope the question is clear to all of you, all right? So Psy says option, okay? He has retracted his message, no worries. Guys, take your time. I mean, it's very important to get the initial problems right once the momentum is set in. Then things become pretty easy. So here the parameter which is changing is C because it is said that point C is moving along the Y axis. Yes, guys, any response? Okay, so first of all, let us figure out what is the slope of the line AC. The slope of the line AC is minus of C by a right, right? So slope of the line L would be A by C, correct? So equation of the line L would be Y equal to A by CX, correct? And since point P lies on the line L, I can say that K is equal to A by C times H. Let me call it as the first equation, correct? Okay, so Rohan has given Rohan, Psy, Meher, Atmesh. They all go for option number three, okay? So let's see what is the correct answer. Yeah, second thing that we all know that the equation of the line BC, the equation of the line BC can be written as X by B plus Y by C equal to one, correct? And since H comma K satisfies it, since H comma K satisfies this, I can write the expression H by B plus K by C equal to one. Now remember, I don't need C in my expression, right? Because C is a parameter. Whenever we are finding the locus equation, we should strike a relationship between H and K by getting rid of the parameter. So C has to be rid, get rid from here by using the fact that from one, C will be equal to AH by K, okay? So let me replace this over here. So it will become H by B, H by B plus K by C. K by C will be AH and K will go up. That is equal to one. Now multiply with H throughout. So you'll get H square by B plus K square by A is equal to H, right? Now if you generalize this, if you generalize this, you are going to get the answer as X square by B plus Y square by A is equal to X, right? So which of the following options say that? Of course, option number three says that. So three becomes the right answer in this case. I think the first one to answer this correctly was Rohan, okay? So Rohan was the first one to answer this correctly. Again, guys, the moral of the story is if you are studying coordinate geometry, you cannot escape locus. So locus can come in various forms and shapes and you should be all ready to deal with it. Is that fine? Any question with respect to this? Can we move on? Great, so let's move on to the third question of the day. Yeah, so here we have a question. AD, BE and CF are altitudes of a triangle ABC whose vertex A is the point minus four comma five. The coordinates of the point E and F are given to you as four comma one and minus one comma minus four, respectively. Which of the following is the equation of BC? Which of the following is the equation for BC? Again, for the benefit of all, I'll quickly sketch the graph so that let's say I call this foot as D. So what is given to you is the coordinates of A. This is CF, this is BE. So CF, F coordinate is given to us as minus one, minus four and this is four comma one, yes. Now we have to get the equation of BC, okay? Atmesha has already given his response. Anyone else? Guys, first have a roadmap in your mind. What are you going to find if you want to find the equation of BC? Of course, you need a slope and you need at least one point on BC or you can find any two points on BC. Let's say if I ask you, can we get the equation of CF point, CF line? For that, I would need the slope of AB, correct? So we know the slope of AB is y2 minus y1, that's minus four, minus five by x2 minus x1, that's minus one plus four. That's going to be minus nine by three, that's minus of three. So the slope of CF, so the slope of CF will be one third. Correct? So the equation of CF line would be nothing but y plus four is equal to one third x plus one. Correct? Okay? Which is nothing but three y plus 12 is equal to x plus one. Okay, so three y minus x plus 11 equal to zero. This is the equation of CF line. Okay? Can I also get the equation of BE? Of course we can get the equation of BE because BE will have a slope which is negative reciprocal of slope of AE. So slope of AE again will be y2 minus y1 by x2 minus x1. That's going to be minus four by eight which is minus of half, okay? So slope of BE is going to be two. So equation of BE would be y minus one is equal to twice x minus four. That is y minus two x plus seven is equal to zero, right? Now, if I solve these two simultaneously, will I be able to get the coordinates of the orthocenter? If yes, please tell me the coordinates of the orthocenter, those who have solved it. So you just have to solve these two simultaneously, okay? So let me multiply this with two. So it'll be six y minus two x plus 22 equal to zero and we have already y minus two x plus seven equal to zero. Let's subtract it. So five y plus 15 is equal to zero. So y is equal to minus of three, okay? And if y is minus of three, putting it in this one, I will get x as two, x will be two. So orthocenter point is, this point is two comma minus three. So h is two comma minus three, correct, right? Now, if I simultaneously, if I get this orthocenter point, okay, can we get the slope of AH? Slope of AH, slope of AH will be again, y two minus y one by x two minus x one, okay? So that will be again minus four by eight, which is minus of half. One second, let me just, oh, I'm so sorry, I'm so sorry, my mistake. Slope of AH will be, yeah, five plus three, and minus four minus two, which is eight by minus six, which is minus of four by three. That means the slope of BC should be three by four, right? Now, we just have to see which of the following has three by four. I think both one and three qualify. So of course, two and four cannot be your answer. Two and four cannot be your answer, correct? Can we somehow find the coordinates of C? Okay, to find the coordinate of C, we just have to simultaneously solve the equation of the line AE, which I think I have already found out, the equation of the line AE, which is y minus two x plus seven equal to zero, and equation of the line FH, the line connecting F and H, okay? So the line connecting F and H, I need to make some space for myself. It's too cluttered now, yeah. Equation of FH will be minus four, y two minus y one by x two minus x one. That's going to be minus one by minus three, which is one by three. So it's one third, x minus two is equal to y plus three. So three y plus nine is equal to x minus two, okay? Now we have to solve these two equations simultaneously. If we solve these two equations simultaneously, I will get the coordinates for the point C, okay? Oh yes, equation of AE. Oh, we found B not AE, okay, so my mistake. So this will be not be the equation of AE. So equation of AE would be the slope would be five minus one, which is four by minus eight, which is minus half. So minus half x minus four is equal to y minus one, okay? So that's going to be, this is going to be minus two. So two minus two y is equal to x minus four, yes. So that's going to be, that's going to be x plus two y minus six equal to zero. Yeah, now we have to solve these two simultaneously. So when you solve this, you will get the coordinates of C. We can get the coordinates of C as eight comma minus one. Right, so eight comma minus one would be the coordinates of C. So once I know the slope of the line and one of the points on it, we can always write down the slope point form of the equation of the line. Now what I will do is, since I have already recognized either of option one and three would be the answer, I would quickly check which of them satisfies eight comma minus one, okay? So let's check over here. So 24 plus four minus 28 equal to zero, yes, it satisfies. So your first option becomes the right option. That's option number one is correct. Again, this question was simple, but it actually made you find a lot of things. So we had to find, first of all, equation of the two altitudes. Then we had to see the point of intersection of these two altitudes, which is at H, right? Then we have to find the equation, the meeting point of FH line and AE line, which is point C. And once I know C, and I know the slope of BC, then only I could write down the equation of the line, okay? So easy question though, but yes, was bit of time consuming. This is what I call as calculation intensive questions, okay? So guys, hope this was clear. So we can move on to the next question. So here is the question. If let P and Q be any two points on the lines represented by two X minus three Y equal to zero and two X plus three Y equal to zero respectively. If the area of the triangle OPQ, where always the origin is five, then which of the following is not possible, right? Read the question very carefully. Which of the following is not the possible equation of locus of midpoint of PQ? So which of them is not a possible locus equation of the midpoint of PQ? So I'm just trying the figure for you. So P and Q are two points on these lines, two X minus three Y equal to zero and two X plus three Y equal to zero. And it is such that the area of the triangle OPQ always remains five, okay? So P and Q are varying in such a way that area of the triangle OPQ always remains five. Area of the triangle OPQ always remains five units. Then you have to find the locus of the midpoint of, you have to find the locus of the midpoint M of P and Q, which I'm calling as H comma K. Yeah, that's what Rohan, it is asking you which of the following is not the possible equation of the locus. So ideally you should be getting two answers, right? So I'm sure two of them would be there in the options and the third one would be then your actual answer. Are you getting the point? So B is possible, you're saying, B is a possible locus, right? So you're saying A and B are possible locus. So C would be your answer because the question is, which of the following is not the possible equation of the locus? Okay, so I'm getting answers from Purve, Quisist, Sondarya, all of them go with the third option. Okay, guys, it's pretty simple, you know? I think Rohan also would have got it, but I think he's not understood the question properly. See, let's say the point P over here is A comma two A by three, right? It's better to choose a parametric point if you want to choose a point on this line. Similarly, I could choose Q as B comma minus two B by three. Okay, so now what would be the area of the triangle OPQ, that would be half determinant of the points. Let's say I take the origin first, zero, zero, one. X one, Y one, one, X two, Y two, one. Okay, this is equal to five, correct? Now, if you clearly simplify this determinant, you would be getting something like the expansion of this determinant with respect to the first row, expand this with respect to the first row. So you'll be getting minus two AB by three, two times is equal to 10, mod of it is equal to 10 because we're dealing with the area of the triangle which has to be a positive quantity, correct? So from here, I can say four AB could be written as plus minus 30, okay? Let me call this as the first equation, four AB is equal to plus minus 30. Now, what would be H? H would be nothing but the midpoint of, sorry, the arithmetic mean of B and A, so which is H plus B, so two H is equal to A plus B, okay? Similarly, two K would be equal to two A minus two B by three, correct? In other words, I can say A minus B is three K, right? Now, from these two equations, let me call it as second equation and third equation. I can say four AB could be written as A plus B the whole square minus A minus B the whole square, correct? That means plus minus 30 could be written as A plus B the whole square which is nothing but four H square and A minus B the whole square is nothing but nine K square. So if I generalize this, I will get four X square minus nine Y square is equal to plus minus 30, right? I can clearly see that option number one and option number two both satisfy this but I don't want the right option. I want the option which doesn't satisfy it. That means your third option becomes the right option, okay? So option number three here becomes correct option because they're asking you which is not the possible equation of the locus. Is that fine? Especially Rohan, is that clear to you? Okay, any doubt, any question with respect to this? All right, so let me make a move on to the next question now which is the fifth question of the day. So the question says if A and B are two points on the line three X plus four Y plus 15 equal to zero such that OA is equal to OB is equal to nine units then the area of the triangle OAB then the area of the triangle OAB has to be found out. Again, guys, please be accurate as far as possible. Don't be in a hurry to answer. Always in coordinate geometry have the habit of drawing the problem. Just catch the problem on the coordinate accesses. So let's say this is the situation where this line equation is three X plus four Y plus 15 is equal to zero. Then you have to find out OA and OB both are 99 units, right? So we have to find out the area of the triangle OAB. Okay, so Rohan is saying option number two. Okay, Rohan, I acknowledge your response. What about others? I think this problem is to be solved with geometry rather than using coordinate geometry. What you require here is the use of geometry. So think in that line. It's not a difficult problem. Yes, so far only Rohan has been able to answer this. What about others? Guys, I'm expecting this to be very, very easy question. Okay, Sai also says option two is correct. Others, Visist, Sondarya, Sanjana, Shweta, Shruti, Atmaish, Purvik. Okay, so even Sondarya says option two is correct. Now guys, very simple. As you can all see that triangle, triangle OAB is an isosceles triangle, right? This is an isosceles triangle. Isosceles triangle means if I drop a perpendicular from O on AB, it is going to even bisect AB. That means this two will be equal, correct? Let me call it as M, okay? So we can all find the distance of origin from the line OM. We all know the distance of a point from a line. That's going to be mod 15 by under root of three square plus four square. That's going to be five by 15 by five, which is three. So this is going to be three, correct? So AM could be found out by using the Pythagoras theorem. AM would be under root of nine square minus three square. That's under root of 72, correct? Under root of 72 is six root two, correct? So if AM is six root two, AB would be 12 root two, right? Because AB is double the length of AM, right? So once we know AB and once we know OM, all I need to do is use the area of the triangle formula as half base into height. So half three into 12 root two. That's going to be 18 root two units or square units. That's option number two is correct, right? Should not have taken you more than, I would say 50 seconds to solve this. So it was an easy question. See, that's what happens when we go prepared with all difficult types of questions and let's say the examiner throws at us an easy question. We start doubting the simplicity of the question and we start thinking in various directions, we missing out the obvious things, okay? Great. So can we move on to the next one? Let's move on to the sixth problem of the day. So here is a problem which says the number of points on the line three X plus four Y equal to five, which are at a distance of sequence square theta plus two cosecans square theta, where theta is any real number from the point one comma three, from the point one comma three. This question is framed in a very tricky manner. I mean, it may confuse a lot of you. So try to understand the requirement of the question. The distance is sequence square theta plus two cosecans square theta, where theta could be any real number. So it has been framed in a very sly fashion. Any idea anyone? Guys, I'm assuming that everybody is trying. Once you're done, please post in your response in the chat box. Yeah, you have to play with the limitation of this entire expression. So what is the limitation that this expression, that means what is the maximum value that this can have or something to do with the limiting case, correct? Say you are correct. Your thought process is aligned to what is required to solve this problem. Yeah, sure, take your time. No, for a particular theta, let's say if you choose a particular theta, how many such points could be there? Now I understand if you vary your theta, you will get different different points. But let's say you decide to take a particular value of theta, how many such points could you find? So treat theta like a parameter. So for every parameter, how many points will exist? That's the question. See, it could be either zero, it could be either one or it could be two, right? So your answer cannot be infinite. So fourth option is ruled out. See guys, I think you did not understand the principle behind this question. The question is trying to say that if I choose, let's say this point, okay? Let's say point P. Is distance from this point? Let me call this point as Q, okay? So this could be expressed as secant square theta plus two cosecant square theta for some theta, for a particular theta, okay? Now, what I'm trying to say is that, let's say if there is no theta existing for which you could express this distance, then there would be actually no point, the answer would be zero in that case. Okay, Shweta is saying option two. See guys, let's try to understand this question first before we comment on the answer. See, what is the minimum distance of three comma one from this line? The minimum distance of three comma one from this line will be three into one plus four into three minus five mod by five, correct? That's going to be three plus 12, which is 15, 15 minus five, 10, 10 by five, which is two, okay? Now, if this expression secant square theta, I'm saying if, I'm not saying, this will always happen. If I'm saying this is less than two, right? What does it imply? That means there will be no point, okay? If secant square theta plus two cosecant square theta is always two, then there will be exactly one point. And if secant square theta, please note I'm writing if, okay? If secant square theta plus cosecant square theta is greater than two, then there will be two points possible, okay? So these three can be answered no point, one point or two point, okay? Now, hold on, let me finish off this. So if you see this expression, secant square theta plus two cosecant square theta, okay? Let me do a simple activity. Let me try to complete squares over here. I can write this as one plus, this I can write it as tan square theta and this I can write it as two plus two cot square theta. That is nothing but three plus tan square theta plus two cot square theta, which actually I can write it as three plus tan theta, I can write minus root two cot theta the whole square plus two root two, correct? Now, this expression is always three plus, let me write it with white ink, three plus two root two plus some perfect square, plus some perfect square. Now, this will always be greater than equal to zero, right? Means this expression will always be greater than equal to three plus two root two, right? Which is clearly greater than two, right? So this distance, which is the perpendicular distance is two and this happens to be always greater than, this distance is always greater than, this distance is always greater than three plus two root two. That means there could be only two points, let's say P and M I call it, for which you can have this as the distance from this line one comma three. So option two becomes correct in this case. So there will be only two such points. So any theta if you take, they will always be two points possible, right? So every theta that you can take or that belongs to real number, you realize two, two, two, two points will come out. Only two points can come out for a particular theta. That's what I want to convey over here, right? Is this okay? So this question was actually paying on your understanding of the maximum value or the minimum value for that matter. Here it's the minimum value of this expression. So minimum value of this was three plus two root two, which is greater than two. That means it'll always have two such points on either side of the shortest distance, okay? No question with respect to this. So let's move on to the seventh question now. So seventh question says A, B, C is an equilateral triangle whose centroid is origin. Base B, C is along this line. Then which of the following option is correct? Area of the triangle is numerically equal to perimeter. Area of the triangle is numerically double the perimeter. Area of the triangle is numerically three times the perimeter and area of the triangle is numerically half the perimeter. It's an equilateral triangle, A, B, C. Looked like an equilateral one. So just let me quickly plot an equilateral one. So this is the centroid, which is actually origin. Now from this diagram, I think things will be pretty clear. You should be able to solve it pretty fast through the help of this diagram, okay? So Sanderia says option one, area of the triangle is numerically equal to the perimeter. Fine, what about others? Do you feel the same? Okay, which is also back set up with option one. Guys, I need one more response. Okay, so let's discuss this. So let us say the height of this equilateral triangle, let's say AM is H. Let's say AM is equal to H, okay? So I can say this is, I know the distance of the centroid from the vertex to the distance of the centroid from the opposite side is in the ratio of two is to one, right? So this is in the ratio of two is to one, correct? So this distance, let me call it as GM. GM will be equal to H by three. And AG will be equal to two H by three, correct? Now, what will be GM? GM is the distance of the origin from this line. We all know the distance of any point from the line. So it'll be 122 mod by under root of 11 square per 60 square. By the way, very less people know that 11 square per 60 square will form 61 square, okay? They form a Pythagorean triplet. So 122 by 61, that is going to be two. So this is two units, correct? So if this is two units, that is the distance of, this is equal to two units, that means H is going to be six units. So entire AM is going to be six units, correct? So if I know the altitude is six units, what about east side? We know that, let's say I call this as A. We know H by A, H by A is equal to sine of 60 degree, right? So A can be written as two H by root three, right? So I can say area of an equilateral triangle, area of an equilateral triangle is going to be root three A square. If I'm not wrong, A square will be this by four. So four, four gets canceled. This will become H square by root three. So H square by root three will be equal to 36 by root three, right? Okay. Now what about the perimeter? Perimeter is going to be three A. Perimeter is going to be three A. So three into A, A is two H by root three, right? So it's going to be six H by root three, which is again 36 by root three. That means numerically speaking, the area of the triangle is equal to the perimeter of this equilateral triangle. So option number A becomes the right option in this case. So the first one to answer this correctly was Sondarya, well done Sondarya, very good. Okay guys, this was simple. I don't see a point, you know, not being able to do this problem. Fine, no questions with respect to this. Can I move on? All right. So we'll move on to the next question now, which is question number eight for the day. If the distance of a given point alpha comma beta from each of the two straight lines, Y equal to MX, okay? Through the origin is D. Then fine, I think this should have been a gamma. Where does gamma come in from? This should have been Y, I guess, not gamma. Yeah, again, sorry for the typo. Let me write this again. AY minus beta X, in fact, alpha Y minus beta X whole square is equal to which of the following, okay? Now, why they're saying two straight lines? Just they have given one straight line is because they probably feel that there are two lines of this nature, Y equal to MX, whose distance from the origin, see if the distance of the given point alpha beta from each of the two points through origin is D. So the distance from this point is D for two such lines. So there'll be two such lines like this of this form to be more precise. So it's important to understand the question first. So there are two lines through origin of the form Y equal to MX, okay? Whose distance from alpha comma beta point is D. So you need to find alpha Y minus beta X whole square expression is which of the following. Is the question clear to all of you? So M is like a parameter, even to be very frank over here. So this is something with the question is indirectly telling you at boss treat M as a parameter because M can change, M is a parameter. Okay, Purvik says option two. Okay, Vishesh also says option two. One more response I need before I start solving this problem. Okay, so this is alpha comma beta point and this distance perpendicular distances from this point from these two lines is always D, okay? So the very first thing that we can actually state from this given condition is that beta minus M alpha by under root of one plus M square mod is equal to D which is nothing but beta minus M alpha at square both the sides. First let me take the under root one plus M square on the other sides and squaring it. This becomes this, okay? Now, since M is a parameter, M always changes but M will always be this, correct? So let me replace, let me replace M with Y by X, okay? So it becomes beta minus Y by X alpha square is equal to D square one plus Y square by X square, okay? Multiply throughout with X square you will find X beta minus alpha Y whole square is equal to D square X square plus Y square which is actually the same as saying alpha Y minus beta X the whole square is D square X square plus Y square which clearly option number two is correct. So I think the first one to answer this was Purvik. So Purvik and then Vishist was the, they're the only two people who could answer this question. Okay, so guys, again, I understand the question was actually slightly difficult to understand but once you know what to do, it was a simple process. All right, so we'll move on to the next question which is question number nine for the day. So read this question carefully. So the question says the base BC of a triangle ABC is bisected at the point P comma Q and the equations to the sides AB and ACR, PX plus QY equal to one and QX plus PY equal to one. Find the equation of the median through A. Okay, seems to be an easy question. Guys only handful of you are answering every time I want everybody to participate. It's you versus you, nobody else. Yeah, we have to find out the equation of the median through A. So a point is known, okay? P comma Q. Now, only thing that is missing is the slope. If at all you are taking that route of using the slope and the point to get the equation of the line AM. Okay, so Purvik says the third option which is the most ugliest of all. Okay, Purvik, was that a hunch or like you actually solved it? All right, so Rohan also follows the suit. He also says option three. Okay, so Andhari also backs it up. Okay, very good. So guys, of course, we cannot solve it by using the fact that, oh, you actually got it, very good then. So we cannot solve it by using the fact that we know a point and we should find the slope. Here we use the concept of family of lines. Right, here we can use the concept of family of lines. So as you can see, AM is a family member, right? Which is produced by the intersection of A, B and A, C lines. So I can say let AM be a PX plus QY minus one plus lambda times QX plus PY minus one equal to zero. Okay, now how do I get this lambda? Very simple, we know that P comma Q lies on AM. So P comma Q will satisfy this equation. So let me substitute X as P and Y as Q. So that will give me P square plus Q square minus one plus lambda times PQ plus PQ, that's two PQ minus one is equal to zero, which implies, I can say lambda is equal to minus of P square plus Q square minus one by two PQ minus one, right? Now, let us substitute this lambda over here. So when I substitute this lambda, you can all see, I'm not writing all the steps because you all are smart enough to know what I'm trying to do. So I'm putting lambda over here, so that will become QX plus PY minus one times P square plus Q square minus one. And yes, option number three clearly satisfies this. So the first person to answer this question again was Purvik, well done, good. Okay, guys, please work on your accuracy first. Accuracy is the priority for you. Okay, so no question with respect to this. So let's move on to the next one now. Let me raise this and let me take you to question number 10 now. All right, so read this question carefully. Seems to be a lengthy question. Yeah, so the question says, there's a point A, X1, Y1, B, X2, Y2, C, X3, Y3 are three vertices of a triangle ABC, LX plus MY plus N equal to zero is the equation of the line L. Okay, if the centroid of the triangle ABC is at the origin and the algebraic sum of the lengths of the perpendicular from the vertices of the triangle ABC on the line L is equal to one, then the sum of the squares of the reciprocals of the intercept made by L on the coordinate axis. Oh, wow, it's a lengthy question. So please slowly try to figure it out through making a proper diagram. If you feel it is necessary. Yes, anyone? Any idea? Okay, so Purvik says option two. Okay, so Sai says option three. I think Purvik has retracted his message. Okay, Sai is going with option number three. So does Atmesh? All right, so three people have already responded. So let us first look into the requirement of the question. Read the last line carefully. We need the sum of the squares of the reciprocal of the intercepts made by the line LX plus MY plus N with the coordinate accesses. Okay, so what is the X intercept? X intercept is when you make your Y expression as zero X intercept will be, okay, minus N by L. Okay, what would be your Y intercept? Y intercept is when you put your X as zero, that's going to be minus N by M, right? So what is required is L square by N square plus M square by N square. That means I need L square plus M square by N square expression. Okay, and this is what is required, okay? Now, what is given to me, let's read one by one. First of all, the centroid is at the origin. That means X1 plus X2 plus X3 is zero and so is Y1 plus Y2 plus Y3 because then only the centroid will be at the origin, right? That's the first set of information. Second set of information is the algebraic sum of the lens of the perpendicular from the vertices, okay? On this line, L is equal to one, right? That means LX1 plus MY plus N by under root of L square plus M square. Now, when I say algebraic sum, I would not account for mod. Algebraic sum is the one which will sign. It can be negative as well, okay? So don't take the mod over here because then only it will be referring to the algebraic sum, correct? So this is given to you as one. Correct? So if I take the denominators on the other side and take L common, so I can see LX1 plus X2 plus X3 plus MY1 plus Y2 plus Y3 plus 3N is equal to under root of L square plus M square, correct? And we know that these two terms would be zero because your X1 plus X2 plus X3 is zero and so is Y1 plus Y2 plus Y3, right? So which will make this as L square plus M square as nine N square if I square both the sides because I'm trying to get to this expression, correct? So if I divide by N square, it's very obvious that my answer will be nine which is option number three is correct. So the first person to answer this correctly was Psi Meher, well done Psi, very good. Is there any question, any concern, any doubt, any clarification that you need with respect to this solution? Please type it in your chat box. If no, then we can move on to the 11th question for the day. So we have a question now in front of your screen which says a straight line passes through the intersection of X minus two Y minus two equal to nine and two X minus BY minus six equal to zero and the origin, okay? Passes through the intersection of these two lines as well as origin, okay? Then the complete set of values of B for which the acute angle between this line and Y equal to zero is less than 45 degrees is less than 45 degrees. Y equal to zero is the X axis, okay? So don't get confused, Y equal to zero is the X axis and slope should be somewhere between minus one to one. That's what it means when it says the slope is less than, sorry, the angle of inclination is less than 45 degree. So this angle is less than 45 degrees, lesser than 45 degree. This is Y equal to zero line. Yeah, they're correct. X minus two Y minus two equal to, let me check why they have given minus two and zero. Oh, I'm really sorry, guys, I'm really sorry this has to be zero. So I'm rewriting this X minus two Y minus two is equal to zero. I think again a typo error. I'm really sorry, guys, please make that change. Yeah, I'm really sorry, that's X minus two Y minus two is equal to zero, not nine. So I was wondering why they have given two constants. Okay, so Atme says option three, okay? So I'm getting wishes, Atme says option two now, then wishes four, Sime hit two, Atme is two, Sondariya four. Okay, guys, let's discuss this. Of course, when they talk about there's a straight line passing through intersection of two lines, the first concept that comes in our mind is the concept of family of straight lines. So I'll say let the desired line be this. And since this passes through origin, that means origin should satisfy this. Okay, origin must satisfy this equation. So minus two, minus six lambda is equal to zero. So lambda is minus one by three. Right, correct? So if I put the value over here, I get minus one by three means three times x minus two y minus two minus one time, two x minus b y minus six equal to zero, okay? So let us group up x terms together. So x, y terms together, so I will have b minus six, and I have minus six plus six, which is going to be zero, okay? Now, this line has a slope of minus one by b minus six, correct, which is actually one by six minus b, okay? Now, this slope, one by six minus b, should lie between minus one to one, right? Which is like getting two inequalities from here, that is six minus b is greater than one. Okay, let me write it like this. Minus one is greater than one by six minus b, and one by six minus b is less than one, okay? Yeah, so I need to solve this inequality now. So let me clear up some space because I need some space to write on. We clarify this, in fact, let me raise this diagram as well, yeah. So one by six minus b plus one is greater than zero, which means six minus b, one plus that will be seven minus b would be greater than zero, or you can write this as b minus seven by b minus six is greater than zero, okay? So this means b should be either greater than seven or b should be less than six. So using the other one now, one by six, which is the ink, one by six minus b, minus one is less than zero, which means b minus five by six minus b is less than zero, which means b minus five by b minus six is greater than zero, which means b is greater than six or b is less than five, okay? Now, when I combine these two conditions on a number line, let's see, what do I get? So I have five, six, seven. One says it should be less than six and greater than seven. Other says it should be greater than six and less than five, okay? So overlap is happening, let me raise this part. So overlap is happening where they say x should, sorry, b should be less than five or b should be greater than seven, okay? Now, many of the students would actually mark option two correct, but the actual answer, which is right is option number four. This is your right option. Why this is the right option? Because I have to exclude four from my interval because when your b is four, these two lines will actually become parallel lines. They will never intersect them. Then there is no point of intersection happening, isn't it? See, what will happen? One equation will become, one equation will become x minus two y minus two equal to zero. Another equation will become two x minus four y minus six equal to zero. That's actually two parallel lines. And if it is two parallel lines, they will never intersect. So you have to remove the value of four from the interval. That means option four is correct. And the first person to answer this correctly was Vishisht. So mostly all of you, and of course, Sondaria was another one to give the right answer, but mostly all of you got carried away by the presence of option number two, okay? But option number four is correct in this case. Again, this was a googly question. Is that fine guys? Any question, any concerns with respect to this? Okay. So now let's move on to the next question now, question number 12. So the question reads like this. The locus of the foot of the perpendicular from the origin on each member of this family is which of the following options? Again, locus-based questions. So remember here A is a parameter, which is actually changing and thereby generating different members of the family. So A is playing the same role as that lambda used to play while we used to write down the equation of a family of lines, okay? Rohan says option four. Guys, there are various ways to solve it. You may take a longer route, you may take a shorter route, it's your call. Yes guys, I'm waiting for a response, only Rohan has responded so far with option four. Vishis Purvik, Sai, Atmesh, Sondarya, Shweta, Sanjana. All right, so guys, if I were you, I would have just taken one of the family members, like I would have just put A as zero, okay? If I take A as zero, I get a three X minus Y, minus one equal to zero, right? That's a line like this. Okay, okay. I'm sorry, it has to go down, okay? Now, what I would do is I would now sketch up perpendicular from the origin onto this particular line, okay? So we know that this line slope is three, so this line will have a slope of minus one by three. So Y is equal to minus one by three X will be this line, okay? So now we'll solve them simultaneously. That means I would solve one third X is equal to three X minus one, that is one is equal to 10 by three X, okay? So X is equal to three by 10, and Y is equal to, if X is three by 10, Y is equal to minus one by 10, correct? Okay, so I would see which of the options satisfy three by 10 comma minus one by 10, okay? Let me put in the, since many of you are saying fourth, let me first put in the fourth one. So two X, which is six X, six by 10 minus one square, plus four times Y minus one is minus 11 by 10 square. Let's see what is the value that I get from here. I'll get minus 10, which is minus four, it is 16 by 100 plus this will give me 121 into four, which is 484 by 100, okay? So let's see what does this give me? It gives me 500 by 100, which is actually five, yes. So four satisfies this particular point, that is three comma 10 and minus one comma 10. So option four is the right answer, okay? So if I were you, I would have solved this question like this, but if I were me, that means as a teacher, if I have to tell you the concept, then I would solve it in the normal way. So let us try to understand how we can solve this in normal way. Is this method clear, guys? I mean, do you need any explanation with respect to the solution? If yes, please ping me, else I'll be clearing off the screen and giving you the normal approach to solve this problem, right? No doubt, everything is fine. Clear, is it clear? Just type CLR if it is clear. All right, so I'll give you the normal approach to solve this problem. So see, guys, when you talk about a family of lines as I told you, you just need to get the parent lines from it. That means I just separate out the A from each other. So I can see four X, A times four X and you will have minus Y and minus two as one of the lines plus three X minus Y plus one equal to zero as another line. So these are the two parent lines which are making this particular line. So three X minus Y, oh my bad, there's a minus one over here. Okay, so these are the two lines which are making this line, right? Now, first of all, I'll see what is the intersection of these two lines. So where do these two lines intersect? So let's subtract them. When you subtract them, you get X minus one equal to zero. So X is one and if X is one, Y is going to be two. Okay, so let's say this is a family member and this family member will have a point one comma two on it. Right? Okay, so let's say the foot of the perpendicular here is H comma K. Okay, now I can say that the slope of the slope of let's say OP and the slope of OQ would multiply to give you minus of one, correct? So what is the slope of OP? Slope of OP is K by H, right? And slope of PQ will be K minus two by H minus one and this should be equal to minus of one. So in plain and simple words, we actually got the condition for the locus that is H H minus one plus K K minus two is equal to a zero, right? Okay, now if you see this term, this is H square minus H and this is K square minus two K equal to zero, correct? So what I'll try to do next is I will try to complete the squares. So I'm just, let me write on top over here. So if I complete the squares, I will write it as H minus half the square minus one fourth plus K minus one whole square minus one equal to zero. That means two H minus one whole square plus four times K minus one whole square is going to be five. That is nothing but option number four again. So this is a longer route to solve the problem, but this helps you to understand the concept okay, in a better way. Is that fine guys? Guys, is this fine? Is this approach fine to all of you? Can I move on to the 13th problem? Okay, is it fine if I don't give you a break? Just type F-I-N-E, if it is fine if I don't give you a break, nobody's writing fine. Okay, if you want a break, type B-R-E-A-K break. If you want a break, oh, mostly people are writing fine. Great, that's the spirit. So let's have the 13th question. All right, so 13th question reads like this. If the line passing through one comma two making an angle of 45 degree with the x-axis in the positive direction meets the pair of straight lines x square plus four xy plus y square equal to zero at A and B find the product of P and PB. So this has been picked up from the concept of pair of straight lines. Yes, guys, any response? Sure, sure, take your time. Okay, so, Atmesh is getting some other answer which is not there in the options. What about others? Do you feel the same? So I've just drawn the figure for you on your screen. So Atmesh, let us see if somebody is able to find one of the options as correct. Else, let me verify whether there's no option correct. Oh, you already identified your own step, okay, great. So please work on your mistakes and see if any one of the option is correct. Guys, here, the distance form of the equation of a line will be helpful because they're talking about P A into PB. So let's say this is R one, this is R two. This is R two. Okay, so Purvik and both Saim here think option number three is correct. Freeing planes at the end of the day for the non-confusion on that one if it's a six or it's a seven, right? So again, going towards a little bit more to the point of understanding today, we are today at the one line, stuff showed that this chapter we study the effect of electrician. Okay, so try using the distance form of a line. I'm sure that will be helpful to you in solving this problem. I hope all of you remember the distance form of a line, right, where if you know the angle of inclination of the line and you know one of the points on the line, you can actually get to any point if you know the distance from that given fixed point, okay? So let's say I assume the equation of the line to be of this form X minus one by cos of theta is equal to Y minus two by sine of theta, okay? And this is actually equal to R, right? We all know the distance form of a line. So if one comma two is this point and I want to get to any point X comma Y whose distance from this fixed point is R, we follow the distance form of a line, right? So I can write the parametric equation of any point as one plus R by root two. So this is one by root two essentially and Y is equal to two plus R by root two, right? Now, so as a point, I got one plus R by root two and two plus R by root two as the points, as a parametric representation of any point lying on this given line. Now, this point should actually satisfy this, okay? So this point should actually satisfy this. So X square means one plus R by root two square plus four XY, XY is this, okay? Plus Y square will be two plus R by root two square and this should be equal to zero, right? Now, this would leave me with a quadratic in R. So let me figure out what is that quadratic in R by simplifying this stuff. So I'll write this as R plus root two, the square, I'll get a two in the denominator and here I can write it as R plus root two, R plus two root two and this is again R plus two root two, square equal to zero, okay? Okay, so R square, four R square and again R square would leave me with, if I'm not wrong, it will leave me with, I think it's going to give me six R square, right? Okay, now the terms containing R, the terms containing R would be, from here I get two root two, I'm just collecting terms and just hand picking them, okay, without actually writing them. So I'll get two root two from here and I'll get two root two plus two root two, which is three root two, three root two and this is 12 root two and from here I will get four root two, correct? So that makes it 18 root two R, if I'm not wrong and the constant terms if I see, constant terms would be a two from here, it will be four into four 16 from here and eight from here, that's going to be 36 if I'm not wrong, sorry, 26, 26. So this will be 26 equal to zero, correct? And if I want R one into R two, which means I'm looking for the product of roots in this. So R one into R two is the product of roots, which is actually 26 by six. So 26 by six is going to be 13 by three, if I'm not wrong, which is option number one is correct. Guys, I'm actually surprised that none of you got this option correct, okay? Okay, correct? So this could be solved so easily by using a concept called the distance form of a line. Guys, don't underestimate this form in equation of a line because it helps you in finding answers to those situations where distance from a point is required, right? So this is like a learning for you, just like we had a session yesterday which where we recall that we had forgotten so many things like conditional identity, et cetera. Again, this form has to be recalled and reused in problems. Okay, no problem. We have another question coming up. So I hope there's no question with respect to this. Are we good to go to the next question? All right, great. So we'll move on to the 14th question now. Again, coming from the pair of straight lines, I believe. So we have a question over here. Let y equal to x line is median to the line, a median to the triangle, OAB, where O is the origin, okay? Equation AX square plus 2HXY plus BY square equal to zero represents combined equation of OA and OB. A and B lie in the ordinate X equal to three. And if the slope of OA is twice the slope of OB, then the greatest possible value of, greatest possible value of, so read the question, greatest possible value of A plus 2H plus B is. I want you to know that, so A is the line, okay? So I could have used a picture, it's still here. Because I didn't know about it, you know, if now it comes, you want to draw it by a line. So the simple thing, you go down the slide over here, A lines move from, you know, outside straight. And when you come out of it, they go from south to north. So you can see here, you want to match, the lines are intersecting, you know. Sure. Yes, guys, any response? So let me just quickly sketch this, so that things are clear in your mind. So there's a triangle OAB, okay? A and B lie on the ordinate X equal to three. So basically, there is a line X equal to three. So this is your line X equal to three. And A and B are on this line, okay? Slope of OAB is twice the slope of OB. Slope of OAB is twice the slope of OB. Slope of OAB is twice the slope of OB. Okay, so this is your A and this is B, correct? And we have Y equal to X as the median. So this point is going to be definitely the line Y equal to X. That's going to be three comma three if I'm not wrong. So these two are equal. Yeah, ordinate is basically a line which is parallel to the Y axis. Lie on the ordinate means it's a line, okay? So Y, a line parallel to the Y axis is also sometimes called the ordinate. Of course, Y coordinate is also sometimes called ordinate. Epsis and ordinate, okay? Now, let's say, since this is the midpoint, let me choose this point as three comma three plus alpha and this has three comma three minus alpha. So alpha up and alpha down I chose, okay? So what is the slope of, what is the slope of OAB? Slope of OAB will be equal to three plus alpha by three and slope of OB is three minus alpha by three, okay? Now we know that three plus alpha by three, okay? And this slope is twice of three minus alpha by three. That means three plus alpha is equal to six minus two alpha. That means three alpha is equal to six, sorry, three alpha is equal to three, correct? Which means alpha is equal to one, okay? That means this point is three comma four and this point is three comma two, okay? Now, what is exactly the equation of OA? OA equation will be, if I'm not wrong, four x minus three y equal to zero and OB equation will be two x minus three y equal to zero. Correct? Now, if I multiply OA and OB, I should be able to get the equation of this pair of straight lines, sorry. So if I multiply them, four x minus three y and two x minus three y equal to zero, they should be actually be mapped to A x square plus two h x y plus P y square equal to zero, which clearly implies A is going to be eight, B is going to be nine and h is going to be minus six, minus 12, which is minus 18, right? So if I need the value of this, it will be A, which is eight. Two h is minus 36 plus B, which is nine, okay? Oh, sorry, two h is compared to 18. So h is going to be, h is going to be minus nine, right? So two h is going to be compared to minus 18, so h is going to be nine. So yeah, so I'll change this value here now. This will become 18, eight minus 18 plus nine, that's going to be minus of one, that's option number three is correct in this case. Option number three is correct in this case. Is that fine, guys? Yes, Atmesh, correct. So option number three becomes the right option in this case. So if you have understood this, you can quickly move on to the 15th question for the day. Again, coming to you from pair of straight lines. If line y equal to mx bisects the angle between the pair of straight lines, then which of the following option is correct? I think this should be a simple one for you to solve. Okay, what will you choose? What will you choose if many lines are going in and coming out, even from half and down, excuse me, from up and down, inside the line and coming out. Yes, so Shweta says option two, how about others? It's pretty simple. You just have to go from the concept of equation of bisectors. So we know that the equation of bisectors of a pair of straight lines is given by x square minus y square by xy is equal to, you can just write a minus b is equal to xy by h, correct? So this should be known to you. So guys, please recall these formulas. They're very useful. They save a lot of time. Now y equal to mx happens to be one of the bisectors. This happens to be one of the bisectors. So this happens to be one of the bisectors means y equal to mx will satisfy this equation. So what I'll do, I'll put y equal to mx in this equation. So I'll get something like this, cancel out the factor of x square, it'll become something like this which becomes h times one minus m square is m times a minus b and yes, option number two becomes the right option. So I think Shweta, Rohan, Vishis, Purvik, Shweta is the first one to answer this correctly. So option number two is the right answer. It's a very simple question should have solved. Everybody should have solved this. The idea is to recall the equation of the bisectors of a pair of straight lines passing through the origin. Any question with respect to this? So we can now move on to the next question for the day. That's question number 16. So x plus y is equal to seven and a x square plus two hx y plus a y square equal to zero are three distinct lines forming a triangle. Okay, then the triangle is isosceles, kaolin, equilateral or right angle triangle. Simple question guys, should not take you more than a minute or so. Just look at this equation very, very carefully. It's a x square plus two hx y plus a y square. So this is repeated over here. A is repeated. There's no B. Don't misread this as B. So Purvik is coming up with the option four, right angle triangle. Anybody else? I just got the response from Purvik. So Atmesh thinks it's option number one. It's an isosceles triangle. I need just one more response. Okay. Vishis says option three. Rohan says option one. Okay, guys. Now, I don't know how many of you have figured out the fact that this pair of straight lines will be symmetric about y equal to x because even if you change y with x, nothing happens to these pair of straight lines. Isn't it? That means I can say y equal to x is very symmetrical, symmetrically placed to these two lines. So let me just change the color over here. So let's say the green ones are the pair of straight lines. Okay. So the green ones are the pair of straight lines and the blue one is y equal to x line. Okay. So the green ones are pair of straight lines. Now, y, x plus y equal to seven would be a line like this. Okay. So this will be your x plus y equal to seven. Correct. Now, from the figure, it is very obvious that these two angles would be the same angle. Correct. That means these two lines would be of same length. That means it has to be an isosceles triangle for sure. So only one option could be correct. So option number one becomes correct in this case. Okay. Even if it is an equilateral triangle, all equilateral triangles must be isosceles. So equilateral triangle is a special case of an isosceles triangle. So we have to mark the most exhaustive option that is given to us in the option. Okay. So guys, very simple. So we'll move on to the next question, which is question number 17. So we got, first time we got a question on circles. P A and P B are tangents to the circle S touching it at point A and B. C is a point on S in between A and B as shown in the figure. LCM line is a tangent to S intersecting P and A at points L and M. Then the perimeter of the triangle PLM depends on... If you're there, you know what to see. No, A plus B should be 0. If A is equal to minus B, then the angle between them is 90 degree Purvick, not when A is equal to B. When A plus B is equal to 0, then the angle between them is 90 degree. Is that fine? Okay. This is P and C only. Okay. Shweta also backs him up. Option 3. Anybody else? Aatmesh says option 2 is correct. Okay. Even Sondarya feels option 2 is correct. Okay. So mostly people are, you know, divided between option 2 and option 3. Okay. So let me just resolve this conflict. All right. So see here. We all know that AL will be equal to LC. Right. AL will be equal to LC. Correct. Right. And we know CM will be equal to MB. Correct. Yes or no? So instead of saying PL plus LC plus CM plus PM to be the perimeter. So let's say this is the perimeter. Okay. Can I just simply replace LC with LA. So it becomes LA. And I can replace MC with MB. Correct. So this will become the perimeter, which is clearly PA plus PB. So your perimeter of the triangle is just depends upon P, A and B. Right. It doesn't depend upon C actually. So the answer will be P, but not on C. So option 2 is going to be correct. So I think Aatmeesh was the first one to answer this correctly. So Aatmeesh, Sondarya, Vishis, Sanjana, Sai Mehir, Rohan, Shweta. Okay. So option number 2 is correct. Well done. Simple question, but again, people got tricked. Is that fine? Good. So let us move on to the 18th question for the day. Yeah. 18th question. Two equal quads, A, B, and A, C of the circle are drawn from the point A. Another quad, P, Q is drawn intersecting A, B, and A, C at points R and S respectively, such that A, R is equal to S, C equal to 7. And R, B equal to AS equal to 3. Find P, R by Q, S. So you have to draw it because there's so many information dumped at you. So please draw the given diagram. So let me just help you out in drawing it. So let's have a pink circle. This would make girls a little happy. So I've drawn a pink circle here. Okay. So let's say two equal quads, let's say A, B, and A, C, they are equal in length drawn to the circle. Of course, the circle equation is X square plus Y square minus 6X minus 8Y minus 24 equal to 0. And this point A is given to us as a root 33 plus 3 comma 0. Quite a weird looking point. There's another quad, P, Q. Let me draw another quad in blue color. So there's another quad, P, Q. Okay. P, Q. Intersecting A, B, and A, C at point R and S respectively. So this is R and this is S. So A, R is equal to S, C. So this matches with this matches with this. And this is equal to 7 also given to us. Okay, 7, 7. And R, B is equal to A, S. So this is 33 given to us. Then we have to find the value of P, R by Q, S. So P, R by Q, S. By the way, from my figure, it seems to be 1 actually. That's the benefit of actually drawing a very accurate graph. Okay. I mean, it's a feeling, it's a feeling. Oh, A may not lie, right, right, right. A may not lie on the circle. Okay. So let's not consider A to be lying on the circle. Oh, but can we verify this before we take that call? So root 33 plus 3 squared, 0 plus, sorry, minus 6, root 33 plus 3, minus 24. Does it become 0? Let's check. So this becomes 3 plus 9. That's going to be 42 minus 6 root 33. And this also becomes minus 6 root 33 minus 42. Sorry, this will be plus. Yeah, this will be giving you 0. Okay, so yeah, so we just initially even I have felt that. But now as I can see, A lies on the circle. But this you have to verify. So it lies on the circle. Sorry, Shweta was the one who said A doesn't lie. So A lies on the circle. Now, from diagram, it seems to be one, but don't go by that feeling. Try to solve it mathematically as well. Yes, anyone? Guys, I want to just know from you that are you all aware of this property that there are two secant lines that say p, q, r, s, t. Okay, do you know this property that pq into qr, pq into qr is sq into qt. How many of you are not aware of this property? This is something which you have been studying since class 9th or 10th for that matter. This is called the secant law in which you, if you, there are two chords which actually intersect. Now it may intersect within the circle or outside the circle, it doesn't matter. pq into qr, okay, will be sq into qt. Now, if I use this particular law, just look at this expression. Let me call this as A. Let me call this as B. Let me call this as C, okay. So look at the intersection of the tangent pq and ab. So from intersection of pq and ab, can I say A into, that is pr into rq. rq is B plus C. That will be equal to ar into rb, that is 3 into 7. Similarly from intersection of pq and ca. So I am not focusing on these two chords, okay. So here I can say, sorry, C times B plus A is equal to again 7 into 3. Now from these two equations, let me call it as equation number one and equation number two. I see the right hand side is the same, okay. And here I can see this B is same. Only A has been replaced with C and C has been replaced with A. That means A should be equal to C ideally. If A is equal to C means pr should be equal to qs. That means the ratio is going to be 1. Is that fine? See the benefit of drawing the curves accurately. So when you draw it accurately, things will automatically turn out for themselves. Especially in objective based questions, you don't have to justify how you got that answer. So you can always solve this by making an accurate diagram. Is that fine guys? Now do you realize this question was actually easy? I would have told you this in my case when I was writing my JE. We got three questions on coordinate geometry and I had no clue how to solve them. I just made diagrams for them and I could end up solving two of the three questions. Great. So we don't have much time. We have around just 12 minutes left. I think sufficient enough to take one more problem. Okay so let's take the question number 19 now. Point P outside a circle with the center at C. Tangents P and PB are drawn. Such that 1 by CA square plus 1 by PA square is equal to 1 over 16. Then the length of the chord AB. Again draw the diagram. Since your request is draw the diagram, you will come to know many things from the diagram. So I will recommend to you that the first finger is B. Next finger is C. C is C. Then the I is A. Please tell me that the force is either A or B. A and B. Capital B is the index three of this quantity inside the three. That's right or not? That's right. This is the middle finger. The force will be toward A. Am I right? Which magnitude is the first one? The magnitude is either right or not? The other is the second one. The center part of the key is the right one. So the rule is that the external magnitude is four. The right one is the magnitude is two. I don't mean the right one is the right one. The whole space is actually... The whole space is actually... The whole portion is... The whole portion is because of that. I just said the portion. The number I did not want to tell you is one more line to remember your brain. But in the case of the... In the case of the syringe carrying wires, it can... So same direction syringe, opposite direction syringe. So that's why it is not more clear to answer. Okay, so Rohan thinks option two is correct. What about others? Yes sir. Guys, let me tell you the questions that we have done so far doesn't involve difficult concepts on these topics. They are just based on your understanding of geometry. For that matter, geometry along with trigonometry. That's the beauty of J equation. They will either ask you locus-based questions or they will ask you concepts which you can actually solve very easily using geometry in association with trigonometry. So you don't have to break your head remembering all those typical concepts like director circle, pair of tangents, limiting points, radical axis. You should know them. I mean, it's for your awareness. And if at all they ask you such questions, you should be ready with it. But mostly they try to see your geometrical understanding of the problem. So again, let me quickly sketch a circle over here. So this circle has the center at C. And there's a point P outside the circle. Let's say P point is over here. And from P point, tangents are sketched. From P points, tangents are sketched to this circle. So let's say these are the tangents. And these points happen to be point A and point B. This would be your P A and P B, C A and C B. So what is given to us is 1 by C A square. If I'm not wrong, 1 by C A square is 1 by radius square only. Plus 1 by P A square is going to be 1 by 16. Now let's do one thing. Let's consider this angle here to be theta angle. Let's consider this angle to be theta angle. Now from this diagram, I can say that R will be equal to, let's say this is R, R will be equal to P A cot of theta. No, tan of theta. So 1 by R square will be like cot square theta by P A square. Now if I substitute over here, I get 1 plus cot square theta by P A square is equal to 1 over 16. Right? And 1 plus cot square theta is like cosecant square theta. So cosecant square theta by P A square is equal to 1 by 16. Which implies P A sin theta is equal to 4. And you would be surprised to know that P A sin theta would actually be this length. This length is actually P A sin theta. So this length is actually 4. So this actually gives me the length of the cot which is 4 plus 4 which is 8. So option number 2 becomes correct in this case. Is that fine? I'm surprised that only one of you have been able to solve this. So guys, what we learned from this session was the application of the concepts that you had learned in straight lines and circles and how geometry and trigonometry must also be kept in mind while solving problems. So today actually I was very optimistic and I had sorted out 36 problems for you. But unfortunately do 19 of them. So what I will do is I will send the PDF to you on the group. Try to solve the remaining 17 questions. And I think we had some problems from the last session as well. And do post me the solution if you can. And if you're stuck somewhere, we'll take it as a doubt session. Meanwhile, if you want, I can do one more question with you. But let's call it a day because you should solve it and send the solution to me. So from 20 to 36, please solve it as a homework-based question and do post it to me on the group. Is that fine guys? I think most of you would be having your board exams starting from Monday. Sorry, I mean the lab exams starting from Monday. Or is there no exam in your school? Okay, you have your pre-boards and board practicals. Anyways guys, all the best for it. Over and out from my side. Thank you so much for coming online today for the live session. Bye bye. Take care.