Transcript for:
Fundamentals of Coordinate Geometry

for for e hello good evening good evening my Ninja Warriors what's up what's up so I can see a few people have joined in darani what's up Sadia good evening sadya hi Som hi NADA hi a good evening good evening welcome to the session and you know what what you have to do as soon as you join the session you guys have to like my session and share the session with your fellow J aspirants right okay all right like it right away okay all right good evening good evening keep joining keep coming okay so today's session is going to be all about straight lines and as you have I'm not sure whether you have seen the thumbnail or not so today's session is going to be all about about basics of coordinate geometry that is basics of qu distance formula section formula right along with that we will progress forward and we will learn about what are the important centers related to the triangle and then we will talk about Locus related problem and shifting of origin so today's session although it is titles titled at as straight lines but today we are not going to touch any form of straight line that will be the second session as you can see I have already uh scheduled one more session for straight line where in we are going to talk about the rest of the remaining portion of straight line which is various forms of straight lines and pair of straight lines family of straight lines and so many things okay so today's session I hope the like uh the agenda is clear that we are going to talk about basics of com coordinate geometry uh section formula distance formula and related to Triangle Locust shifting of origin as well okay all right I am absolutely good darani how are you how are you good gun I am seeing you after so many days okay so yes for all of those who are new to the channel any student who is new to the channel or new to my session this is my brief introduction my name is n and I have more than 10 years of teaching experience in both online and offline as I can clearly as as you can clearly and I have produced top result top selections in iits nits and various reputated colleges as well okay all right any session Saturday go are you not seeing any any uh session scheduled is it Saturday Co any session you are not seeing Saturday Co I will schedule it if it is not schedule I will schedule it okay the Saturday Co again we are going to do straight lines because straight lines is not very small chapter and you know what straight lines is very very important chapter yes let me just talk about straight lines a bit straight lines actually if you understand straight lines nicely now then let me tell you what are the chapters this chapter is going to make an impact on so your circles chapter will become easy all the parabola ellipse hyperbola po conic section whole conic section parabola ellipse hyperbola right is depending upon basics of straight lines and I would say straight lines and circle both these two chapters are basic building block of complete conic section along with that most of you do not know that 3D geometry which is there in your class 12th right so 3D geometry has a major thing from straight lines yes so it is going to impact 3D geometry by the way 3D geometry is one of the high weighted chapters of overall maths not even 12th of overall 11th and 12th so I will suggest that straight lines even though that I feel that one question you can expect from this chapter but you know the importance is not limited to one question or four marks importance of straight lines is much more than what do you see in your J main question paper because without straight lines circles you won't be able to understand without straight lines conic section you won't be able to understand and without straight lines 3D geometry is also a difficult nut to crack okay all right so are we ready are we actually ready for the straight lines okay so I will start with very basic and you know what today's session is all about Basics only we are going to touch Basics only in today's session so today's session will be a confidence booster session okay and the next session will be highly important session okay all right so let us start let us start let us start what is mod mod modulus good evening nothing nothing to say aish Shri ready H and also you have to uh bring your notes as well uh do you like do you when I teach you in this series in this Roar series you know what do what do we do in Roar Series in Roar Series in one short we cover the whole chapter okay so what you need to do is you need to bring your pen and paper and then along with me you have to solve it especially in mathematics come on without pen and paper J mathematics is not at all possible okay so today's session how long so we'll try to wrap up the session within like 2.5 or 3 hours okay so today's session is not going to be that long like 6 six hours hours long because we have divided the straight lines in two different portions okay all right good evening s Good Evening Sun J what is aod1 application of derivatives part one maybe you are saying CH now let's come back to this particular portion okay all right so coordinate plane so you know coordinate plane right coordinate plane this is your x-axis this is your Y axis this is your minus x-axis this is your minus y AIS yes or no okay so in the first quadrant so this is your first quadrant and this is your second quadrant right this is your third quadrant this is your fourth quadrant now I'm going to tell you what you already know in 10th class in 10th class you have already studied this right binomial theorem right after this session like right after straight lines I'm planning to uh schedule okay nothing to say CH first quadrant may you are going to tell me now you are going to tell me tell me in first quadrant X and Y both are positive or negative X and Y so over here in first quadrant clearly this right hand side of X is positive X left hand side is negative X so in first quadrant clearly positive X is always positive and y y is always again positive in first quadrant so in first quadrant both are positive right so if I say 2 comma 3 is a point so you will directly jum jump onto this point and tell me that ma'am 2 comma 3 will lie in first quadrant now let us come back to Second quadrant now the second quadrant me clearly we are towards the negative x-axis so X will be X will be negative and Y will be still positive because we are above this point right so Y is positive so if I say - 2A 3 what you are going to tell me tell me - 2A 3 okay - 2A 3 - 2A 3 ma'am second quadrant because X is negative and Y is positive right I can see NADA okay NADA is doing PG mathematics thank you thank you NADA thank you so much all right third third quadrant H so in third quadrant oh third quadrant May X is also negative and Y is also negative okay so both negative -2a -3 the moment you see this coordinate you will be able to tell me that ma'am third quadrant you are talking about third quadrant exactly and if I talk about fourth quadrant my dear students fourth quadrant it is very obvious X is positive and Y is negative that means minus + 2 comma -3 is that point which will lie in fourth quadrant very basic right very basic okay all right now distance so this is your coordinate system I hope that in class 10th you have already studied this right this thing to you already have studied right now one more thing which you already have studied we are just brushing up which is distance formula okay now what is distance formula let us understand so distance formula May uh basically what we will talk about is distance between two points okay so distance between two points we are going to talk about let us take that first point is this let us take P point and the second point is this which is your Q Point okay P point and Q point now I'm going to assume I am going to assume where what is the aod M mean value are not talking about that H straight lines aod will come then we will talk about it right now we cannot talk okay so P point I am assuming that P point is X1 comma y1 okay and Q point is X2 comma Y2 correct and we are looking for distance between PQ so let us first join this PQ correct PQ join and now PQ distance let me just clearly write down the Formula First and then talk about the derivation so PQ is nothing but X2 - X1 square plus Y2 - y1 Square okay PQ is nothing but X2 - X1 square + Y2 - y1 Square now you have to think a little bit how to prove it ma'am how did you get it exactly so if you clearly see this is actually looking like a Pythagoras Theorem wherein this PQ is hypoten hypotenuse is square root of perpendicular square plus base square right so in here also we are going to try to do that first of all let us complete the triangle okay so this is my triangle this is my triangle p q let us say this is your R okay now what we want in triangle pqr if I want to figure out the PQ length I need PR length that is base and I need RQ length that is tell me tell me RQ length will be perpendicular okay so I am looking for this distance the this this distance now just tell me one thing tell me so this point is x1a y1 what does that exactly mean X1 comma y1 if any point is having X1 comma y1 what does exactly mean that means this P point I have got so if I go X1 distance horizontal so this length this length is X1 these two length this this length is also X1 and this length is also X1 okay so X1 comma y1 means X1 means always horizontal distance so we have woged horizontally X1 and we have Ved vertically y1 y1 this y distance is what vertical distance so this is your y1 okay this is your y1 so this is your y1 clearly and this is your X1 so now can you just tell me what is this point will called so on x- axis Y is always zero and the x coordinate will be distance from the origin so this is your X1 if this is X1 comma y1 this is X1 comma 0 similarly if this is X1 comma y1 this is 0a y1 because on y AIS X is zero right yes D what is doubt what is your doubt yes ask me uh root have I what happened the why sad smiley okay all right so I hope that you know this this distance is X1 and this distance is X2 right the complete distance this distance is X2 and I am actually interested in so this distance is the bigger distance is what X2 this is X1 so tell me the distance PR so PR distance will be the bigger distance X2 minus X1 this is your PR okay what this is your PR and what about QR so QR will be this bigger distance minus smaller distance bigger distance is what so this is complete Y2 can you see this is Y2 so this distance is your Y2 y because this coordinate is Y2 that means the vertical distance from this axis is your Y2 and if I talk about X2 so yeah so this is Y2 complete and this is y1 and if you are looking for QR Y2 minus y1 so this is clearly Y2 - y1 so this is your perpendicular and this is your base so perpendicular and base are right here so hypotenuse is equal to squ < TK of p² + b square correct or not so that's why we have applied okay that's why we have applied p² + b sare p is what Y2 - y1 Square this is your perpendicular square and X2 - X1 Square means base Square okay so perpendicular square plus base square is your PQ that's how we have figured out the distance formula B clear or not if distance between two point points given shall we use plus or minus sign ma'am I am actually is this for first year or second year so this is for J Main and uh I believe that for like uh your school if you're talking about in comparison with school so this is for first year right okay first year Pythagoras Theorem yes Pythagoras Theorem Pythagoras theorem in the Triangle so shall we start and solve in some problems on distance formula D and then let us understand then let us understand that what is your doubt okay danani let us solve some basic problems the distance between these two point is what we are looking for okay so this is distance with this is point P this is point Q all we need to know is just distance formula so PQ will be what square root of X2 - X1 s + Y2 - y1 Square you know this is X1 y1 this is X2 Y2 correct okay so PQ distance will be what tell me X2 - X1 that means 7 - -3 7 - -3 so 7 - -3 square plus Y2 - y1 will be - 6 - 4 squ okay all right so PQ distance will be what 7 + 3 10 10 square is 100 plus again 10 square 100 because square is there the minus sign will not make any sense so this will be 200 K root 200 root will be 10 < tk2 units 10 < tk2 units option b any doubt any confusion any doubt any confusion till now just tell me I am right here because in straight lines what happens is beginning is very very easy okay and then gradually you will not know that exactly when so it is like this level one level one level one level one level one level two level two level two level three bounce Okay so so that I will make sure that nothing is going to be bouncer okay all right okay now again one more formula one more question on distance formula but with a little bit knowledge of trigonometry let's see how many of you remember the values how many of you are practicing self-study how many of you are doing selfstudy so distance between these two points again this is X1 y1 this is X X2 Y2 correct so PQ will be now I'm not going to write the formula okay now you need to apply it directly let's do that to X2 - X1 Square to 0 - a sin 60° square plus Y2 - y1 Square to cos 30° a COS 30° - 0 square all right clear so PQ will be what tell me so this is a sin 60° you know what is the value of 60° sin 60° it is < tk3 by2 so this is going to be a < tk3 by 2 whole sare plus a COS 30° cos 30° sin 60° both are same aunk 3x2 whole s clear so this is actually going to be 3x 4 this is also 3x 4 so 3x 4 a² multiply 2 because twice it is written 2 1 2 2 2 are 4 so this is going to be a square will come out so it is going to be a < tk3 by 2 so so so so so a roo tk3 by2 option b again easy peasy okay all right no no no no option b this is wrong this is wrong this is wrong wrong wrong wrong wrong wrong wrong wrong wrong wrong wrong wrong okay why wrong because option C Is Right now the moment you take a square out of square root it will become modulus my dear students okay everybody is saying B now option me you can select now distance is never negative have you seen distance negative okay so please remember please remember square root of a square is equal to modulus a will you forget it okay these very easy things Jungle Book is not available you guys can help Juju thoughts because uh Jungle Book is available to everyone maybe sometimes U you can try with your another device try once with another device uh maybe it is not getting opened on your phone maybe okay you can check with your like uh parents phone or uh another phone friend's phone once okay all right clear clear clear option C one more question is there on our plate okay CH now can you see the beginning question says this now please understand this what is this this is your second quadrant Alpha lies in second quadrant okay for visualizing in this is your second quadrant now distance between the points a distance between the points so again PQ direct formula can I apply are you comfortable now right so this is going to be X2 Y2 and X1 y1 0 - tan Alpha whole Square so that is going to give you tan Square Alpha + 1 - 2 square to 1 square okay then PQ will be what 1 + tan Square Alpha is SEC Square Alpha again you might do the same mistake again you might do the same mistake my dear students PQ will be equals to actually mod SEC Alpha right mod SEC Alpha mod SEC Alpha so obviously option A is wrong and B is also wrong either of these two depending upon what is happening okay so mod SEC Alpha now just tell me one thing distance to will never be negative H now I can see so many options C are already there SEC if I talk about SEC SEC is actually negative SEC Alpha is negative why why because Alpha is in second quadrant and second quadrant may only sign or a second quadrant only sign is positive so SEC is negative now so if SEC is negative just tell me mod SEC Alpha will become minus SEC Alpha so many of you giving me the incorrect answer option D is absolutely right my dear students what happened to you what happened to you dhani again just tell me one thing you might be thinking Ma'am why are you saying D why are you saying D because you will say ma' distance you told us now distance cannot be negative then why are you like taking the negative answer you have to take the positive one now so it should be C why it is D who is going to tell me why is it D so just tell me one thing AA let us pick any angle which you are comfortable in do you know 120° can we take it or 135° so let's take Alpha is equal to 135° or we can say 3 < by 4 here cos Alpha is - < tk2 negative obviously so SEC Alpha will become what uh sorry this is - 1 byun2 no h let me just explain - 1 by < tk2 because second quadrant right so this is going to be - < tk2 by 1 correct this is your SE Alpha so what you are saying is what you are saying is ke if you if C option is correct CH let us for an instant for of one time let us see that SEC Alpha this SEC Alpha is the answer if this is the correct answer then tell me the value of SEC Alpha will be what a negative value or a positive value obviously a negative value why because if Alpha is in second quadrant SE Alpha will be a negative value that means option C when I put when I'm putting the value although it is looking positive it is looking plus SEC SEC Alpha but actually it is a negative quantity and distance cannot be negative my dear students okay SEC minus SEC Alpha is actually a positive number tell me why because SEC Alpha is itself negative the value of SE Alpha will be minus of some positive Co like -2 negative < tk2 type so minus < tk2 if you plug in the value over minus minus will become plus that's why D option is correct so how do we understand ma'am first of all this understanding is clear that out of c and d d is positive actually if I tell tell you about the process if I tell you about the process mod SEC Alpha so modulus how does a modulus function work one student was asking also how does a modulus function work so just give me a little space in your brain can you give me a little space in your brain okay all right so mod of X will be X Mod X Mod of X will be X if your this function this this expression is positive if this is negative then you write Min - x okay so mod of any quantity will be equal to either SEC alpha or minus SEC Alpha depends on what depends on this SEC Alpha if your SEC Alpha is positive then do not worry about it just go ahead and write mod SEC Alpha is equals to SEC Alpha but if your SEC Alpha is negative because Alpha in second quadrant the SEC Alpha will come negative definitely then you have to write it minus SEC Alpha minus SEC Alpha so that's why minus SEC Alpha in this case will be the correct option okay all right clear CH now let us do one more question now let us get better with each and every question a triangle with vertexes this this and this is what a triangle with vertexes this this this and this what so for that let us take the help of distance formula because what will happen is the moment all the sides are equal then it will become an equilateral triangle right and the moment two sides are equal then it will become isoceles triangle and so on the moment it will satisfy Pythagoras Theorem it will become right angle take the distance formula will give me anything everything so p q and R now let us calculate PQ can I calculate directly okay are you comfortable now directly PQ right so -1 - 0 square plus sorry -1 - 4 Square now X2 - X1 we are doing to -1 - 4 Square then - 1 - 0 square this is your PQ so this is 5 Square 25 the root 26 okay all right PQ then we are here for QR let us calculate QR okay QR so this is going to be 3 - - 1 square Q R 3 - -1 Square okay and 5 - -1 Square clear so this is actually going to be what 4 Square 16 and 6 Square 36 so 16 and 36 this is going to be what 52 AA last QR then p r or RP this this thing and this thing distance between PR can I do that okay so to 3 - 4 Square X2 - X1 and 5 - 0 square so this this is going to be 1 + 25 < TK 26 okay first of all let us eliminate equilateral triangle either it is isos or right angled isoceles why because clearly PQ and P are equal H so if I just do this PQ and P R both are otk 26 < TK 26 and your QR is what < TK 52 so last thing which we need to check is for the right angled so if I check the right angled if I check the right angled then what will happen this is the longest side longest side can be the hypotenuse hypotenuse so hypotenuse Square should be square of other two sides so 52 should be equals to 26 + 26 which is actually satisfying that means it is isosceles plus right angle option D is absolutely right and uh yeah most appropriate answer is D okay all right yes serella yes supti correct clear any doubt any confusion let me know okay so that's how the application of distance formula so this is what it is okay now the last question no actually this this we will do in uh when we know the equation of lines okay so let let us now directly jump towards internal division formula section formula okay so section formula now let us understand from the very basic first of all let us take a line A and B let us take two points A and B where a is having X1 comma y1 as the coordinates and B is having X2 comma Y2 as the coordinates okay and now in the middle of A and B There is a point P can you see the P point P point now P points P Point divides line joining a and b p divides line joining A and B internally that means that between A and B exactly okay and in the ratio m is to n so the ratio is m is to n what does that mean that AP upon PB AP upon PB is nothing but M upon n okay all right AP upon PB is M upon M so now we need to know the coordinates of B did you understand the question number one the problem statement is suppose you are having a and b two points and there is a point P which divides the whole line segment in the ratio m m is to n then you need to calculate the coordinates of P okay so there is a trick okay there is a trick I'm not going into the proof but there is a direct trick to calculate or you can say the formula trick is there okay so what is that formula trick let me just show you you need x coordinate of p and you need y coordinate of P right okay so x coordinate of p p will be crisscross method so there is a crisscross method so this m will get multiplied with X2 M X2 and now the n will get multiplied with X1 right so M multiplied with X2 n multiplied with X1 so MX2 + nx1 upon m + n this is the formula for your x coordinate of P Point okay yes D you know the formula and now similar if you remember the X formula then you you have to remember the Y formula just replace X2 and X1 by Y2 and y1 so m M Y2 M Y2 cross method okay crisscross M Y2 plus NY1 upon m + n okay distance formula section formula all right any doubt any confusion so yes distance formula say uh sorry what did I say no distance formula right section formula so section formula is MX2 + nx1 upon m+ n now there is a special case of it there is one special case of it that is midpoint okay so special case says midpoint midpoint means now A and B are two points but this p is nothing but a midpoint of A and B what does that mean that means AP and PB are equal or we can say that the ratio is 1 is to one these two portions are equal here m is to n actually is equal to 1 is to 1 so what we need to do is we need to replace M and N both by 1 okay so X2 + X1 upon 1 + 1 2 so what is your formula X = X2 + X1 upon 2 and Y = Y2 + y1 upon 2 all right yes clear any doubt any confusion do let me know okay so this is your midpoint formula now internal Division if it is clear now then external division what is external division now the situation is same A and B are two points but this time P point is not in the middle right P point is not in between A and B rather this P point is actually lying outside externally to the line joining A and B right so the formula for this is very similar to internal division but just a sign change you have to do so if you have to calculate the coordinates of Point P what you need to do is uh uh uh so yes m is to n is the ratio so again AP is M and PB is n right so the formula is what again that crisscross form fora so x coordinate will be MX2 - nx1 remember so what was the original formula plus was there just replace this minus sign upon M - n and y will be M Y2 - nx1 upon M - n okay all right take you will understand external division better once we start solving the problems and I'm here for that right problem solving CH let's start okay yes let us start find the length of median from vertex a of a triangle ABC whose vertexes are a b and c okay a so basically there is a triangle ABC whose vertex is given to you in straight lines the thumb rule the thumb rule is to plot the situation not even in straight lines but also in circles Parabola ellipse hyperbola and 3D geometry all these chapters will actually test your visualization skill okay and the more you uh paint the picture of this question let's paint the picture of the question the more you the situation will be clearer the question will be clearer the high chances that you will be able to solve that question so although even though you know that triangle ABC looks like this but plot plot plot plotting is the key right so a b c a Tell Me -1a 3 B is what 1A -1 and C is what 5A 1 this is your triangle right length of the median from the vertex a now you do you know the median anybody who can tell me the like meaning of this median the median is nothing but from the vertex a you drop a line towards the base B opposite base from the vertex a because they are telling us now from the vertex a you are dropping a median median means median mid median means actually mid so basically this line joining midpoint of BC and a so basically median will be what median from a is nothing but what line joining a and midpoint of BC understood what is your median now okay so line joining a and midpoint of BC okay all right now to get the length of that median we need to know both the points suppose this is your X right X is your median so we need to get calculate length of median that means we are looking for ax now a coordinates we have right the coordinates of a we have if someone can tell me the coordinates of X then my question will be done done yes or no now X is what x is nothing but the midpoint of BC so to calculate X we need to apply the midpoint formula what is midpoint formula right so that is going to be so the coordinate of X is what x will be 1 + 5 by 2 and Y will be what -1 + 1 by 2 are you getting my point X1 + X2 by 2 y1 + Y 2 by 2 okay so this is going to be what tell me so this is your clearly 6x2 that is uh 6X 2 is 3 so X is coming out as three and this is coming out as zero so this is going to be 3 comma 0 is your midpoint right so now what you're looking for is your a is clear and your so a point is -1 comma 3 and this point is 3 comma 0 this is your triangle this is your a this is your X now apply the distance formula in this okay square root of 3 - -1 and 0 - 3 correct so this is 4 Square 16 3 Square 9 oh my goodness 5 is the answer ma'am one formula is wrong aan which formula which formula are you talking about five is the correct answer everybody understood any confusion any doubts so far all right now okay let us solve the another type of question okay again Basics we are doing Basics still now mid points of the sides of triangle are given find the coordinate of its vertex oh my goodness so there is a triangle okay all right so we can name the triangle right let us take the triangle is a b c and for every side midpoint of the side of triangle are given so this is midpoint of ab let's say so 2 comma 1 is p point which is midpoint of ab this is your Q -1a -3 and this is your R right that is going to be 4A 5 what we are looking for is coordinates of a b and c coordinates of a b and c in Y part formula is wrong here oops huh so this is y1 huh H take understood corrected CH now we need to look for a b and c okay so let us assume something let us assume that coordinates of a will be what A1 B1 this is A2 B2 this is A3 B3 correct okay and now let us apply the formula let us do that okay all right so let us apply the formula first of all let us apply the formula over here take 2 comma 1 is midpoint of ab correct or not 2 comma 1 is clearly midpoint of ab it is given in the question so can we just write A1 + A2 by 2 is equals to what 2 and B1 + B2 by 2 is equal to one right can we do that just the it is just the game of equation solving now A1 + A2 is = 4 and B1 + B2 is = 2 this is first set of equations right and now second similarly now -1a -3 is midpoint of tell me tell me ac correct or not so what we can do is minus1 will be equals to so A1 + A3 by 2 is equal to -2 can I write directly -2 and and and and a B1 + B3 B1 + B3 will be - 6 right upon so basically B1 + B3 by 2 is what - 3 multiply two and we are going to get it okay all right how are we getting it so so 2A 1 is the midpoint of so we are directly applying the formula A1 + A2 by 2 is this are you getting or not CH so this is your second equation similarly now about this can I just write it directly A2 + A3 tell me A2 + A3 is what 8 and B2 + B3 is what B2 + B3 is 10 okay so we have got how many equations just copy the these all these equations on the next slide can I do that A1 + A2 is coming out as [Music] 4 right then A1 + A3 is min -2 and a A2 + A3 is 8 now how to solve the system of equations how to do that so what we need to do is we need to add all the equations so if I add I'm going to get twice of A1 A1 twice A2 A2 twice A3 A3 twice so A1 + A2 + A3 all the values twice will be equals to 4 + 8 12 12 - 2 10 so clearly A1 + A2 + A3 is what 5 that's how we solve the equation now what we need to do is A2 + a A1 + A2 we can plug in the value over here in this equation A1 + A2 is what 4 so can I replace A1 + A2 by 4 so 4 + A3 is 5 okay ma'am this implies A3 is 1 okay all right and now if you talk about A1 + A3 if in the same equation you put A1 + A3 so A1 + A3 will be 2 so 2 + A2 will be 5 this will give you A2 is equal to 3 and similarly using this equation A2 + A3 you can replace it by 8 so what will you get A1 + 8 is = 5 so A1 + 8 is = 5 A1 will be equals to Min -3 so now you have got all the three values A1 A2 A3 okay and now what you need to do is same thing you have to repeat for B1 B2 B3 are you ready for it so B1 + B2 is clearly 2 ma'am yes B1 + B2 is 2 you are correct then B1 + B3 is- 6 okay and B2 + B3 is 10 tell me what to do add everything B2 + B3 is 10 correct 2uh add everything to B1 B1 coming out twice B2 B2 also coming out twice B3 B3 also coming out Thrice 10 - 6 + 2 4 + 2 6 so 6x2 3 so B1 + B2 + B3 is equal to 3 if you use a A2 value A2 value this isus 2us 5 + 2 7 sorry s s A2 value s A2 is s ah it is ly what see I am doing subjective approach I'm writing also H do not worry it is not that lengthy A2 is 8 huh eight A1 + A3 is min-2 right huh -2 -2 if I put -2 there plus s s seven huh right so up just put B1 plus B2 if you put two over here to B3 1 B1 + B3 to B2 9 and B3 B1 sorry B1 uh- 7 coordinates -3a - 7 is your a coordinate A1A B1 a2a B2 is 7A 9 and finally A3 comma B3 is 1A 1 let me check let me check let me check yes adika absolutely on point can you see adika answer is matching H okay yes dju okay I will not use Hindi at all 1A 2 VRI why 1A 2 it should be 1A 1 right right oh 1 comma 1 clear CH let's talk about okay last question on distance formula hopefully then we will move on to our next topic last question on distance formula are we ready for it all my Warriors are you ready for it let the angular opposite points of a parallelogram so this time they are talking about parallelogram okay ma'am let us plot a parallelogram okay now in a parallelogram they are saying angular opposite points angular opposite points that means 3A 4 and 1 comma -2 are angular opposite points so this is what we call angular opposite opposite angles right these two are opposite angles so that's why angular opposite points okay now coordinates of remaining two points are 6A 1 and X comma y so the this point is 6a 1 this point is X comma y so if you just want to name it PQ r s okay and we need to figure out the coordinates of X comma y that is that is s how will we do it anyone can tell me think about it okay think about it you need to think also think think think anybody can tell me in the chat box okay so it is a bad idea actually if you just calculate PQ PQ though we can calculate I know but it is very bad idea if you want to calculate PQ and the length if you want to equate the length no it is not going to solve your problem okay it will make your problem a little bit lengthier so we will use a parallelogram property which we already used in uh 10th class right midpoint property so if I join this PR this is your diagonal right so the midpoint of diagonal will be let's say this pqrs let's say midpoint is X okay X is midpoint midpoint of diagonal is X can I calculate this midpoint by the way so to calculate the midpoint what we need to do is 3 + 1 4 by 2 right so 2 comma 4 - 2 by 2 4 - 2 is 2 2 by 2 is 1 yes already VI Shri shria absolutely correct SRI viya no finding all the distance is a bad idea you have to think about Innovative methods because you don't have time okay everybody will will actually uh think about that only the very first thought distance equal no I am telling you now stop thinking like that okay so this is midpoint now if I draw one more uh diagonal then one more diagonal has also has a midpoint and that is also X right so basically diagonal number two basically midpoint of P midpoint of p r will be equal equal to what midpoint of of Qs now midpoint of PR will be what tell me we just calculated 3 + 1 by 2 comma 4 - 2x 2 this is midpoint of PR correct this will be equal to midpoint of Qs Qs so 6 + x by 2 comma 1 + y by 2 correct so this is going to be 4X 2 so 2A 1 will be equal to 6 + x by 2A y + 1 by 2 now equate 2 will be equal to this so 6 + x by 2 is 2 2 2 4 x = -2 and over here y + 1 is = 2 to Y will be y + 1 is = 2 y will be equal to -1 so -2a -1 is the answer who is telling me - 2 comma -1 which three - 2 comma minus one no sorry n two sorry plus one plus one plus one plus one yes she is absolutely right from the very beginning H okay give me a big thumbs up if you are understanding this all right and now let us talk about one question on external and internal division okay all right so in this question they are talking about in what ratio does The X AIS in what ratio does the x-axis divides the line joining these two points okay so first of all I have already told you that whenever you see straight line question plot the situation plotting the situation will help you a lot okay so this is your x- axis now 2 comma - 3 2 comma minus 3 will look like this 2 and minus 3 will be somewhere over here okay and 5 comma 6 will be here so this is your a this is your B now line joining AB so let us join this AB now let us read the question again at what ratio does the x axis x axis so this is actually that point P so Point P divides the line segment in which ratio they are talking about now listen to me carefully this question I have kept it especially very basic question to let you know that never ever assume a ratio as m is to n do not do that mistake what you need to do is you need to assume Lambda is to 1 you know Lambda is not important basically the second m is to n May the second ratio should be one and the First Rate ratio should be that variable m is to one also you can take X is to one also you can take okay the Lambda is to 1 is the ratio of is the ratio which we are assuming but you will say ma'am how can you say that like like the second MS2 and like the ratio will be 3 is to 1 4 is to 1 how can you say how can you say I am actually not saying that what I'm saying is you put ratio Lambda is to one because any day one variable is easier to solve and Lambda itself will be that value I will show you how okay oh okay okay so now we need to apply that now Point P just tell me one thing the coordinates of Point P will be what x comma Y no the coordinate will be X comma 0 because on x a very important on x- axis y will be equals to zero on x- axis on x-axis x-axis p y is always zero now you just apply that formula a point is what 2A - 3 B point is 5A 6 and clearly now crisscross method Lambda * 5 so X will be equals to now we don't have to do that X we are not going to apply that formula why because I have only one variable apply that y y coordinate will be equal to Lambda 6 + 1 * - 3 upon Lambda + 1 what did I apply and this is zero because y coordinate of the point p is zero my dear students so I just want to know the value of Lambda H and I have two options to match to match what match the x x coordinate the section formula we are applying MX2 + nx1 upon m + n that will not lead me anything now y coordinate because it is zero now it will give me the answer right so this is if you solve it Lambda is going to be 1 by 2 so what is the ratio tell me 1X 2 ra 1 now can you see automatically it happened okay so this is going to be 1 Rao 2 1 ratio 2 answer is 1 by two the which three from the beginning only which she is saying yes yes ma'am vishi huh Vish is saying that 1 is to two ma'am 1 is to 2 1 is to two so this question had actually two takeaways what are the takeaways from this question just let me know in the chat box number one takeaway is whenever a ratio is asked or whenever you have to take a ratio never ever take m is to n always take Lambda is to 1 that is one takeaway second takeaway what is second takeaway just tell me second takeaways on x-axis Y is always zero that question will not give you but you have to be alert you have to show your presence of mind okay all right now let us move on to the next topic which is area of a triangle are we ready for it area of a triangle okay area of a triangle ABC so for that we need to have a triangle H triangle can you clearly see on your screen a is x1a y1 B is x2a y1 C is x3a Y3 now there is there are actually two methods one by one we will see so area of triangle is half now you have to make that determinant W approach you have to make a determinant now in First Column you have to write X1 y1 in second column you have to write X2 Y2 in third column X3 Y3 and and fourth column again back to X1 y1 okay ma'am so whenever three points are given I need to make four columns H X1 y1 X2 Y2 X3 Y3 and one more back to like cyclic order Start From Here X1 y1 in First Column X2 Y2 in second column X3 Y3 so this a b c a so this is a b c a now what you need to do is you need to do crisscross criss cross criss cross can you see so how to do the crisscross have a look X1 Y2 minus y1 X2 right so let us open it crisscross X1 Y2 - y1 X2 plus plus now next again X2 Y3 minus Y2 X3 plus next now what we are having is huh here it is already written can you see X1 Y2 and X1 Y2 and y1 X2 what about this X2 Y3 and X3 Y2 next is this X3 y1 and X1 Y3 if you if you rearrange what you will get is half X1 1 Y2 - Y3 X2 Y3 - y1 and x3 y1 - Y2 it can also be written down as this way what is this this is a determinant way so this is uh the method number one yeah similar to determinant and exactly determinant now another method so area is also equal to half First Column will be all the x coordinate X1 X2 X3 second will be all y coordinate y1 Y2 Y3 third column will be one the both will give you same results okay all right but you must be wondering K ma'am actually this whole if I tell you used met H used to Method number two you are you may be used to Method number two now you must be thinking ma'am actually both the methods are not that difficult and this is a direct formula so like at what point J main can trap us exactly at what point in this concept because if if they give you all three X1 y1 X2 Y2 X X3 Y3 if they give you all three vertex you are going to put the value you are going to get the area okay now just ask me this question ma'am okay this is a very straightforward question at exactly what what is that thing where J main or J Advance can trap us what are the where are the chances of making mistake ma'am so mistakes will happen when they will actually not give you the complete vertices instead of that what they will give you is they will give you area area of triangle will be provided to you okay and coordinates X1 y1 will also be provided X2 Y2 will also be provided X3 Y3 may they will actually hide the Y3 so they will give you Y3 as any constant A and B any variable basically any unknown so one unknown they will give you from any of the vertex everything rest everything including area will be given so now the chances of making mistake will be what so area for area you have to apply the external modulus okay so in this question modulus is important if you forget the modulus now H majority of the students will forget the modulus in Reverse question do you understand what what is reverse question wherein they will give you the area but they will hide any Vortex and if you forget to apply the modulus then you going to make a major mistake that we are going to do it okay all right clear this is a very basic question this is a very just N Things okay so this is a very basic question first of all let us understand like how to apply the formula then let us just talk about how to defeat how to beat the N okay so area of the triangle so area will be what half now 32 so this is X1 y1 first row will be 3 2 1 this is X2 Y2 11 8 1 X3 Y3 8 12 1 okay and now we need to apply modulus because area cannot be negative right okay so area will be what modulus of Half of Half will be there now just start expanding the determinant you know how to expand the determinant so this is going to be 3 8 - 12 8 - 12 is -4 then - 2 11 - 8 11 - 8 is 3 + 1 last so this is going to be uh 11 into 12 so that is going to be 132 - 64 okay so area will be equals to half will always be there take and this is -12 - 6 + 132 - 64 will be 36 + 32 that is going to be 68 68 minus 18 so 68 - 18 will be 50 by 2 to 25 square units will be the correct answer 20 20 why 20 25 25 now 25 square units 20 50 by 2 now have I done any mistake 36 let me just check 36 and 32 is 68 68 minus huh so actually uh I have also uh done like I have to apply now it's fine H so very easy like direct formula based question now now now here comes the real question D okay let's do it the maximum the maximum area area this is area the maximum area of the triangle formed by these three points is what has asked in this question right so maximum area take so we know how to do it area will be equals to 1 by 2 so the first row will be X1 y1 and 1 so 0 0 one very easy first row we are getting one then this is a COS Theta B sin Theta and 1 right and next a COS Theta Min - B sin Theta and 1 take okay so this is your area that is going to be half now expand it so 0 0 so last how to open the determinant just tell me right so 1 1 so this is min - a COS Theta sin Theta and minus a COS Theta sin Theta okay so this is your area and let's not forget let's not forget there is a modulus okay let's not forget the modulus so area will be tell me half okay half and modulus is already there so that's why negative sign we have neglected but modulus is still applied so this is going to be mod of 2 a sin Theta cos Theta correct okay so a will be what tell me do not cancel 2 two do not do that because we need that two 2 sin Theta cos Theta will be what so mod of ab sin 2 Theta okay can I just do that 1 by to mod ab and mod sin Theta okay and now we are interested in maximum value of the area maximum value of the area so to see A and B actually are fixed what is variable Theta is the variable if we vary Theta the value of sin 2 Theta will change we need to ensure that at at that particular value we need to ensure that sin 2 Theta should have maximum value sin Theta is missing missing they're missing sorry don't know sin Theta take so this this should be maximum value of sin 2 Theta is what 1 right so that is going to give you 1 so mod of ab by 2 will be the maximum value of area a Max a Max a Max a Max is mod AB by 2 option C is absolutely right my dear students yes adika all right absolutely correct okay all right yes CH so area of triangle I hope that this is clear H area of triangle is clear now one more important point was there which talks about uh where exactly is that area of triangle only yes area of triangle so area of triangle but now if three points are colinear that means all the three points lie on the same line then what will happen to area of triangle the area of triangle will be zero obviously can you clearly see there is no area which is clearly seen in this colinearity of three points right so here comes the condition of colinearity of three points points yes how to make like if three points are colinear they are telling you these three points are colinear and find the value of something so and so thing so what you need to do is very easy area of triangle ABC will be zero if three points will be colinear so you just have to apply the same thing uh half to there is no no need to write half H so what is the condition just tell me so that will be X1 y1 1 X2 Y2 1 X3 Y3 1 will be zero if you make it zero then the condition ofinity we will achieve what happened the and now area of quadrilateral let us talk about area of let us take a quick 5 minute break yeah water break then we will talk about area of triangle sorry area of quadrilateral right okay mhm so let's take a 10 minute uh 7 minute break okay 7 minute break we can take 7:15 CH 10 minutes we can take let's meet at resume at 8:00 p.m. okay no like bigger breaks like 20 minutes W the session is not that 6 hour long Okay small bra breakes will work for us for for e e e e e e e e e for e e e e spe for e e e for for all right all right ma'am please make detailed lectures detail this is detailed only now vikash this is detailed only are you not finding it detailed H you want more detailed all right area of quadrilateral so remember I was telling you there were two methods right adika are you here I was telling you there were two methods to find area of triangle H one was the proper determinant which we are using and another was uh that like uh if three points are given then we will make four columns X1 y1 X2 Y2 X3 Y3 and then X1 y1 again so in area of quadrilateral there is a reason why I taught you this because that is actually a generalized method okay that is a generalized method okay uh determinants part one PDF in Jungle Book yes I'm going to do WR again WR today because that PDF is in this system which I'm using right now so after that I'm going to use that okay uh do that CH huh so what I'm trying trying to say is whenever you see now generalize not even for quadrilateral but for Pentagon but for hexagon like n * polygon area of polygon you can uh calculate just by this generalized method so whenever you see quadrilateral that means how many points Four Points so you need to make five columns for triangle you made four columns remember let me go back triangle let me go back to the triangle and let me just show you for triangle what did we do we made four columns right column one column two column three column four one extra column right so for quadrilateral right this method is very very important because that's a generalized method so if I talk about quadrilateral then what will happen is quadrilateral will have quadrilateral will have how many points Four Points so how many columns five columns how tell me so ma'am half X1 y1 X2 Y2 X3 Y3 X4 y4 and back to X1 y1 so basically cyclic order X1 y1 X2 Y2 X3 Y3 X4 y4 and back to X1 y1 one additional now how to like open it you know how to open it okay so let us do that first first step did you notice now crisscross X1 Y2 - y1 X2 can you see half half is already there in both triangle quad everywhere Pentagon half X1 Y2 - y1 X2 plus X2 Y3 minus Y2 X3 right are you getting it are you understanding now tell me what would be the next term x3 y4 - Y3 X4 my dear students and uh uh uh last final X4 y1 minus X1 y4 this is your area of pqrs quadrilateral and this is actually can be generalized for pentagon hexagon heptagon octagon nonon deagon do deagon tu tu tuck some are you getting it so area of polygon polygon means n sides so can you see there are n sides right first second third fourth fifth how many rows I will place how many columns sorry so for area of polygon there are n sides now to one extra column I have to make okay so half will be there okay X1 y1 X2 Y2 X3 Y3 tuck tuck tuck tuck tuck X1 xn YN and one extra which is X1 y1 again how how will we do that crisscross tell me that crisscross X1 Y2 minus y1 X2 and we are going to do that keep on doing that what would be the last term last term would be X and y1 minus y and X1 do not forget to apply the modulus okay all right are we ready for a question then CH area of Pentagon we need to calculate so area will be what half ma'am first one one okay second is 721 third is what 7 - 3 4th is what 12 2 and 0 - 3 and one last this one will be same the first and last column should be same and let's not forget there will be modulus okay so area Delta of Pentagon would be half 21 - 7 is 14 plus okay this and this now next 7 3 are - 21 to- 21 and uh Min - 21 into 7 then plus okay so this is the first thing this is the second thing now next 7 2 are 14- - + 36 take and then 36 - 0 to - 36 + - 36 - 0 and the last which is 0 + 3 0 - - 3 will will become 0 + 3 modulus is already there and we just have to solve it take all right okay 1 by two one student is saying Arian questions say to solve to what do you think this question is I'm not solving this question H my ghost is solving this question some Paranormal Activity is going on or my AI avar is solving this question huh Aran what do you say I am solving this question this is me only me writing the 40 okay this is actually - 7 - 7 - 21 into 7 + 14 + 36 36 36 cancel now plus 3 that's it now you just solve it out modus let's not forget okay so half if I take 7 common to Min - 7 common if I take it will become 22 and 14 + 3 will become 17 H now you can just solve it out take modulus of it and tell me the correct answer also okay all right clear H to correct answer just tell me can you solve it and tell me 68.5 137 by2 all right clear CH now let us talk about yes so the first part of straight lines is clear I guess H wherein we discussed area of triangle we discussed distance formula we discussed area of quadrilateral area of parallelogram we discussed distance formula and a midpoint formula all these things till now we have covered now we are entering in the second phase of straight lines which where we will be discussing about centers related to the triangle what are the centers related to the triangle tell me yeah this is a basic question now we we we were trying to actually apply it okay so we were trying to apply it for application I will pick a basic question okay H so so that now like in future when we go for like majority of question when when it where it comes that okay you have to calculate area of triangle quadrilateral so you don't have to like think about it how to calculate so yes centers of a triangle what are adika is saying Oro Center centroid tell me more tell me more circum Center D singing okay one more one more one more and we will talk about centroid we will talk about circum Center we will talk about orthocenter what is one more pending one more in Center did you guys forget it in Center also we are going to learn okay so one by one because in J main if you just uh go and check the question paper in J main there are a lot of questions from these centers so these centers are very important okay all right pay attention and these centers are important for upcoming like circles and Parabola right that chapter will also combine the question with centers of a triangle okay hi good evening sidhant how are you so CH let's start with the center the easiest of all the Cent uh the easiest of all the uh centers of a triangle okay so now what is a centroid centroid is nothing but so whenever you have a triangle can you see a triangle over here so whenever you see a triangle so in Triangle there are three sides okay now these three sides will have three midpoints respectively suppose AB is having suppose BC is having d as a midpoint right and AD will be called as a median okay what is the median I told you already median is nothing but line joining vertex with opposite side midpoint so vertex a and midpoint of BC if I join them together ad will become a median similarly this time B and midpoint of AC which is e right so B and E if I join it is another the second median and the third median if you can see that is going to be CF f is the midpoint of ab correct so if you join all the three if you like uh draw all the three medians their point of concurrency or their meeting point is exactly one this meeting point which we call as point of concurrency what is point of concurrency that means meeting at single point so that single point you are seeing is nothing but G which is centroid okay all right clear H now what is important so first thing which is important is uh the coordinates of SE centroid coordinates of centroid so the coordinates of centroid we can actually calculate okay so coordinate of centroid G car we have to calculate how do I calculate just tell me so for that what we need to do is so first of all let me just tell you G let us suppose this is X comma y yeah so coordinate of centroid will be given by what x coordinate will be sum of all the x coordinate X1 + X2 + X3 by 3 this is your x coordinate of the triangle right centroid and what are the what is the uh y coordinate same pattern is same template is same that is going to become y1 + Y 2 + Y3 by 3 okay so this is your y coordinate of CID very basic this everybody should understand everybody should learn now one more important point which you are already seeing on your screen is this centroid divides the median in ratio 2 is to 1 centroid divides the median in what ratio 2 is to 1 is that ratio okay what does that mean that you have to remember very very clearly what does that mean so basically centroid divides the median so median is AD so G Point divides ad in which ratio to this is twice and this is one times 2 is to 1 remember this portion is always bigger and this portion is always smaller and this also BG is 2 G is also 2 is to 1 and if I talk about this this will be two and this will be 1 is it clear okay any doubt any confusion okay all right so centroid divides the median in which ratio 2 is 1 now let us do a question right let's do a question based on this so here you have a question which says if the centroid of the triangle formed by a comma b b comma C and C comma a is the origin okay so this time this time what they are saying is centroid of the triangle is origin okay so what does that mean centroid that is g is nothing but Z 0a 0 right what they are trying to ask is a CU plus b Cub plus C Cub value value of a CU plus so I want I want you guys to try this out first uh one uh student Vish is already saying VRI is already telling me that the answer is B let's see let's see I'm not saying he's correct let's see let's see what what about others every time VRI is giving me answer where are you D huh CH centroid is 0 comma 0 so let us apply that formula which we have understood right away okay so x x coordinate will be what X1 + X2 + X3 by 3 A + B + C by 3 is equals to 0 and next also will give you same result by the way b + C + a y1 + Y 2 + Y3 by 3 is also zero okay that means both things actually all the two both the things are giving me same thing which is basically a + b + C is nothing but zero and now what are they asking a cub + B Cub + C Cub value this is so very obvious those people who remember the condition from ninth class this was the in 9th class then A + B + C is zero then a CU + B CU + C Cub becomes 3 * a b c and if a + b + C is0 then a CU + B CU + C CU becomes 3 a c so clearly like this is a question from uh current centroid but it is superimposing something which we already studied in our grade n h okay hi drk gaming how are you hello hello let's do one more question are we ready for it take uh dju yeah I understand you're looking for advanced questions so this session we are not going to do Advanced question so I'm going to take one more session on straight lines so this session is all about clearing the basics because when I teach you Advanced question when I give you Advanced question then you won't feel like so the second question is going to be a really intense session I already told you in the beginning so this session is all about building Basics okay so that your Basics are rock solid and then we will go towards the second session straight line so this is all about coordinate geometry we are introduction to coordinate geometry okay but this is also in your syllabus by the way okay so two vertices of a triangle are given okay to x1a y1 is given to you which is tell me -1a 4 and x2a Y2 is already given to you which is 5A 2 and x3a Y3 is unknown and centroid that is X comma y centroid coordinates this time is given which is what tell me 0a -3 now we just have to apply the formula H we just have to apply the formula let's do that h x will be what X1 + X2 + X3 so X will be X1 why am I doing that and now directly I can apply 0 will be equals to -1 + 5 + X3 by 3 okay and - 3 will be equal to tell me 4 + 2 + Y3 upon 3 correct X3 value and Y3 value you can clearly calculate so X3 will be so this is 5 - 1 X3 will be -4 so tell me tell me this is wrong this is wrong and now from here Y3 + 6 = - 9 no no no no no no keep keep keep keep keep keep H from the current slide H okay so this is actually going to be what Min - 9 H - 9 so this is minus 6 - 9 is option a option A all right correct Mundo is option a option A is correct Mundo okay so centroid is clear I I guess H centroid May so there is this much in centroid now centroid will get combined with circum Center oroc Center to make beautiful questions right now in Center what is in Center just tell me one thing tell me who remember what is in Center who knows it what is in Center so in Center is nothing but point of concurrency of point of concurrency of or meeting point of inter internal angle bis sector my dear students internal angle bis sectors okay so again if I draw a triangle triangle ABC right and internal angle bis sector so that means I need to draw all the angles right so angle a angle a CO bis sector will look like this a is what angle bis sector that means if this angle complete angle is let's say 30° 15 15° so this line will divide the angle in two equal parts this is how angle bis sectors look like okay now again similarly we will do with angle B and angle C and now you can clearly see this point is nothing but in Center my dear students in Center okay and now the formula to calculate in Center is what the formula to calculate in Center for to calculate in Center we need more than the vertex to calculate centroid is very easy we only need the vertex H centroid May what we need to do is X1 + X2 plus X3 by3 y1 + Y2 + y Y3 by3 but this time we need length of all the sides as well so we need BC that is side opposite to vertex a will be denoted as small a that is BC side opposite to capital B vertex is nothing but small B B and side which is opposite to this is small C and if you see the formula of in Center that will be nothing but X this I in Center x coordinate will be a X1 understand this a X1 that means length MTI vertex opposite plus b * X2 plus C * X3 to C * C * X3 upon sum of all all the sides of the triangle right if you understand this then similar thing you have to apply for 4 44 just tell me y also so this is going to be what a y1 + b Y2 + C Y3 Upon A + B + C okay all right now similarly so there are some things uh we need to remember about in Center okay what is that let us understand so first of all this H so if I have to calculate AI upon iix let's say AI upon iix that means this upon this it was very easy in centroid 2 is to 1 was the ratio but hey in Center it is a little bit trickier but I will tell you how to remember it so AI upon iix will be nothing but C + B Upon A now you must be like ma'am hold on how are you even getting it how will we remember it so let me hold on let me tell you that actually AI a i upon iix is what C + B Upon A yes so how are we getting it so just tell me AI so just add left and right left and right or you can say the adjacent length so that is going to be B+ C can you see B+ C in the numerator b + C and the remaining thing will be in denominator okay okay did not get it do it with one more thing do it with one one with me also H do it with me also H so a I by iix now tell me bi upon iix who is going to tell me bi upon iix so ma'am you told us bi H bi is this so what are the adjacent just tell me C and A so what you need to write in the numerator A and C and whichever side is left you have to write it over here okay ma'am CI upon not iix actually this is i y no b i i y this is y and I Z CI upon i z now tell me CI so CI upon sorry this is CI this is CI and I Z this is Zed this is y huh CI upon i z to ma'am c i so what is the B + a ma'am b + A A + B what is remaining left out will be C okay all right okay remember this now one more thing one more just last thing I know that this I've also faced a lot of thing how to remember it but then I realized but I have to remember it now so I made it easy for you you have to remember it now one more ratio you have to remember in in Center okay which is this BX and XC so BX upon XC and it is also not very difficult to remember H BX upon XC okay so just if it is BX now just go towards the left do this lift do you know how we like whenever we are on road how will we like stop someone do you know the symbol if you want to like take a lift on the road although it is unsafe but if even if you want to take it what is the symbol for someone to stop like drop me somewhere it is like this so same we have to do it XC upon x c upon x c upon BX no no flumo say BX upon XC H you can make it XC upon BX also but this is fine H so yes BX upon XC so just tell me BX BX is actually going towards BX is towards C can you see BX is towards C and XC is towards B to C by B C by B understanding understanding or not so basically if BX upon XC you are liking about BX upon x c now to left right so left C upon b h c upon B now similarly if I ask you what is ay upon YC just tell me ay upon YC ay upon YC so it is going to be C Upon A H to we are actually a y upon YC is like this a y so lift M C and A C Upon A C Upon A so you just have to understand BX upon XC is this only C and B the C by b h can are you able to visualize it h BX upon XC is C upon b h and then a y upon YC is what C Upon A C Upon A and last shall I ask you a z upon ZB now you are going to tell me you are going to tell me you are going to tell me tell me tell me tell me what will be the answer sir to you Ma'am okay okay ma'am AZ upon ZB to AZ upon ZB now AZ this is the thing to B Upon A B Upon A yes Shiva Kumari absolutely right B Upon A adika absolutely right take this is the in Center take now in Center is is actually related to a circle do you know that in Center is nothing but so in Center so in Center is related to a circle which we call it as a in circle so what is in circle in circle is nothing but that Circle which is drawn inside the triangle but which will touch all the sides of the triangle that means the largest possible Circle we need to draw H largest possible Circle we need to draw with the center in Center the center would be in center right the center would be in center right and this this is your in Circle H just the terminologies so you have to understand the terminology as well now very basic question on the in Center I have in my plate are we ready for it so very basic question says the vertices of a triangle are 4A -2 - 2A 4 and 5A 5 then we need to figure out it it's in Center in Center you know without plotting the triangle now it is very difficult to solve such questions 4 comma -2 H now then we are having -2a 4 then we are having 5A 5 to calculate in Center we actually need to calculate the length of the sides so this side will be what just tell me to four square root of 36 and 36 square root of 36 + 36 can you just orally calculate this squ < TK of 2 into 36 6 < tk2 so this is 6 < tk2 verify verify the side let us take this is a b and c mhm okay all right so 4A -2 - 2A 4 6 < tk2 square root of 6 correct calculation right now 5 - - 2 7 49 5 - - 2 49 7 Square 49 and 1 the root 50 so this is square root of 50 this length is square root of 50 square root of 50 is 5 < tk2 okay and this AC will be square root of 1 and okay for say again 5 < tk2 this seems to be an equilateral triangle rly is saying first which lesson to watch inverse trigonometry or sets and relations beta sets and relations are very basic it is like your class 11th sets sets basically majority we have studied sets only if uh like it is actually not linked what is linked is relations and functions so you should watch relations and function functions first right trigonometry first and then go towards uh ITF inverse trigonometry function now how what to do just apply the formula you remember okay so x coordinate of in Center will be what a X1 so a X1 5 < tk2 this is your a 5 < tk2 into 4 so 5 < tk2 is a X1 is 4 plus now 5 < tk2 into -2 5 < tk22 into -2 this is your B X2 and finally C X1 so 6 < tk2 into 5 upon we need to add all the lengths okay the 5 < tk2 + 5 < tk2 + 6 < tk2 clearly < tk2 < tk2 < tk2 < tk2 < tk2 is actually getting out < tk2 is gone so x coordinate will be 5 4 are 20 plus - 10 and 6 5 are 30 so 50 okay and this is five or P or P TH or CH 16 so X will be tell me tell me tell me 40 by 16 4 four are 4 10 to 5x 2 is your x coordinate what does that mean is clearly option b is wrong C is wrong D is wrong option A is absolutely right my dear students Hanah okay and actually uh I don't even have to like calculate in this question but even if I calculate can you see same data will get formed even if you calculate y coordinate now just tell me 5 < tk2 into 4 H again see y 5 < tk2 into 4 will get formed right same set will get formed calculation will be exactly same 5 < tk2 into 4 5 < tk2 into 4 y coordinate will be same exactly right and then 5 < tk2 into -2 5 < tk2 into -2 okay and then 6 < tk2 into 5 can you see same exactly X and Y will be exactly same option A is absolutely right easy peasy lemon squeezy question what do you think okay now the third third Center which is Oro Center okay ortho center let us understand the ortho center ortho center is point of concurrencies of all the altit udes we can also call it as perpendicular my dear students perpendicular okay so basically again we need to draw a triangle and then and then what we need to do is we need to draw drop a an altitude from vertex a we need to drop an altitude from vertex altitude means a line which is coming from vertex a but is perpendicular to BC can you see this similar thing we need to do with all the vertex a line from B will be dropped such that it is perpendicular to AC and a line from c will be dropped such that perpendicular to the DC okay okay so over here this is the point H H is ortho center okay Shiva Kumari is saying if this is in Mains paper we will be uh yes Shiva Kumari but you have to be be there in my Saturday session then if you feel that that is also easy a com combination then you can just text me mam yes H periodicity I have already uh rali I have already taught you periodicity maybe you have not seen that uh session completely towards the end we have studied periodicity also okay okay so yes ortho center now formula for ortho center is actually they will directly not ask formula for Alo Center there is a technique by the help of that technique you'll be able to calculate it what you need to remember about orthocenter let me tell you is orthocenter in case of right angle triangle that is a special case so just tell me one thing very important okay so again for a right angle triangle let us try to create all the three altitudes H let us try to create all the altitudes so from point A what I have to do is I have to drop a perpendicular right which is perpendicular to BC okay let us do the same thing for right angle triangle which we just did it for a general triangle so point a this is your point a what we are going to do is this is your site BC we are actually going to drop a perpendicular to side BC but wait a minute wait a minute wait a minute ma'am hold on so if I have to draw the perpendicular but this perpendicular is already done actually if You observe the side AC in this case ma'am side AC is itself the altitude from vertex a yes or no h side AC is the altitude from vertex a which is actually perpendicular to BC okay so this is Altitude number one we need three altitudes right one by one so we need AC okay AC is already one altitude now let us talk about the second altitude and then we will talk about the third altitude now just tell me again from B point I have to drop a perpendicular to AC isn't it from this point if I just drop the perpendicular to AC what will happen so B say if I drop the perpendicular that second altitude the second altitude ma'am ma'am hold on the side BC is itself the altitude from B okay so altitude number one is BC altitude number two is AC and their meeting point or their point of concurrency is nothing but C which is nothing but the oroc center and the third if I talk about Point C so if I drop the like altitude from point C to this this this will look like this and now just tell me ac BC and third altitude meeting point is orthocenter what does that mean can you see orthocenter is nothing but right angular vertex that vertex which is actually getting a right angle in right angle triangle [Music] vertex at right angle vertex at right angle is orthos Center only very very important very very important okay all right clear okay so whenever a question so yeah so what is the trick ninja tip ninja tip time for a ninja tip ninja tip okay what is the time whenever ever in a question if three Vortex are given to you okay and they are asking you about what is the ortho center the very first thing which should come in your mind is wait a minute isn't it a right angle triangle let me check because if this is a right angle triangle that then one of the vertex which is already given in the question is answer and that vertex which is having the right angle triangle which is having the right angle is your answer B is your answer okay all right so ninja tip for the day so whenever a question is all about orthocenter the very first condition you are going to check is is it a right angle triangle or not and trust me 99% that triangle will be a right right angle triangle and you don't have to like apply the bigger formulas or anything you just have to identify which out of these three vertex is that vertex which is at right angle okay all right clear all right now next the last Center which is what do you mean by Ninja ma'am what do you mean Ninja ma'am are you asking what is ninja so my name is nam jinghan and in offline since offline and online Nam jinghan the students were call you students used to call me NJ NJ NJ ma'am NJ ma'am NJ H it sounded like ninja ninja NJ ninja so that's why my name became ninja you guys call me ninja only okay H so the point of concurrency of perpendicular side B sectors of a triangle what is a circum Center understand very basic what is perpendicular side bis sector so suppose you have taken a triangle okay and now you are trying to draw what perp pendicular bis sectors now what are perpendicular bis sector so suppose you take BC okay and if you take midpoint of BC so midpoint of BC is D and from this D you have to drop a perpendicular you have to draw perpendicular and this perpendicular need not to be passed through a we remember this perpendicular we are not dropping from vertex rather this perpendicular we are dropping from the midpoint of sides so this is one perpendicular bis sector now this is another perpendicular bis sector which is passing through midpoint of ab which is f and the third perpendicular bis sector and this very point you can see is circum Center yes this very point is nothing but circum Center all right and if I talk about the circum Center okay to very important point just like in center and circum Center in circum in circum so what is the like relation similarity between in and circum or what is difference in inside in inside so remember that in circle was actually drawn inside the triangle but circum Circle circum Center circum Circle so circum is nothing but outside the triangle can you see that this is your circum Circle okay clear mhm right so Center is what the center of this circum circle is circum Center all right clear now again these two things you have to understand especially ortho center just tell me we just studied that or Oro Center is in case of right angle Triangle oroc Center is nothing but a Vertex at right angle okay now what about circum Center let us see let us see okay ma'am now suppose it is a right angle triangle now if I try to draw if I try to draw a circle circle which is passing through let's try to draw a circle which is passing through Circle let us try to draw the circle a little bit better right H oh nice nice no it's not good but anyway I can't do better than this okay so this is a circle right now this the center of this circle will be what so this is a circum Circle right we are looking for in right right angle triangle case what we are looking for we are looking for center right now just tell me one thing that this line actually is subtending an angle of 90° at circumference remember that theorem which you have already studied in grade 10 angle in a semicircle if you consider a semicircle to angle angle in a semicircle is always 90° did you remember that angle in a semicircle is 90° right so now that what does that mean that AB divides the complete circle into two equal parts so AB is actually a diameter of that Circle oh my God that means midpoint of AC midpoint of AC which is what hypotenuse is nothing but what is nothing but circum center of a right angle triangle and now one important thing over here you have to understand you have to understand the geometry of it if this is the midpoint just tell me these two are equal and and and hold on if I join these two point this is also equal so this this length is also radius obviously because I'm joining the center to the circumference so this is also R radius this is also radius because half of diameter now midpoint Center Center to circumference radius this is also radius so these three things are absolutely equal understand the geometry behind it Hanah yes midpoint of AC circum Center Clear okay now very important o ratio G ratio C OG GC sorry sorry sorry sorry ogc I OG is to GC OG is to GC ratio is 2 is 1 very important very important OG is to GC ratio will be 2 is to 1 now ma'am what is this always you have to remember this huh I will just this is my actually standard way of saying it so whenever I tell you ogc ratio ogc ratio ogc ratio so you will tell me 2 is to 1 ma'am but what is ogc ratio so for ogc ratio to understand this first of all you understand o is nothing but orthocenter denoted by H there is a standard dention o o is orthocenter even if you forget H it is fine but o means orthocenter what is g you know it right what is g ma'am G is centroid standard standard way of representing a centroid and C C what is C ma'am circum Center H so it is actually not very difficult to remember also now OG g c orthocenter centroid circum Center this ratio is nothing but 2 is to 1 first thing is in any triangle orthocenter centroid and circum Center these three centers are lying in one line that is the that line is called as isers line isers line although this is like nobody will ask you what this line is called we call it but this line is called as iser's line H so centroid divides oros Center and circum Center in which ratio 2 is to 1 ratio so centroid divides ortho center and now the very important part most of the student will forget okay there was a ratio which was 2 is to one cent was in Middle but I don't remember which was in like left hand side and which was in right hand side that's why I say remember ogc okay you have that shortcut original OG original OG original H to OG ratio GC that's how you can remember it to centroid divides orthocenter and circum Center into 2 is to1 ratio that is one thing another thing is in isoceles triangle in is this is also very important in isoceles triangle centroid in center and circum Center and orthocenter in isolus Triangle these all these four centers which we have already studied lie on the same line okay anyway in any triangle centroid circum Center and orthocenter anywhere lie on the same line in any any given triangle but when it comes to isoceles in Center also join that line are you getting my point let me just repeat it in generic any triangle scalen triangle any triangle orthocenter centroid and circum Center form a line okay but if I talk about isoceles isoceles May in Center also will like join this in Center also will stay on this line to in any triangle centroid circum Center orthocenter was actually like lying on that line now when it comes to isos in Center also Li on that line and one very important important of all B of all God of all in an equilateral triangle very very important these four points are one and same these four points are one and same so if I give you an equilateral triangle this is centroid this is orthocenter this is circum Center and this is what in Center that means all four points are same only AA okay so Ma'am why are you saying this is very important I mean okay take I mean what is the use of it how can we like apply this thing in question so basically whenever a question says that calculate in Center or calculate circum Center or calculate orthocenter okay and an equilateral triangle is given to you for example then you understand that out of all four centers which one is the easiest to calculate obviously centroid right so instead of calculating orthocenter or circum Center or in Center you will always calculate centroid because that is easier to calculate so you will calculate centroid and you will say that this is centroid this is the or Center this is in Center this is same everything is same only right so you calculate croid and you chipa everywhere H so like that this can help this can be helpful in calculation got it that's how you can take a help now let the orthocenter and centroid of a triangle is given a ortho center is given now remember that o c line o g c correct orthocenter is given -3a 5 and uh G 3A 3 centroid H C is circum Center which we do not know then the radius of the circle having a line segment AC as diameter okay so this is your actually a point this is your B point and this is your C only as per the question right so can I just calculate the C Point obviously by using 2 is to 1 ratio to calculate C point and then what they are asking is if AC is the diameter then you have to figure out radius so calculate AC which is the diameter divide by two will give you radius tell me will you be able to do that yes or no hello the question says that orthocenter and centroid are given orthocenter we know o take centroid is G circum Center is C which they have not given so we know that o orthocenter G is given the 2 is to one is the ratio can we using this ratio can we calculate this C if we if we are able to calculate C then we can calculate AC distance which is a diameter half of it will be radius just tell me is is that question very difficult okay mhm CH let's understand okay so ma'am suppose I mark it as XA y h and let let us apply section formula section formula says that this three is equals 2 2 * X remember that crisscross crisscross M * X1 1 * - 3 upon 2 + 1 is 3 from here actually you can calculate what 3 3 are 9 9 + 3 is 12 from here 2x = 12 can can I write X = to 6 from here directly can I do that yes H this is x coordinate so 6 comma C is coming out as 6 comma similarly y we can calculate right so that is what so three this is three now again 2 into y equal to no sorry 2 into y plus so basically M y1 sorry m y 2 and N y1 so + 5 upon 2 + 1 3 from here if I want to calculate y 3 3 is are 9 9 - 5 is 4 2 Y is = 4 to Y is = 2 just calculate and let me know whether I have done correct or not 6A 2 is point C and A is - 3A 5 so obviously diameter is AC which is what square root of tell me 6 6 + 3 Square so 6 + 3 Square 9 Square 81 plus next 2 - 5 that is - 3 square is 9 81 + 9 right so AC will be what otk 90 H the radius will be what < TK 90 upon 2 now root 90 upon 2 is matching with withth the root 90 is what tell me 9 and 10 so 3un 10 by 2 3 < TK 10 by 2 can anybody body Tell Me 3 < tk1 by 2 okay so again 3un 10 by2 if I rationalize it then can you see this is actually 3 < tk10 by 2 option C Co if you rationalize you will get the answer yes V absolutely right H easy peasy what do you think okay ma'am clear now shifting of origin okay shift the two more topics are pending shifting of origin and Locus so it will take us another half an hour probably so let us take a 5 minute break can we take a quick 5 minute break okay so what is the time right [Music] now what is the time right now Take 5 minute quick 5 minute break and a 5 minute break I don't know what's the time right now then the pending two topics we can do hi sh shumari I saw your answer yes option C sorry huh yes your answers are visible yes yes yes for e e for e e for e e for e e e e e e e e e e e e for pyqs uh there is uh already a series going on huh for pq's a series is already going on we where we are covering all the pyqs by the way okay and these sessions in these sessions also we are covering pyq but you all know that there are so many pyqs in 2024 so here we have to teach also we can't cover all pyq although we try to cover all the important pyqs as much as we can okay but this session will not cover all the pyq that's why a series is going on I don't know whether you know or not but that thing is actually giving you all the pyqs which were asked in J in 2024 chapter wise also okay and uh uh yeah you obviously shivak Kumari yes definitely you're going to get the questions from these topics yes definitely when is sequence and series already finished sequence and series to already done now done and dusted okay sequence and series go and watch that lecture Sun okay tell shifting of origin so what we do is this is your uh the original X's X small X and small Y correct and now shifting of or means that what you do is now X this is your original uh I would say original axis okay all right now in this system there is a point small X comma small Y correct are you getting my point and what I'm going to do is I'm going to make a generalize also so this is your let's say this point is 2 comma 3 correct 2 comma 3 are you getting my point so what we did is Shifting of origin I will introduce but let us understand there is a scenario whereas there is x- Axis the original one y axis original one and there is a point in this particular scenario X comma y so if I just want to like understand the scenario P so this length is what so this length if I talk about now this length will be what tell me so this length from here to here this is X right and this length is Y correct or not this length is y this is your X length this is your y length okay now what we are doing is we are shifting the origin H right now the origin at 0a 0 h i now we are shifting the origin at some other point which is let's say h comma K so this this H comma K and H comma K is suppose your uh new origin so now new origin means new x-axis and New Y AIS okay all right so this is X this is your y okay so if I ask you coordinate of this point with respect to XY axis is what with respect to the original XY axis understand ma'am with respect to original the coordinate is 0 comma 0 this is the origin okay coordinate of this point with respect to new with respect to new coordinate system is what ma'am coordinate of this point with respect to new system is origin right this is not transformation of XIs this is Shifting of origin that is different so this is a 0 comma 0 with respect to capital x and capital Y try to try to understand this okay and if I talk about this point this point is H comma K with respect to the original system okay now we need to establish the relation between small X and capital x okay okay CH so just tell me one thing tell me one thing this point is X comma y with respect to the original system now this point P this position P will have new coordinates obious L capital x comma capital Y yes or no yes wherein this length will be X this length will be y because X comma y with respect to this okay X and Y got it and this is your X this is your Y and if I talk about this is your H this is your K okay now what we need to do is we just need to equate the distances what you are seeing right now if I seeing that this x is actually small X is equals to h plus this length h plus can you see what is this this length is capital x oh Ma'am okay okay okay okay getting it getting it and now this complete y length we are equating just length y y will be equals to k+ this which is capital Y so y will be equal to k + capital Y this is the formula now ma'am what is small X ma'am hold on what is small X just tell me one thing the small X is nothing but the x coordinate with respect to original axis ma'am what is h h is x coordinate of that point where you have shifted the origin what is capital x capital x is final coordinate of the same point basically one point will have two different naming okay one point will have two different type of naming okay one with respect to the original one one with respect to the smaller one so this P point the same point okay same point will have X comma Y and capital x comma capital Y so capital x is nothing but coordinates with respect to new origin now if we solve one question of then you will understand stand all you have to remember is original is equal to h plus final coordinate original is equals to K plus final coordinate okay all right yes CH formula you have to remember then everything will be sorted at what point the origin be shifted that means what they are asking it's h comma K if the coordinate of a point 4A 5 this is your small X small y this is cap capital x capital Y you know the formula just apply just apply one student has a name why and he's writing am I late what I am seeing is why am I late I don't know bro you know why are you late what were you doing huh it started at 6:30 it is almost 2.5 hours you're late by 2.5 5 hours okay okay CH let us apply that formula X = to capital x + H H is what we are looking for small X is what 4 clearly H is coming out as 7 similarly y = capital y + k k is what we are looking for 5 = 9 + k k is nothing but -4 7 comma -4 option a easy easy easy peasy application based very easy now let us Level Up Level Up Level Up Level Up Level Up level up so this is also a song and rap type again another rap level up level up okay axis are shifted to the point 1 comma so H = to 1 and K isal to minus 2 that is the very first information now without rotation that means we are not rotating so there is one more concept called rotation of axis so that's why they are clearing we are not rotating axis so we are just following this the horizontal shifting one student was writing also transformation of axis no we are not doing that we are not transforming we are just translating that is called as shifting of origin H so now the equation this transforms to which equation that means this is the original equation so clearly small X small x equals to capital x + 1 and small Y is = capital y - 2 okay so now the equation is given in small X and small y that is the original equation so now small X can be replaced by capital x + 1 that will give me a new transformed equation right so let us do that no so 2 * x² x + 1 s + y - 2 s - 4 * x + 1 + 4 * y - 2 equals to0 so now what I will do is if I were at your place H if I was a student what we have what I would have done so I would not calculate the X and Y only I need to calculate is if I'm sure by the way only if I'm sure and if I have very less time if I want to save my time now I will only focus on what is the constant term because if you see over here constant term is minus 4 here six here zero here zero okay constant terms both are zero okay so it will eliminate the options first of all what will be the constant terms so here I'm going to get + two here I'm going to get + 4 H + 2 and if I open it what will I get + 4 minus 4 only the constant terms - 8 so the constant term clearly is 2 - 8 that is - 6 is my constant term - 6 in the left hand side will become so equals to 0 - 6 equals to 0 so on the right hand side it will go it will become option b okay anyway this is a short trick but if I open it and I if I match the complete answer let me just do it for you 2 x² + 4x + 2 + Y 2 + 4 - 2 y + 4x + 4 + 4 y - 8 = 0 H sorry it is - 4x - 4 right H these two are minus clearly - 4 say + 4 cancel 2x² + Y 2 2 x² + y s then we are having 4 x cancel - 4x then we are having minus 2 y actually this is - 4 y right - 4 y + 4 y cancel so 2X x s + y sare can you see and the constant term is - 6 okay okay all right now again another question shift the origin to a suitable point so that it will not contain a term in Y and the constant okay hold on hold on hold on shift the origin to a suitable point that means the question is asking about at what point the origin should be shifted at what point the origin should be shifted so that so that the transformed equation related to this will have no constant term and no term related to y very good question in terms of Shifting of origin okay ma'am hold on okay shift the origin to a suitable point so H comma K is unknown and we know the formula X will be replaced by this capital x + h y will be replaced by y + K so let's do that y Square will be what capital y + k squ + 4 capital y + K right Plus 8 * capital x + h - 2 = 0 now to get the value of H and K we need to make the constant term zero and the term related to y0 are you getting my point so let's do that okay term related to Y so we don't need to open it completely although we can but if I were at your place I will only do that part which is required I need a term related to y yes or no so term related to Y in fact coefficient of Y from here and the coefficient of Y tell me not y Square y okay so ma'am from here if you expand this now one term will be y Square One term will be K square and another term will be 2 y k so 2 y k 2 K will be the coefficient of Y okay ma'am now plus 4 y to plus 4 will be the coefficient why I'm extracting that's it that's it ma'am that's it the coefficient of Y should be zero clearly this implies K is coming out as Min -2 okay all right hold on now the constant term should also be zero the constant term let us collect together from here what is the constant term ma'am K Square from here what is the constant term ma'am 4K from here what is the constant term okay 8 H constant term is actually coming very big - 2 = to 0 so basically constant term is zero okay so now we know the value of k k is min-2 right so why not we put it over here and get the value of H Min -2 if I put it over here I will get 4 then if I put it over here it will become - 8 + 8 H - 2 = 0 so this is going to be 8 H right and this is min -4 - 2 = 6 h = 3x 4 okay so the origin should be shifted at have I done anything wrong n you're also coming same 3x 4A minus 2 is that point where we should shift the origin so that this equation have no term related to Y and no constant term okay all right six is it six guys let me just check one time the answer if I check it is 3x 4 comma minus 2 correct H 3x 4A -2 all right now the last topic of today so by the end of this topic Locust so four five questions we will do on this topic as well so we will be absolutely Rock Solid in basics of straight lines right and I am calling it Basics actually but you go and check there are questions from this these topics as well but if you don't pay attention because the straight line is a very big chunk so what we do is we try to neglect these things which are like this is easy this is easy but I realize no if this you won't be able to understand then the next part you won't be able to understand okay so yes Locust is something you go and check from this topic so many questions are there centers related to Circle you go and check so many topics so many questions are there in previous year questions right okay so Locus very very important what is Locus and every time I'm teaching since 10 years now I'm just telling you this exactly this is the topic which teachers find it very easy a Locus very easy very easy but students every student ma'am Locus I'm not able to to like solve Locus question I'm not able to Lo solve Locus question hearing since 10 years okay and let me tell you the secret recipe of locus okay today so yes Locus what is a locust so basically Locus is nothing but a point which is moving a point which is moving in 2D plane okay a point which is moving in a 2d plane but not randomly no no no no no no it is not moving randomly but some geometrical condition is there some constraint is there that that particle has to follow that point has to follow and that path that that that point that which the path follows that path is nothing called Locus so when a point moves in plane under a certain geometric conditions that I'm going to tell you the point traces out a path and that path under that condition is nothing but Locus so suppose if I say there is a point which is moving H moving in such a way that whose distance from a fixed point so this is your fixed point right so a point which is moving in a plane whose distance from a fixed point is always same so basically what I'm trying to say is that okay okay so basically let's say the the a point which is moving in a plane in such a way that its distance from a fixed point is equal to let's say two okay so there is a point whose distance from fixed point is what two so just tell me what are those points which are two units away from my fixed points so it will look like this it will look like this two units 2 units 2 units 2 units so if I try to draw all the path it indicates that it is a circle circle is that thing which whose distance from fixed Point everywhere is same remember that rounder we take that fixed length length is 2 units point is fixed that's what we do if you remember H that what do you call it divider n rounder what do you call so we keep that tip of rounder over here and what do we do we keep the moving point which is pencil moving point is pencil and distance is fix fixed we are not changing it so what we get is circle okay all right so this is like this path is called as Locus in that case it is circle all right so Locus definition I guess it is clear H know now there are certain steps if you follow those steps if you understand those steps then the locus solving will be very very easy for you what are those steps let us understand to find locus of any point the very first step is consider that moving point to be H comma K and point P let's say you can change it also p q whatever but H comma K you have to assign that moving point should be H comma K that is number one step so let us do it h whatever steps I'm going to write I will do it for circles okay transformation of graphs to I have already taught Now Beta no one just go and check H there was one session transformation of graphs CH so this is your fixed point and let's say fixed point is given as 0 comma 0 in this particular case and the problem statement is find the locus of a Point whose distance from a fixed point is one you understood the statement find the locus of a point of a moving point whose distance from a fixed point that is origin is one okay so what we need to do is we understood okay ma'am that uh there is a point P which is let's say h comma K which is a moving point okay which is a a moving point H H comma K 0a 0 now the distance between what they are saying is distance between these two points is one the distance between fixed Point let's say o and point pH comma K is what no no no we cannot assign X comma y janum uh we have to assign it in the very end although for reducing the step you can do it but it can lead to some errors so that's why what am I saying now in the beginning please follow this then you will understand what short trick you can apply okay all right now first step clear we have already done that now what is the Second Step write the given condition in mathematical form so Locus why exactly Locus problems are difficult it is actually not difficult the problem all the student faces is Step number two converting the geometrical condition into a mathematical form is what something which students find it very difficult but if you understand this thing now gometry to algebra basically geometry to equation very easy so now in this question what they are saying is distance from origin of a moving point is what fixed and that is one so that means what they are saying is apply distance formula that's it okay so mathematical condition of geometry square root of H - 0 square + K - 0 square is equals to 1 now this geometry can be like it is not necessary it will be always be distance it can be distance it can be something else also that we will understand okay yeah so op is equal to 1 is your geometrical condition which we have can you see the step number uh like two this is geometry geometry to algebra conversion happened over here and now it is coming out as h² + k s = 1 and if in this whole process because this is a very simple example but sometimes it may happen you have to introduce your own variable to solve the problem we do that take this as a take this as B so we need to eliminate all the variables if we have introduced in this case we have not done that so we don't don't no need to worry now in the last last step is you have to replace H by X and K by Y and this final will give you the equation of locus so equation of circle now what is this that we will have a different chapter Al together but a circle having origin as Center and radius is equals to 1 is the equation of this is x² + y² is = 1 this is the locus of Point P okay okay no one we have not studied tilting of axis okay and uh let me just tell you this is not at all in silabus also in J main uh this is the last topic so we will do three more questions on this topic and we are done okay all right CH are we ready for the questions because the like concept should be clear then you tuck tuck tuck do all the questions okay huh for J Advance we will do no one yes for in the next class we are going to do that H uh transformation of axis take all right indefinite integration yes okay we will do it one by one huh good evening good evening TM playy games all right take locus of a point which moves such that its distance from 0 comma 0 is twice the distance from the Y AIS AA so this again the geometrical condition is there first of all we have to plot the geometry then translate it to algebraic equation that's it okay logus of a point so there is a point whose distance from origin okay so there is a point p h comma K step number one H comma K taken right now the distance of this point from origin H that is op is twice of of the distance from y AIS so this distance let's say o sorry Y axis y a hold on hold on not the x- axis y AIS so this is your y h now so this is 2 * py 2 * py yeah this is the geometry right op is nothing but double of py correct this is the geometry now py how to calculate py what is py let us understand what is py so py is nothing but actually uh horizontal distance so this is nothing but H only x coordinate of this point H whenever we have to find out horizontal distance okay so horizontal this is your H right so op is equals to 2 py now op can you calculate op is what square root of H - 0 square step number two we are doing which is very easy in this case right geometry to algebra algebraic equation is equals to 2 * py what is py tell me h no mod H why because this H can be negative but what they are talking about is distance from y- axis now distance from Y axis is H but H if H any coordinate can be negative also now modulus H will give you the proper distance now Square it x² + k s will give you 4 h² so clearly if you shift everything towards right hand side 3x² - y² = 0 already replaced H and uh K by X and Y also already no no no not this one zero is coming now H so option b yes b b b b indefinite integral yes manikanta we will do it h clear shall we move on to the next problem my dear students mhm the ends of the hypotenuse of a right angled triangle are AA okay ends of the hypotenuse right angle triangle is here ends so 6 comma 0 and 0a or rather I would say let's just do it over here 6 comma 0 and 0a 6 so basically this is your what xaxis that's why 6 comma 0 this is your y AIS correct ends of the hypotenuse then the locus of third vertex so this is your H comma k h h comma K so what what we need to do is so this is a situation now we have to understand one geometrical condition which they have not given it should come right like through this right angle triangle what is the geometry behind right angle triangle tell me ma'am Pythagoras Theorem this is your base ma'am this is your perpendicular this is hypotenuse just apply it perpendicular square plus base square equals to hypotenuse Square let's apply it right away okay okay ma'am so so so so so it will be H so square root square is already there now to h² + K - 6 whole Square h² K - 6 this is your perpendicular plus base so base is what H - 6 square plus K squ this is your base perpendicular base and hypotenuse is this which is what 36 + 36 right square is already there so no need to worry 36 okay now expand it h square and H Square will give you 2 h² k squ k Square will give you 2K squ 36 and 36 will give you what sorry this is actually 72 36 + 36 so here also 36 + 36 is 72 from here you are going to get -2 K this is - 12 H = 72 gone and gone okay so this is finally divide by two right away x² + Y 2 - 6 x - 6 y = to Z is the final Locus which is matching with option b Madam stepwise one step at a time step by step okay all right two more questions to go and congratulations you have made it till the very end right so now two more questions and we will be done now rest two questions are actually very important are we ready for it yes or no CH let's do it the locus of these two points okay ma'am this is to they have not given anything no condition no geometry no moving point what H CH let's see so locus of this point we need to calculate H basically h is your what t² + t + 1 and K is your t² minus t + 1 ma'am what to do now what to do what exactly they are asking us so if we like understand our goal the agenda always the agenda is to have a relation between X and Y or rather I would say H and K yes or no h options always say option is not having any T all the options are having only X and Y that means all the options want us to calculate the relation between H and K that means what does that mean that we just need to eliminate T okay and if you carefully observe I have actually mentioned this point while teaching you steps to find Locus if any if any uh like uh variable is there you have to eliminate it remember or not go back and check so that means the whole thing is to eliminate so now it is purely equation game not at all Locus game purely equation game no no need to do the factors also no need H it is purely equation yeah actually factors say factors will not help so how do you eliminate t is something which is very like uh which will be the decider so first thing is what I'm going to do is let us add these two equations right let us add so H + K will be equal to 2 t² + 2 so 2 of t² + 1 clear that is the first thing now same what we are going to do is same thing we are going to write H = t² + t + 1 and = t ² - t + 1 now what we are going to do is this is K right we are going to subtract subtract H - K so ma'am h- K is actually t sare t Square cancel T minus so this is 2T and 1 one also cancel say H and K right H minus K is 2T okay take okay hold on ma'am we are actually getting it you want to eliminate Tina you want to next class cannot be the indefinite integration manikant I'm so sorry about it why because see we have a lot more to cover there is a gap we have not covered a differentiation how can I teach in indefinite integration nobody I know that you know uh differentiation but uh most of the people are like not very well versed with differentiation so differentiation I will teach then application of derivatives and then comes indefinite integration so sorry but you have to wait for it okay all right and yeah now what to do so you have to eliminate T to why not I do one thing that I calculate the value of T from here and plug in that value here t² so this is H - K upon 2 whole square + 1 okay ma'am can you see how beautifully how beautifully what is happening T is eliminated that's it we just have to open it now okay let's do that so H + K is equals to 2 is already there okay h² + k 2 - 2 k h upon 4 + 1 okay now let us multiply two over here so this is going to give you H+ plus K or right away I will multiply it right away now why to wait for the next step this two will be gone inside so it will actually become what so 2 2 4 no no no no no no four upon two sorry sorry sorry so this is upon two + 2 correct H now just take LCM so 2 H + + 2 K will give you what h² + k 2 - 2 k h + 4 correct so what is the correct answer let us rearrange x² + y sare replacing also at the same time x square + y Square then you have - 2x y - 2x - 2 y + 4 equal to0 okay hold on because y term is not not here clearly option b is absolutely right so Shiva Kumari you are actually doing very fine H every time I seeing you that you are answering like before me only H know yes okay take got it CH now two more questions on our plate and we will be tuck tuck tuck dun dun dun okay CH let's go let's go let's go a r this is a very classic question by the way temp plate question this every every teacher will teach you this question every teacher's note will have this question okay A Rod of length L slides with its end onto perpendicular lines okay so two perpendicular lines it is better to draw those two perpendicular lines we have like legendary perpendicular lines X and Y axis right why not take them and make them perpendicular lines so a rod of length L slides with its ends on two perpendicular lines a rod is there whose ends are actually sliding on two perpendicular this is your Rod okay let us mark it as AB a now the locus of its midpoint is what we need to figure out so this point is your P which is H comma k okay now what we need to do is we need to talk about the geometry about it what is the geometry behind it so geometry is nothing but so geometry behind it is what we need to use the geometry of Pythagoras Theorem so let's suppose this is a to o a square plus sorry o a square plus o squ equal to AB Square AB to we know L AB to we know it is L now o a and OB o a and OB to ma'am a to we don't know we don't know a all we know is point p and p is never over here to how will we introduce H comma K in this geometry this is the main point so what we will do is this thing this we have not utilized this part now this information is already there we need to do something about this information to solve the question midpoint okay so midpoint May point if p is the midpoint now then we need to have a let us assume a comma 0 now we are introducing the variable so whenever in Locus you try to introduce variable keep in mind we need to eliminate in the end but we will do that it is not that difficult very easy h a comma 0 and 0 comma B now midpoint formula let us apply so ma'am yes midpoint formula says that a + 0 by 2 is your X H and 0 + B by 2 is K okay hold on so ma'am actually what is happening a is your 2K 2 H and B is equal to 2 K keep this in mind keep this in mind we are going to eliminate this A and B by introducing H okay now now you have taken the coordinate you this is 0a 0 to you can apply Pythagoras Theorem very easily ma'am o a square o a is nothing but what a square OB Square this is your b square AB Square given in the question L Square now a will be replaced by what ma'am 2x Square done the elimination thing very easy right + 2 K whole squ = to l s so 4 x² + 4 y² = to your l s okay okay hold on which answer which answer option b very easy peasy lemon squeezy okay all right clear one last question are we ready for it yo bro find the locus of a variable I want you guys to do it locus of a variable Point whose distance from this is equals to distance from that array Madam array what is this ma'am now that we have done a lot of questions so this will be a cake walk for us so this is a homework question I will leave here only I will not go beyond this point take this is a homework question very easy not at all very difficult so I believe that we are done with today's session but before you go I have something to share keep keep keep H I have something to share so for that actually you need to just wait okay give me a moment because I want to show you something okay so yes so I want to talk about vanto Pro course I want to talk about vantu Pro course so you can clearly see that the results are in front of you J 2024 results this is not about J main this these are the results which are for J advance so in J Advance 2024 4538 students cleared J Advance using vanto J Pro course okay so basically uh if I just directly go and uh talk about this course okay if you are a dropper if you are J 2025 aspirant class 12th if you are in class 11th then this course is actually for you just uh give me a moment and yeah so let me just directly talk about the course yes if you are an 11th grader then you can clearly see this is a pro classic course what are the like key features of this course first of all this is not a recorded course live online CL course which is going to save a lot of your time especially for j225 aspirants you know that only 5 months are there right less than 5 months to for um yeah for 2025 if you are in class 12th okay the proc classic course is for you if you are a dropper then the po pro classic courses for you obviously both courses are different because in class 12th you are very new to especially the fresh class uh 12th topics and if you're a dropper then your strategy should be very different okay so that's why two different course for J 2025 aspirants are there Pro classic course online live course is there and and uh obviously because you're studying online doesn't mean that all the books will be online digital books no you have to have the offline material also the practice books will be delivered to you at your doorsteps right and the doubts we know the importance of doubts 24/7 live doubt clearing will be going on in this particular course right and if you do not improve you know that if you do not improve in marks we'll refund your fees okay so these are the very very important features of proc classic course if you're a dropper or if you're a 2025 aspirant in case you are a 2026 aspirant that means class 11th then this is the proc classic course for you okay so best teachers will be teaching you in all the three batches all the three courses doubt solving will be there right you know that I'm not sure how many of you know but in class quizzes and leaderboards are something that students are very fond of right so I am also teaching in a batch where students ask that ask for the quiz and leaderboard so right real time you will get that understanding whether I have understood this topic or not okay so many practice questions and the quality practice questions will be there okay so and post class doubt solving will be there previous year question will be discussed in this particular batch but if you're looking for more personalizing this is also a personalized attention thing but if you're looking for a lot more personalized attention so you will be assigned up with a personal teacher if you want to go for Pro Plus Course Pro Plus course is nothing it is same as Pro classic just a little difference is there and that difference make the dish very delicious that is your mentorship right so personal teacher will be assigned to you which will actually monitor you throughout the course this is available for class 11th Pro Plus is also available for class 12th and this option is also available for dropper as well so if you're planning to start your J preparation with us if you feel what you need to do is you need to fill the form which is given in the description box if you fill the form then everything will be explained to you and then you have to decide ball is in your coat you have to decide whether I want to go with this or that or something else okay all right so this is me n on this note we'll sign off now and uh anything else you are saying let me just go back to the chat okay H hold on one student is asking ma'am what about pair of straight lines so Saturday we are meeting again for straight lines yes gor for in on Saturday we are meeting again wherein because a lot more type not even pair of straight lines a lot more topics are remaining what are those topics tell me whole the like various forms of straight lines will be there then we are having family of lines then we are having pair of straight lines condition of concurrency angle bis sectors uh homogenization pair of straight lines yes so many things are pending okay so now I will take your leave this is meam signing off I will see you on Saturday good night bye-bye see you