 So we have now a general second-degree equation and we know that from our conic section experience that a General second-degree equation would have a second-degree term. It will have linear terms also And it can have constant term also. Okay, so this is what we call as a general second-degree equation now This equation other than representing a conic that is circle, ellipse, parabola, hyperbola that we have already seen can also represent A pair of straight lines that means it could be a representative of something like, you know, lx, my plus n1 or you can say n into Another line equation something like this Okay, and from now From this equation, it must be very obvious that they will not pass through origin Guys, if the if any pair of straight lines have to pass through origin, then the C term will definitely be not there Okay, so one of them may pass through but both of them will not intersect at the origin That means the intersection point will be some other point other than the origin Okay, if they have to intersect at origin g f and c terms would all be zero That means it would convert itself to a homogeneous second-degree equation Okay, now we only seen in our past that second-degree equation can represent conics So how can we say that or what condition must we have such that this will represent a pair of straight lines Okay, let us talk about that and then we'll also use that to see what is the intersection point of the two lines represented by this second-degree general equation So do you remember when I started the session of conic with you last year I had given an expression delta Right, do you remember delta expression? And I had given you the fact that if delta is zero then this second-degree equation, let me write it as s d e will represent a pair of straight lines and I did not prove this particular fact that time Now the time has come that I should prove this for you So listen to this prove very very carefully Yeah, that coffee thing I don't want to repeat this, it's getting recorded So this condition, that ugly condition a b c 2 f g h minus a s square minus b g square minus c s square If this is equal to zero then it would represent a pair of straight lines If it is not zero, it can represent circle, parabola, ellipse, hyperbola That's what we had discussed in our last year discussion So now we'll prove it So I'm going to give you various proofs for it I'm going to do various proofs for it The first proof, let's look into this If you look at this equation Can I say this equation is like a quadratic in x Let me write it like this So that you're able to relate that it's a quadratic in x So a x square I've copied And x term, I'm segregating it out So let's say 2x I have h y and g And this is like a constant term you can say for the time being So what I'm doing, I'm treating it as a quadratic What about b y square? Oh I'm so sorry Thank you Richard for pointing that out Thank you so much So I'm treating this as a quadratic in x Now if I have a quadratic in x I can use my Shridhar Acharya formula Or quadratic equation formula to represent x As minus b So b will be 2h y plus g So minus b Plus minus b square That's 4 h y plus g square Minus 4ac c you can write b y square 2f y plus c By 2a By 2a Now all of you please listen to my claim over here If I say If this has to represent a pair of straight lines Then the quantity within the under root That is this quantity Which I am underlining with white This must be a perfect square Would you agree with me on that? And why is that so? Is because if this is not a perfect square I will end up seeing these radical powers on the variables Which is highly unacceptable When you are dealing with straight line equations Have you ever seen a straight line where If you represent your x in terms of y You have some kind of radical powers coming up in x and y? Never So if it has to be linearly factorized This term which is underlined with white Must be a perfect square Yes or no? Anybody who thinks this argument Is not acceptable please let me know Okay So if I am claiming that this guy I am writing it now separately So the term which I have underlined I am writing it separately So I am claiming that this fellow Is a perfect square This fellow is a perfect square Can I say then this equal to 0 will have Equal roots Because equal roots come from Which are perfect square In other words what I am trying to say is that Drop the factor of 4 This equation will have equal roots And this is an equation in y So let me just write it in a proper quadratic in y So y square will have coefficient I have to write it in terms of y So y square will have coefficient x square minus a b It will have 2 g h minus a f y So I am just hand picking the terms Let me know if I am missing out any terms And constant terms would be g square minus a c So this should have equal roots I have got some message from someone Y equal roots Richard if this is a perfect square Then an equation which is equated to 0 So this is something square So won't this have equal roots in terms of y Because equal roots come from Those expressions which are perfect squares For example if I say I am saying this is a perfect square Because it is x minus 1 square So the moment you equate it to 0 You will end up getting equal roots for it Is that clear now Richard? Now in that case I can say the discriminant of it That is b square minus 4 a c I am changing the colour here Should be equal to 0 So b square that is 4 g h minus a f the whole square Minus 4 a c Minus 4 a c This should be equal to 0 Let me drop the factor of 4 from both the sides This 4 and this 4 Let me drop it off Let me open the brackets So here I will have h square g square Minus You will have 2 a f g h And you have a f square Here if I multiply them I will get h square g square Minus a c h square Minus a b g square And plus a square b c Okay If I open the brackets You can well see that h square g square term will vanish off And drop Drop a throughout Yeah through a square That is what I was explaining through That It has to be a perfect square Because if it is not Then it will lead to Some kind of under root expressions on y So it will not represent a straight line Right? Straight line is never known to have Radical powers on y They are all Power of 1 That is why I am claiming it To be a perfect square Does it satisfy your question? Query? Yeah So now if you drop an a from everywhere Okay If I drop an a from everywhere You will end up getting Minus 2 f g h Okay And there is an a square Yeah And there is an a drop from here So you will get a f square There is an a drop from here You will get c h square A drop from here Plus b g square And you will have minus a b c Correct me if I am wrong Okay So if you multiply Throughout with a negative sign You will end up getting A b c Plus 2 f g h Minus a f square Minus b g square Minus c h square Equal to 0 And this is what we wanted to prove Hence proved Sir I have a doubt sir Yeah tell me Sir what is the perfect square Could it also be equal to Some other perfect square And then there will come a constant Then when we subtract it That could have equal roots Sorry So like let's say That 4 thing is equal to 25 the entire expression Okay So then we subtract 25 And say that that equations Must have equal roots Won't that also satisfy That will also work But at the end of the day You are claiming this to be A perfect square in themselves So if you get it to 0 It should have equal roots That is also fine See you are trying to say Yeah you are trying to say p square is some q square right Yes sir Yes and then you are trying to form That is also fine That is also Try it out That should also give you the same result But unnecessarily Using some other variable into it So any scroll up of it I need to take the initial thing I got disconnected Okay you can take a snapshot If you want Okay sir So I will just write down The initial and final thing Sure Ruchi does it Answer your question So even if you take it It should not matter It should not matter But again Why would it be a constant When you have variables Why involve inside it Oh yes sir Okay so try to think in that Angle also Yes sir One thing that I would like I would have Like to hear from you What if A were 0 Okay Now if A were 0 I have a doubt The conditions will not be affected I will take that in the next This thing You have done it Whoever was copying Are you done with the copying part? Yes sir can you scroll up To the top of the page Oh sure Sir? Yes sir tell me Rahul you are asking something No sir I just wanted to copy it So I didn't understand Why the claim Why is that thing a perfect square The one underlined in white I am explaining this For the third time See have you ever seen something like X is equal to root Y plus 1 By something As a let's say equation of a straight line Correct No sir Do you have something like this As a equation of a straight line I mean some Y function Let me not just write Y Just some Y function This function of Y let's say Plus some constant 1 or something So this is not possible That's why if this term Within the under root sign If this doesn't come out to be a perfect square You'll end up getting all radical Powers on Y Or there would be such an expression Which will have something like You know a quadratic under root Isn't it? So imagine as if you have been Given a equation of a straight line like this Let's say Does it look like a straight line equation? It will not This guy has to be a perfect square It has to be a linear term in Y That's why I claimed here That this expression should be a perfect square So the final expression was The general form rate For the The final expression which it got The ABC plus 2 FGH Right That's the general This thing That's the general condition For any second-degree general equation To represent a pair of straight lines Okay So if that condition is met We say That that second-degree equation Which has been provided Would represent a pair of straight lines Okay Getting the point Now If E is equal to 0 You would realize That your equation AX square Plus 2 H XY Plus BY square Plus 2 GX Plus 2 FY Plus C equal to 0 This term would vanish Correct And you would be only left with BY square Let's say 2 Y HX plus F Plus 2 GX plus C Now I'm treating it as a quadratic in Y Yes, kind of Correct Correct here Okay That means one of the equations Actually doesn't have an X in it Okay Now Oh God Are you there with me guys? As I told you The power went off And it has come back now Sorry for the inconvenience Now here also If I use The second way Doing the proof Yeah Second Before second way We'll have to clear off the fact that What if A becomes 0 Will that condition change? If not What will happen? What will be the new condition? Let's talk about that Okay So if you see now I have made it as a quadratic in Y I have made it as a quadratic in Y Okay Again by using the Shridharacharya formula I can say Y is equal to minus B Plus minus under root B square Minus 4 AC By 2B Yes or no? Now just like my previous argument Can I say this has to be a perfect square Because we have never known Any kind of a linear equation of a line Which has got under root of a quadratic in X Isn't it? That means This equation Should have equal roots By the very same argument Hello are you there? Yes sir Okay So This equation I can say X square coefficient will be H square Okay X will be H F Minus 2BG And constant terms would be F square minus BC Okay So here also the discriminant Which is B square minus 4 AC Must be equal to 0 So B square That is 4 times H F minus BG Square Should be equal to 4 AC Is that fine? No it's sufficient and I have done very It can last Okay So what we do is now We cancel out the 4 factor So on both the sides Expand it So it gives me H square F square Minus 2 H Or you can 2B FGH Plus B square G Square And this side I will have You can Let me write it like this Minus BCH Square So cancel off Cancel off H square F square for both the sides Cancel one factor of B From both the sides So I end up getting minus 2FGH Plus BG Square Right Plus C H square equal to 0 Multiply with a negative 1 So you end up getting This condition Now what I claim here is that The condition has not changed From this You just have A Becoming 0 everywhere Check it out guys So this condition still holds here Is just that now your A Is becoming 0 So this term has vanished This term has vanished So only this term will survive So there is no change In the condition whatsoever Now many people ask Sir what about if B is also 0 Because here you have B also Correct? That could also be a Possibility Yes or no So yes, that's the question Vikas has asked What if B is 0? Very good question So now we will take a situation Vikas where we have A and B both as 0s In that case Can I say Your pair of straight lines Equation that is This equation will not have This term Will not have this term That will only have X, Y term X and Y So I will have an Equation of this nature Now here both conditions are happening together That's why I am taking it separately I know the first condition Can take care of the fact that B is 0 So now I am taking care When both of these happen together Okay Now listen to this here I will multiply throughout I will multiply Throughout By H by 2 I will tell you why I am doing that So when I do that I get H2XY I get HGX I get HFY And I get CH by 2 Okay Now Let me send this CH by 2 On the other side So I will make it equal to Minus CH by 2 Okay Let me add An FG term On both the sides FG term on both the sides Okay Now from these two terms I will take an HX common So I will get HY plus G And from these two terms I will take an F term common I will again get HY plus G And on the right hand side I get FG minus CH by 2 This is factorizable as HX plus F In times HY plus G Is equal to FG minus CH by 2 Okay Listen to my claim here If this way To represent a pair of straight lines Then this guy should have been 0 Can somebody tell me why You can unmute yourself and speak What will happen if this term were not 0 What would it represent then Anybody See if you see this Take an H common Take an H common Let's say this is a non-zero term Let's say some value K If K is not equal to 0 Do you realize that This expression Will actually look like something HY is equal to a non-zero value Okay Let me call it as C and C is not 0 So when you took H common Why did you put a Y under the X Sorry There is a Y under the X Shouldn't that be a X My bad Okay So if C is not 0 What does this equation represent Hyperbola Correct It would represent a rectangular hyperbola It would not represent A pair of straight lines Okay So what should happen is that This term should have been 0 That means 2 FG Should be minus CH Should be equal to 0 2 FGH minus CH Square should have been 0 And you can very well see that This situation actually emerges From the very same condition That we started with Where you have put your A and B both as 0 So if A and B both as 0 These terms will vanish off And only 2 FGH And CS square would be left off And that's why even if A and B both as 0 The condition still Persists to be true Okay Now let's move on To the second proof of the same thing For the second proof I would use the fact That a homogeneous Second degree equation Mind you homogeneous second degree equation Would always represent a pair of Straight lines passing through origin Correct, this is what we had learnt In our before break session Correct Let this particular Equation that we have Okay Meets at X1 Y1 Okay Some point X1 Y1 The pair of lines which is Which is represented by this equation It meets at It meets at X1 Y1 Correct Shreya, correct Now what I will do I will shift my origin I will shift my origin To that point X1 Y1 Why am I doing that? See, look at this figure carefully This was the pair of Straight lines represented by this equation Correct Now what I am doing is my origin was somewhere And I am shifting my origin to this point That means I am Shifting my origin to X1 Y1 So 00 has gone To X1 Y1 In that scenario How would this equation change? Would the constant term disappear? It should disappear And moreover A coefficient of X should also disappear And coefficient of Y should also disappear Isn't it? If your origin comes to this position Let me write it in white If your origin comes to this position It would convert itself to A homogeneous Correct, but that is not my question My question is What would that new form of this equation be? What should I replace my X with? X1 X1 This is an all time confusion That everybody has If your origin shifts to H, K Your old X becomes New X plus H Old Y becomes new Y plus K This has to be committed to memory Please, please, please Keep this result very, very handy And ready You can't miss out on those So what will happen to this equation? It will become X minus X A X Let me write it as capital X Plus X1 whole square B Capital Y plus Y1 Whole square Plus 2H capital X plus X1 Capital Y plus Y1 2G Capital X plus X1 2F capital Y plus Y1 Plus C equal to 0 And I claim that This equation, let me name it as Equation, let me call it as 1 And let me call it as 2 I claim that 2 is A homogeneous Second degree equation That is The coefficient of X Coefficient of Y And the constant term Must all be 0 Agreed Agreed my dear or not If no, please stop me And ask questions So the capital X is The old center The new X The new variable of the equation Okay So this will result What I am claiming is that This would result into something like this Sorry for using Capital A and B also But this would result into A second degree homogeneous Equation like this, that is what I am claiming Is my claim correct? Yes sir What will happen if the coefficient Of X and Y is not 0 Coefficient of X and Y will be 0 under this condition That is what I am going to write next According to the coefficient of X and Y And constant term being equal to 0 Your X1 Y1 B, G, H, A Etc would be In such a way That it would result to be 0 for sure Okay Because we have established this as a theory In the previous discussion Okay Yes sir Can you all help me with the coefficient Of X first What will be the coefficient of X over here I claim it is going to be 2A X1 From this guy This guy will give me 2G So let me write 2G also There will be no X here There will be an X over here So that will be 2HY1 Just let me know if I am missing out On any term Okay Which further implies that AX1 HY1 G would be equal to 0 Let me call it as 3 Okay Now all of you please help me to get the coefficient Of Y from this expression You don't have to expand it Now let's not Waste our time expanding Let's handpick terms This will never give me a Y This will give me 2BY1 Okay This guy will give me 2HX1 Okay and This will give me 2F So can I say this is as we are saying HX1 BY1 Plus F equal to 0 Let me call it as 4th equation Okay Now constant terms should also be 0 Now help me out with the constant terms I will have I will have This will give me BY1 I will write that later on Okay let me write it here BY1 I will have from this term 2HX1 Y1 Okay And from this term I will get 2GX1 From this term I will have 2FY1 plus C equal to 0 Okay let me call this as the fifth equation Now all of you I would like you to pay attention over here Very very important step This fifth equation I am going to write in such a way that I include 3 and 4 within it See how will I do that This is AX1 square correct So I would write it as X1 into AX1 Okay and I am going to introduce this entire term over here Okay and I will see what extra term do I have to incorporate to complete fifth So you can see AX1 square is already taken care of HX1Y1 is partially taken care of So I need one more HX1Y1 So what I will do is I will introduce Y1 HX1 and this entire term I will write over here Now pay attention AX1 square is taken care of HX1Y1 means two HX1Y1 is taken care of BY1 square is taken care of Okay and GX1 and FY1 is taken care of So one more GX1 and FY1 scope is there and of course C So this fifth equation I have just rewritten like this Is this understood This is a very very important step and see whether you have understood it Then only I will move forward Understood Any doubt, any kind of void in understanding Let me know Now people are wondering Why did I do this Because I could safely write this as 0 and this as 0 from 3 and 4 Isn't it? So which finally leads me to The fact that GX1 plus FY1 C is equal to 0 Let me call it as the sixth equation Okay Now the time has come that I Summarize here If you see that Third, fourth and fifth Sorry, third, fourth and sixth These three equations Seem to suggest This and this guy They seem to suggest that AX HY plus equal to 0 HX plus BY plus F equal to 0 And GX plus FY plus C equal to 0 Represent Represent Concurrent lines Why concurrent lines My dear Because they are all passing through They are all passing through X1, Y1 Correct And if you had learned Your condition for concurrency In your straight lines chapter I am sure most of you Would have an idea about determinants In physics you must have done it Then the coefficients here Must be 0 Yes or no This is your condition for concurrency If not you can open your RD Sharma And you will find this condition For concurrency There in your RD Sharma book also Present everywhere Any standard book you pick up You will get this condition for concurrency For this statement Or alternately You can solve these two Put it in the third one You will end up getting the expansion Of this determinant And what is that ABC Plus 2FGH Minus AF squared Minus BG squared If you expand this You will end up getting this expression Getting my point So either you use condition for concurrency If you don't know that Solve the first two or solve any two And put it in the third You will end up getting this expression There is another way to look at it But only when you have done Your determinants chapter I will tell you There is something called non-trivial solution So if a system of equation Which is like this Is called a homogeneous system of linear equations If it has A non-trivial solution This determinant will come into picture But that is something which you will not understand Right now we will wait for a determinant Chapter to revisit this concept Okay So are you convinced with the second proof also? Okay now Here comes one more understanding Which probably emerges from third and fourth So all of you I would like you to look Third and fourth equation one second Something which actually is very interesting It says that The point of intersection Of the two straight lines Which were represented by that pair of equation Second degree General second degree equation Can also be obtained By solving three and four simultaneously Correct So x1, y1 Right Can be obtained By solving Three and four Agreed Yes or no So you must be thinking why three and four Why not sir is talking four and six Okay But three and four is something which you can obtain in this way All of you please pay attention Okay Go back to this equation Call this expression as S for the time being Do you remember the S term which we were using Every time in our conic section S stands for second degree Now I would like you to do This operation Do s by do x equal to zero Now people who are wondering what is do s and do x It means Partial derivative with respect to x With respect to x Partial Derivative With respect to x What does meaning of with respect to x Means only x has to be treated as a Variable Rest all the terms would be treated as constants Getting my point everybody So do a partial derivative Of s with respect to x and tell me the result Won't you get two a x I am writing it over here Two a x By differentiating this guy What is the derivative What is the partial derivative of By with respect to x Zero What is the partial derivative of this guy Two h y Two h y What is the partial derivative of Two g x Can I say these two will again Zero Can I say that it will end up Giving you I am sorry x plus h y plus g equal to zero Plus g equal to zero You see that the same equation The same equation I had Got over here also You see this Yeah In a similar way you would realize that The second one that is your Partial derivative with respect to y If you do a partial derivative With respect to y And check it out over here You will end up getting two b y You will end up getting Two h x Okay And you will end up getting two f Equal to zero That means it will give you h x plus By plus f equal to zero That is the same thing that we had written over here Or here whatever you want to see So what I am claiming is that This x and y x one and y one In this section Would be obtained by Solving these two Equations together If you would recall the same thing We had done in conic section where Solving these two would actually give you The center of the conic section So you can apply this to Find the You can also apply this to find the Center of a conic So you can use it for a circle Hyperbola ellipse Getting my point In the present case It is helping us to get the Point of intersection of the Two straight lines Is the idea clear? We will take a quick exercise on this So that the concept is clear in your mind My question is Find the Point of intersection Of the two lines Whose combined equation is 12x squared minus 10xy Plus 2y squared Plus 11x Minus 5y plus 2 equal to 0 For that we need to look into The partial derivative itself The concept of partial derivative itself Which is beyond our scope right now Yeah You have to do some reading on Partial derivatives I can suggest you A very good book for that Shanti Narayan partial derivative You may find the soft copy There is a lot of books You can avail easily on z-lib.org If you feel you want some High level book or some Book which is not prescribed Not available in the market normally You can find this on this online Z-lib.org Shanti Narayan Differential equation Differential calculus book But again do you want To waste that precious time of yours You are in classes right now You can do a lot of activities Before After your JEE exam Okay now two ways You can solve this question One by Getting the lines themselves And solving them Simultaneously But remember this is a waste of time Okay And the second way is You simultaneously solve These two This is time efficient Okay I'll do both the ways and show you See Param is already ready With the answer I'll do both the ways and show you By the way many people ask me Sir when you do partial Derivative like this does it give you The two lines which represent The straight lines The answer is may or may not In fact it may not Most of the time It just means that There exist two such lines Whose point of intersection is the same As the point of intersection of these two It doesn't mean they have to be the same lines themselves For example let's say These are the original lines They meet at some point x1, y1 There can be some other two lines also Which meet at the same point x1, y1 So these two lines are obtained From do s by do x equal to 0 And do s by do y equal to 0 It needn't be the original lines guys Let me tell you So what's the condition for the two lines Being the original lines For the two lines What is the condition for two lines To be the original lines There is no such condition You have to solve them actually You have to solve the equation How do I solve it? I will tell you that But from this I am saying that this may not give you In fact I should say I will not say the word may not Because I have not seen a case So far in my solution I can say this will not give you Okay So first I will solve These two first Because the second one is Easy to get Then I will check with the original lines So do x will give you 24x minus 10y Plus 11 equal to 0 Correct Yes I know And do s by do y equal to 0 Will give you Minus 10x Minus 10x plus 4y minus 5 Minus 5 equal to 0 Okay Let's without wasting my time Sketch them on the Okay I don't want to solve them simultaneously I want to save your time And my time as well By the way the people Who are new to this class I normally use Some graphing software To plot the lines Desmos is one of them So we have 24x Hope you can see my screen Minus 10y Plus 11 equal to 0 Okay and the other line Will be minus 10x Plus 4y Minus 5 equal to 0 Okay x comes 1 by 2 Okay So where do they intersect Where do they intersect So I got x is 1 by 2 Yeah I think it's Minus no I think it's minus 3 by 2 See this is the intersection point Yeah I got 3 by 2 Minus 3 by 2 Minus 3 by 2 and minus 5 by 2 Yeah I got the signs This is the intersection point Yeah I got y as Minus 3 by 2 minus 5 by 2 Oh I took one of the Minus 3 by 2 Yeah one second one second What I'll do is x is minus 3 by 2 y is minus 5 by 2 Yeah x is minus 3 Yes it's minus 3 by 2 and minus 5 by 2 Yes it's minus 3 by 2 and minus 5 by 2 Yes it's minus 3 by 2 and minus 5 by 2 Right I will get the same result Sorry I'll get the same result By getting the equation Of the two lines Now this itself is an exercise Okay all of you please follow here Because you may get also this question What are the two lines Which represent a pair of states With that method also Okay So all of you please pay attention In solving such cases We first try to Focus on the second degree terms Remember I had done a similar question In the beginning of today's session So take the second degree Equation remember line parallel To the two lines and One comma one In the same way I will factorize this guy I'm sure this is easy to factorize formula also for that so this can be broken as minus 6 minus 4 x y minus 4 x y that's 2 y square take a 6 x common so you get 2 x minus y take a minus 2 2 x so you get 6 minus 2 y I'll put some gap over here and 2 x minus y and I'll put some gap over here now this gap is because I know there are other terms also correct that is 11 x minus 5 y plus 2 so there has to be some constants present over here so I'm claiming that this pair of straight lines is made up of product of this plus now how do I get L and how do I get an M okay so all of you please pay attention in order to get L and M because as a question yeah but how do you find the point of intersection themselves let's say I don't want to use dou s by dou x method partial derivative method because yeah so the problem is we can't use the partial derivative if I want to get the let's say I don't want to use this to get the equations anyways now how do I get L and M all of you please focus in the multiplication of these two what will be the coefficient of x let's see when these two multiply and when these two multiply I will get an x that means 2l plus 6m is the coefficient of x compare it with the coefficient of x over here second compare the coefficient of y so this multiplies to M and this multiplies to minus L minus 2m will be equal to minus 5 correct okay let's solve for L and M let me multiply this with a 2 so 2 let me write it separately so 2l plus 6m is equal to 11 and minus 2l minus 4m is equal to minus 10 M is 1 by 2 M is 1 by 2 we can say L is going to be and 4 is going to be 4 4 correct so my two lines that I will get would be respectively 6x minus 2y plus 4 equal to 0 plus 4 equal to 0 and the other one would be 2x minus y and M is what 1 by 2 M is 1 by 2 now without waste of time I will just check whether this point here satisfies them or not put a minus 3 by 2 over here what do you get minus 9 put a minus 5 by 2 here you get a plus 5 plus 4 equal to 0 satisfied yes sir put in the second one minus 2x sorry 2x will be minus 3 minus 5 will be 5 by 2 plus half so 3 minus 3 0 that is also satisfied correct but these two are not the same lines that we had obtained while we were doing the partial derivatives as you can see okay so at least not like that time consuming also it is not time consuming I mean here we were lucky that it got factorized and LMN values were like coming from very simple simultaneous equations but partial derivative like did I? partial derivative is like directly end up getting these two lines and the section you get is that fine? yes sir so what does the partial derivative mean here sir? see it tells you that if you keep your y constant how does your x change in a 3D structure right you can keep any one of the variables constant and the other two varying correct so s is like your z and your x if you have kept x as a variable and y as constant you are doing a partial derivative with respect to x if you are differentiating that what's the point? okay yes sir how does that doubt can we consider that the equation to be a quadratic in y when a variable x is a part of it yes we can yeah we can take it as a quadratic in y when there is an x in it in that case yes the coefficients will be variables themselves not constants okay Karthik has asked me to go to the previous page okay Karthik I am going to the previous page do you want me to drag a breakdown let me know so let's move on to the next concept next concept as we had taken up in our homogeneous pair of straight lines equation angle between the two lines angle between the pair of lines represented by the general second-degree equation not a homogeneous in this case but a general second-degree equation guys now just want you to answer this will g f and c play any role in the angle no sir no sir will not we had just now seen that in the previous example sorry I missed out in the previous example we saw that the second-degree equation contains the knowledge about the slope these terms g f and c arise because of the constants involved in those two equations remember l and m we had taken in the previous example those l m n where is l and m were responsible for this g f and c yeah so my theory here is plain and simple the angle of intersection will still hold to be the same expression that we had derived in the homogeneous equation you can also say that even if you shift your origin and bring your origin to the intersection point these three terms are not going to be disturbed and if these three terms are not going to be disturbed your your angle between the two lines is still going to hold the same formula is that fine is any question here no sir so I didn't understand like the last part which you said like if you shift the origin how exactly the same thing yeah in this case let's say we have two lines and the angle between them is theta and this is x1 y1 okay if you shift your origin to this point let's say you brought your origin to this point what will happen this equation the first three terms will remain as such only this g f and c terms will make changes because if you try doing this operation let me show you that operation if you do x plus x1 square 2h x plus x1 y plus y1 and b y plus b y plus y1 square if you just focus on x square coefficient it will be a only if you focus on x y coefficient it will be 2h and if you focus on y square coefficient it will be b only rest only you will have more you know some changes happening to this guys will not change whatever we are going to get they will all be 0 0 only right yeah because you're shifting your origin to this position yeah so the angle doesn't change on shifting of origin that's what i'm trying to say okay yeah let's take up this question okay before we move on uh i would like you to just derive the formula for point of intersection also and keep it i'm sorry i'm asking you to derive this because we had already learned a method very beautiful method to find the point of intersection but let's have a formula for this ready because i have seen many questions which require you to use this in this point of intersection formula for this general second degree equation in fact i'm going to give you this formula and let's quickly derive it so the formula is the point of intersection x1 y1 is given by bg minus hf by h square minus ab comma af minus gh by h square minus ab so this is very easy to derive uh we can always follow our old method okay now you must be wondering if the original equation is given to us we can follow this directly and not deal with abh gf and c terms there you will have numbers in in place of them numbers are always very convenient to deal with but let us use this and get this formula and i'll also tell you the way to remember it why i'm asking you to remember it because i have seen there are many questions where they use these kind of expressions in the options okay and in that in those cases you need to be aware of it so partial derivative will give you two ax i'm dropping the factor of two here and this will give me uh two h y so h y and here i will get a g and partial derivative of y will give me from here i will get two h x so i'll write hx only two by so i'll write by only and here i will get an x okay so i can use your cross multiplication method that we had discussed a little while ago to solve this so ab minus h square will come down over here for minus y gh minus af will come oh sorry af minus gh will come i've already taken a negative sign here so af minus gh will come and for one ab minus oh have i done a mistake over here the other i'm sorry this is hf sorry minus hf minus and this will be ab minus so from here you can get x as just take this term and the last term take it up switch the position of the denominator so you'll have bg minus hf by h square minus ab that's what we have over here take the last two over here these two and you can say y is equal to af minus gh by h square minus ab okay so that's how we get up this time is that clear yes now the way to remember this formula is very easy you don't have to derive this every time do you remember i had written a determinant while i was deriving the condition to represent a pair of straight lines i had written this determinant sorry i had written this correct while i was deriving the condition for determine the pair of straight lines okay this is what i have written just write down the first two rows i'll tell you how to remember ah hb gf okay then repeat the first over here okay now do a cross multiplication like this so cross multiplication here will give you ab minus h square cross multiplication here will give you hf minus bg cross multiplication here will give you gh minus af you know how to do cross multiplication right it is always left top with right bottom first so this operation will happen first then this operation will happen okay now divide throughout with ab minus h square so if you divide this by this will become a one this will become hf minus bg by ab minus h square and this will become gh minus af by ab minus h square basically this and this becomes your x1 and y1 done as you can check with this this is actually this term and this is actually this term you can switch the position of the terms in the numerator and denominator this is for your memory aid so now we'll take up a question if you want to note this down please note it can we take can we take up a question now ready one to the next page so let's take this question so if this equation represents if this equation represents a pair of straight lines which are parallel now when I say they are parallel I don't mean to say they are coincident okay so here parallel means they are not coincident they are something like this then prove that h would be either root ab or minus root ab and g under root b will be equal to f under root a or g under root b will be equal to minus f under root a and then after we are done with this we will talk about the second part that is the distance between them let's have around two to and a half minutes for doing the first part type done once you're done with it then I'll discuss it if you have no clue type no clue so that I know you are you know you're not moving ahead and I can start the discussion again I'm repeating parallel lines here does not mean coincident ever comes in your mind right out so Ruchir has done with the first part okay yeah tan formula gives you the first part that is h square is equal to ab but how would you get the second part second part as in how would you get this guys these ways this and this these two you'll get from tan formula correct yeah you can also think it like this that the point of intersection is infinity so denominator should be undefined so h square is equal to ab so h is equal to plus minus root of ab that is also a way to look at it guys mind you the previous derivation there was h square minus ab term in the denominator okay so if that term is zero and if the numerator is not zero then you can say the lines are meeting at infinity means they're parallel to each other we'll have a good discussion about it don't worry that's correct me that would help you to solve the first part of the first part that is h square is equal to ab but how would you solve the ones which I've shown with this zigzag underline that's the question guys one more minute and then I'll begin with the discussion for the first part sir do you in some way use the point of intersection formula for that oh not necessarily okay so would it work if we use partial derivatives then try to solve it okay there are so many speculations going around let me just let me deal with this question at least the first part okay then second part I'll leave up to you to do it see many of you have claimed that I would use the condition that they are parallel so you know if you go back to the previous slide if you go back to the previous slide for them not to meet basically you are saying h square is minus ab should be zero okay but by the way that is true when your denominator is not zero okay if both numerator and denominator are zero they will be coincident lines we'll see that in some way okay so this clearly solves the first of the first part that is proving h is equal to under root ab and this so if they are parallel this guy must be zero as a necessary condition okay as a necessary condition this must be true so these are the outcomes from this particular result correct no doubt about it but how would I solve the other part that is how do I prove g under root b is f under root a or g under root b is minus f under root a for that I have to start from scratch what I do here is if I am claiming that this represents two parallel lines can I assume the equation to be lx plus my plus n and lx plus my plus n1 remember since they are parallel I am keeping these two as same in both the lines and I am only changing them in the constant terms okay so remember two parallel lines only differ in their constant terms you can always make the x and y coefficients same okay now comparing the coefficients a is l squared b is m squared okay any doubt about this okay what about c c is nn1 correct what about the x term 2g if I'm not mistaken 2g will be n1l plus nl so I can say ln plus n1 2f will be what 2f will be what m times n plus n1 what will be the h term 2h term would be 2lm connect me if I'm wrong now we know that the condition for it to represent a pair of straight lines is what what are the condition for it to represent a pair of straight lines abc plus 2fgh minus af square minus bg square minus cx square equal to zero okay guys this may not be used right now I'll use it in some time now use the fact that h is equal to under root ab in this so I can write this as abc okay this will be what 2fg under root ab okay minus af square minus bg square and this will again become ab correct abc abc gone correct now look at the remaining terms you can multiply throughout with a negative sign also I claim that this term on the left hand side of the equation is actually under root af plus sorry minus under root bg whole square agreed which clearly means f under root a must be equal to g under root b isn't it what we wanted to prove it f under root a is g under root b so first part is proved okay if you took h as negative under root ab then see what will happen again look at this equation abc will remain unchanged let it be abc 2fgh will become minus 2fg under root ab correct minus af square will remain changed minus bg square will remain unchanged minus cx square will again become minus abc correct isn't it so it become plus whole square right so if you if you take a negative sign common throughout and remove it you end up getting actually this expression which happens to be a perfect square of f under root a plus g under root b so that means f under root a is negative of g under root b so you end up getting the second part that is your this part is that fine now please go ahead and solve the second part hint is you would require these for part two uh let me know once you're done try done once you're done okay so shreya has a question partial derivatives and compare coefficients of a1 yeah okay can you it it must work shreya because i think the condition will not change did you try that out shreya so so i got the lines with partial derivative and then but then the distance between them is coming to the video by the way if they are parallel shreya it will not be equal to c1 by c2 see parallel means coefficient of x will be same as coefficients of y but the it will not be same as the ratio of the coefficients of constant terms uh i think there's a slight error in that uh see if you say a1 x b1 y plus c1 equal to zero and a2 x b2 y plus c2 equal to zero are parallel lines a1 by a2 should be equal to b1 by b2 should not be equal to c1 if the lines are coincident that it becomes equal to c1 by ruchir uh can you repeat what did you say just now which is so yeah the partial derivative will give you the line right the two lines okay but will it give you both the conditions that we have over here on the plate yes yes anybody who is done with part two other than ruchir park that's only for perpendicular lines so the parallel line stand should be equal to zero by zero so done so parallel lines at square should be equal to here i'm not saying zero by zero line for condition zero by zero will be for the points of intersection not for the angle uh here if i'm able to convey this to you when you're dealing with two coincident lines in the formula of the point of intersection both numerator and denominator would be zero okay sir but then um how would uh the tan theta formula change like is there any way using that we can figure out that if the lines are parallel tan theta is necessary but not a sufficient condition this only tells you that the line are parallel so it may be parallel it may be coincident okay this condition is a necessary condition but not a sufficient condition to tell you whether they are parallel or they're coincident what my point okay sir so in the tan theta if a plus um if a plus b zero then we can only say for sure that they are perpendicular nothing else right nothing else okay thank you sir so most of you have got it i'm giving you 30 seconds more to wrap up things then i'll discuss it so instead of lx plus my plus n and lx plus my plus n one can we take the lines as root a x plus root b y plus c one and root a x plus root b y plus c two root a into x plus root b into y plus c one now that you know that comparison you can do anything you can take no worries shall we'll discuss this enough time has been devoted to this see guys let us recall that when two lines are perpendicular parallel okay so i'm just making a fresh diagram over there when two lines are parallel okay what is the distance between them when you know their equation to be lx plus my plus n equal to zero and lx plus my plus n one equal to zero mod of m n plus n one by root of a square plus b c yeah it's actually mod of the difference of their constant terms by under root of l square plus m square correct now from this comparison l square and m square is already known to us as a and b correct so we can directly make the change let me just drop the divider over here so we can write directly say the denominator part will become under root a plus b now what about the numerator part i have to play with it so first of all i can write this as under root of n one sorry n plus n one the whole square minus four n one n n one correct okay now n plus n one let me pick it up from this let me see what we want to prove we want to prove g square minus six so let me pick it up from the expression over here which i'm showing with the arrow sign okay yes so n plus n one is two g by and n n one is already known to us as c so let us put it over here i'll end up getting this as four g square by l square minus four c okay let me just put it up a bit by under root of by under root of a plus b okay so let me do one thing let me write everything under one under root symbol two can be taken out g square minus l square c by l square a plus b what is l square what is l square a a a so can i not write this as two under root of g square minus ac by a a plus b and that gives me what i needed that is this formula but this also opened me another root or another way to get the same distance had i used my n n n plus n one from this equation that is n plus n one is two f by m then how would this formula change let me show you so in that case again i'm going back to this step it would have become d is equal to under root of two f by m the whole square so let me just write it directly four f square by m square minus four c by a plus b okay again performing the same operations i can say under root of so f square minus ac by d into a absolutely absolutely so you'll end up getting two under root of f square minus ac by b a plus b so there's a small change here instead of g we have got an f and instead of a a plus b in the denominator we have got b a plus b is that fine by the way this is going to be b c not a c this is b c because m square is b slip off so both are the distances between the two parallel lines okay so both can be used okay now one more analysis is pending over here if i claim that your equation that is your second degree equation represents coincident lines if they represent coincident lines can i say that g square will be equal to ac f square will be equal to bc and f square will be equal to a b simultaneously yes or no guys are you there yes sir yes why is this so this term will become zero because they have become coincident see a coincident okay they will be overlapping on each other correct now from these three expressions can you show me that because they are parallel so this is a necessary condition and apart from that you need these two conditions to show their coincident so just stating h square equal to a b can imply two things the line may be parallel or coincident now i want you to show me that bg will be equal to hf also in this case and af will be equal to gh so from these can you derive these two guys also if you are if you are not able to show this exact thing you show b square g square is equal to x square f square which is obvious if you do b square g square you end up getting b square ac by using the first expression and h square is ab and f square is bc isn't it giving you a b square c is equal to a b square c that means yes this will also be true in a similar way you can show that af will also be equal to gh now here is an interesting thing that comes up from this entire exercise do you realize that the point of intersection was bg minus hf by h square minus ab comma af minus gh by h square minus ab it means for a pair of straight lines which are numerator should be zero those will be zero yeah suggesting that it will take an indeterminate structure that means no such point exists no all points sir it is taking an inter indeterminate indeterminate form indeterminate form as you would have learnt in your limits zero by zero so it can take various forms depending upon your situation okay so here the situation is like no such point exists just means that they will meet themselves both these lines will meet at infinitely many points okay okay okay yes okay shall we take some uh yeah uh this is the concept that we have done and now we're going to talk about equation of the pair of bisector lines for pair of straight lines given by the general second degree equation so remember we had done for homogeneous one what was the formula for the homogeneous one so if if this was the line what was the equation of the uh pair of bisectors for what in quality yeah x per minus y square divided by e minus b x by h now tell me can i use this this fact to get the equation of the bisectors of this guy given that i know their point of intersection so let's say these the lines which are representative of this meet at x1 y1 can i use that fact absolutely here correct shifting of origin but how can you guide me on that you can unmute yourself and speak out what do you want so you already had let's say a situation where this was origin then you know what are the equation of these pair of bisectors correct so i want i want to use this to get the pair of bisector for this fellow how will i do it so you substitute x and y somebody is speaking something so do you substitute x and y in terms of the shifted origin so you take the current equation x square and y square and you reverse shift it to the origin okay right now we are at this stage i want to go to this line i want to go to these pair of straight lines so where should i shift my origin so you you have to shift it to minus x one minus absolutely param is absolutely correct guys origin needs to be shifted to minus x1 minus y1 then only this point will behave as x1 y1 correct yes or no see originally there was a point which was origin okay my mistake i drew this line okay if you want if you want this point to now become a comma b where should this guy go this guy should come at minus a minus b not so if this becomes your origin then only this point will become a comma b so if i want to make something if i want to push something one unit away from me i have to move one unit away behind from that point isn't it so he's absolutely correct if you make these changes if you make this change in the equation of this homogeneous second degree you will end up getting this particular pair of straight lines and same changes you have to bring in this equation to get the new pair of bisectors in other words all you need to do was replace your x with capital x minus x1 y with capital y minus y1 guys please revise shifting of origin concept i find many people getting muddled up here so this equation will become capital x minus x1 square minus capital y minus y1 square given that i know x1 y1 so i'm assuming that i have been able to successfully find x1 y1 by using my partial derivative approach so all i need to do is just transform these equations like this okay by the way we are not used to writing things in capital letters so you can bring back your capital x and capital y in terms of small x and small y so this would be your equation of the pair of bisectors is that fine okay so nothing great here you just have to know your new you can say you just have to apply the shifting of origin concept to the previous equation that we have learned and we are two so can you explain that shifting part once more just yeah sure see we are right now at a position where my point of intersection is origin correct yeah we are we also know their bisector equations these are your bisectors okay i want this guy to become so what transformation should i apply you will say you have to make this point x1 y1 so where should origin flow such that this point is called x1 y1 the origin should go at minus x1 minus y1 isn't it okay because everything is counted yeah yeah okay so if if you shift your origin from zero zero to minus x minus y how will this guy react how will the bisectors react the bisectors react yeah yeah and then this the same pair of straight lines will now become your bisector equations yeah okay at the point yeah okay so let's have one quick question on this yes share x1 y1 is the point of intersection so let's take this question let's say the very same line that we had 12 x square minus 10 x y plus 2 y square plus 11 x minus 5 y plus 2 equal to 0 okay so consider this pair of straight lines now i have two questions one being find the angle between the lines which constitute this pair angle between the lines which constitute this pair and second is find the equation of the pair of bisectors of those two lines my life is actually easy here because we know that the point of intersection was minus 3 by 2 minus 5 by 2 this is actually the old question that we had taken up so that we don't waste our time finding so let's have one minute for the first part of the question that is very simple find the angle that's correct rochir rochir saying correct he is also correct ah vikas silly mistake check through also silly mistake check yeah yeah of course your answer will only be in terms of tan inverse only you can state tan inverse correct rahul correct param okay guys first part very easy it only depends upon these guys the last three terms do not have any role to play so tan of theta would be 2 under root x square minus ab by a plus b you can put a mod so 2 okay now what is your h here is minus 5 so this is 25 minus ab is 24 by mod a plus b is mod 14 okay that gives you 1 by 7 so theta is tan inverse 1 by 7 next is find out the equation of the bisectors now just right done if you're done i don't want you to write that down okay so for bisectors all you need to do is just have this framework in your mind ready this was the basic framework for the equation of the bisectors all you need to do is replace this guy with x minus x1 so this will become 3 by 2 okay minus y plus 5 by 2 by a minus b a minus b is 12 minus 2 and this will be equal to x y x y by h h is minus 5 okay you can simplify it at your own end i don't want you to waste time i have one more concept to quickly wrap up in the last 15 minutes so guys just need your attention over here i know you all are very tired online for our session is slightly tiring okay one last session and we'll almost complete this chapter the last concept that you are going to talk about is the concept of homogenization is the concept of homogenization homogenization listen to this idea this concept says that if you have a second degree equation curve so let's say this is a curve whose equation is a second degree so it can be a conic is it conics have second degree equations so it could be a circle it could be a ellipse it could be a parabola hyperbola and let us say there is a straight line let me call the straight line as lx plus my plus n equal to 0 okay now this concepts of homogenization says if from the origin you draw two lines which connect origin to their point of intersection then the combined equation of these pair of lines so you can see those green lines l1 and l2 so the combined equation of this two green lines l1 and l2 can be obtained by homogenizing homogenizing lx plus my plus n equal to 0 with the second degree equation over here now guys let me tell you this second degree equation is a conic okay it could be a pair of straight lines also getting my point okay so how does it work and how what is the process of homogenization what do i mean by this homogenization means you have to convert this equation into a homogeneous second degree term that means every term here should become a second degree term you must be wondering how it will become a second degree term let me show how so what i'll do here is i will write this equation first like this lx plus my is equal to minus n which can further be written as i have some question here no ah yeah karthik we're trying to find out the equation of l1 and l2 but not separately as a combined equation we want of course there are other ways to do it people will say sir i will sit and find their point of intersection then i will construct these two lines then i will multiply it etc etc yes you can do it like this but the method which i'm going to tell you is a very very handy one and you will get many locus based questions on this particular concept okay so i'll i'll write it over here this concept is useful for locus questions now see how it works so here what i do is i obtained one like this so let me call this as equation one now what i do here since i'm homogenizing it the second degree terms i will not disturb so whichever terms are second degree i will copy them as such now here if you see there's a first degree term this is a degree one term so what i'll do is i'll put a one over here this is also a second first degree term so i'll also put a one over here and see the zero degree term so i'll put a one square over here now this one that i have put over here i will replace i'll replace my one with lx plus my by minus n okay so if i do that i end up getting something like this as you can see here every term in this equation every term here would be a second degree every term would be a every term would be a second degree term okay so we know that a second degree equation a homogeneous second degree equation like this will always represent a pair of straight lines intersecting at origin or passing through origin this is only what i wanted right we wanted these two lines isn't it a pair of straight lines passing through origin so again i'm using the concept that a pair of a homogeneous second degree equation will actually represent a pair of straight lines passing through origin okay so this is your equation of l1 and l2 combined and let me tell you it looks very ugly in this present form but when you are actually applying it you'll be able to do it very very conveniently let me show you in a question uh anybody who is willing to copy something should i go to the next sheet sir i have a doubt sir yeah oh sorry yeah tell me ruchir sir you said that this ax square plus 2 hxy could also be a pair of straight lines but if it's a pair of straight lines won't there be four points of intersection see i'm talking about something of this nature cut and these two points are connected to all this okay sir yes yes this is your lx plus my plus n line and this is the conic which is represented this is a spare of straight line represented by that second degree yes thanks so what's the proof that these straight lines will intersect at those points what is the proof that they will meet at origin no uh what's the proof that the lines will intersect in those same points that the first line in the conic intersect because you aren't you simultaneously solving it the same point which will satisfy the line will also satisfy the conic the homogenization actually is taking care of that isn't it see let's say here if you had a point x1 y1 won't it make simultaneously the x y of the line and x y of the conic zero simultaneously over here yes okay see only only term that would have you know missed out was this term but this is also taken care because x and y will satisfy line equation also right yes sir so it takes care of the fact that these lines l1 and l2 will pass through the very same points where the blue line is meeting the conic okay let's have so initially when we um take the first line with l and we try and homogenize it that that line is the equation of that line is for which line uh once again come again uh Dhruv like um during homogenization we start off with an equation for a line right it would be given to you yeah but which line does that represent is it the line that intersects the conic yeah this is the line which intersects the conic and the one we are finding it are these white lines okay so when we homogenize the line that intersects the conic we get the other two l1 and l2 absolutely absolutely we are homogenizing this guy with this guy to get this guy okay sir okay so here is a very simple question for you uh prove that the pair of state line joining the origin to the intersection of the curve by the way if you are not able to see this it's actually our ellipse equation by this line are coincident if this condition is met okay let me help you out with this because this is the very first problem you are solving on this concept so akshit is done anybody else varam is also done now let's try to understand oh sorry guys uh load shedding happening am i audible yes sir yes sir what i'm going to do is i'm going to homogenize this with this line so these two terms they are already second degree i will not disturb them what is not second degree is this guy one okay you could have you could have brought the one on the other side also doesn't make any difference to our approach but this one i would have to homogenize it by using the fact that one could be written as lx plus my by minus it okay so when we do that we end up seeing something of this shot okay now what i'm claiming is that this represents a pair of state lines which will be not only passing through origin but also coincident so can i say this is coincident my h square should be equal to ab first of all okay origin is already taken care of because it is a homogeneous second degree it's a homogeneous second degree equation okay now what is my h what is my a and what is my b here let us try to get those terms from this equation so x square coefficient here is one by a square minus l square by m square if i'm not wrong y square coefficient is one by b square minus m square by n square and minus two l m x y by n square will be my the remaining term the first thing where we said x square isn't it l square by n square yeah i'm so sorry yeah thank you for making that correction correct now let's try to apply the formula h square is equal to ab it should be l m by n square minus okay let's do one small thing since there's an n square over here i'll get rid of it by multiplying throughout i'm just using a simplification over here it's not a mandatory step and here i'll get minus two l m x y okay so h square is like l square m square okay and ab will be like n square by a square minus l square and and n square by b square minus m square correct let's see okay we need to simplify it even more so why why do you take the condition n square equal to ab because they're coincident so coincident the angle must be zero oh okay the parallel condition is the angle between zero is a sufficient condition also because they're passing through origin and they are parallel they have to be coincident yeah yeah correct so that's why i use that okay let me square this up okay i'm just simplifying this you could have done this at your own and also cancel off this l square m square from both the position both the sides okay you could even drop n square term also conveniently isn't it so you can write it as uh let me write n square by a square b square is equal to m square by a square and l square by b square now finally touching uh the uh just final touch up here so l square a square l square b square m square should be equal to n square a square b square i'm canceling from the denominator is this what we wanted to prove yes this is what we wanted to do okay so this is a very good example where you could use your concept of homogenization any question here anybody i didn't know why we use the condition of coincidence because it is given in the question here akansha so but in index lines are coincident look here yeah anybody else so we have to prove that they're coincident because we took the condition that they are coincident okay see coincident if this and vice versa will also be good okay fine so if they are coincident then this will be satisfied and if this is satisfied they are coincident okay sir at square is equal to ab isn't that the condition for them to be parallel and not coincident this is didn't i write this is a condition for both the situations parallel and coincident yes sir but but using the parallel condition would have said only parallel and not coincident but it's homogeneous right so homogenous yeah homogenous parallel will automatically become coincident yeah if homogenous if homogenous okay okay guys we are done with this chapter probably we'll have to solve more questions i'll be requesting you to solve the level wise problems on the rank booster people who are just newly added i'll send it once again to you okay so please complete that before you attempt your module questions especially concept of homogenization etc they're very useful in locus base so try to apply them for solving locus base questions so next class that we meet we are going to take up a theory of equations chapter and then we'll get back to the concept of the left over part of phonics