 Hello and welcome to the session. In this session we will discuss a question which says that show that the lines joining the midpoints of the sides of the triangle to the vertices opposite to it are concurrent. Now before starting the solution of this question, we should know our result. And that is, if it is possible to find three constants lambda, mu and mu to be 0 such that lambda into even x plus b1y plus even the whole plus mu into a2x plus b2y plus c3 the whole plus mu into a3x plus b3y plus c3 the whole is equal to 0. Now let us name it as 1. Then the lines even x plus b1y plus c1 is equal to 0, a2x plus b2y plus c2 is equal to 0, a3x plus b3y plus c3 is equal to 0 are concurrent. Let us name it as 2, 3 and 4 and particularly when lambda is equal to mu is equal to mu is equal to 1. Then the line 3 plus b1y plus c1 plus a2x plus b3y plus c2 plus a3h plus b3y plus c3 is equal to 0. Now let this be p, this be q and this be r. So the line is equal to 0, q is equal to 0 and r is equal to 0 where p, q and r represent the linear expression in x and y is equal to 0. Then this result will work out as a key idea with the solution. Now in this question we have to show that the lines joining the midpoints of the sides of the triangle to the vertices opposite to it are concurrent. Now let us take a triangle a, b, c in which the coordinates of a, r, x1, y1, bits of b, r, x2, y2 and the coordinates of c, r, b, e, cf and ad are points of the sides of a triangle to the vertices opposite to a, b, e are the medians. Therefore d, e, b, c now point formula of dr, y3, y2 as d is the midpoint of bc so its coordinates are given by midpoint formula. So by midpoint formula of er, x3 by 2 and y1 plus y3 by 2, x2 by 2 and y1 plus y2 by 2. So the two points x1, y1 and x2, y2 then the equation of that line is given by this formula. Therefore the equation ad, ad by 3 by 4 coordinates of b and these are the coordinates of a. The equation of ad by 2 point form will be y minus y1 is equal to 2 plus y3 by 2 minus y1 whole upon 3 by 2 minus x1 double into x minus x1 double. That is y minus y1 is equal to y2 minus y1 over x2 minus x1 into x minus x1 which further implies y minus y1 is equal to plus y3 whole upon x2 plus x3 minus 2 x1 double into x minus x1 double. Now by past multiplying this implies y1 whole into x1 whole is equal to y2 plus y3 minus 2 y1 whole. That implies y into x2 plus x3 minus 2 x1 whole minus y1 into x2 plus x3 minus 2 x1 whole is equal to x into y2 plus y3 minus 2 y1 whole minus x1 into y2 plus y3 minus 2 y1 whole. This further implies y into x2 x1 whole minus x into y2 plus y3 minus 2 y1 whole minus y1 into x2 plus x3 minus 2 x1 whole plus x1 into y2 plus y3 minus 2 y1 whole is equal to 0. And let us explain this as how this one is the equation of the median A D. The median B E is 2 the whole minus x into y1 plus y3 minus 2 y2 the whole minus y2 into x1 plus x3 minus 2 x2 the whole into y1 plus y3 minus 2 y2 the whole is equal to this as equation number 2. Now y into x1 plus x2 minus 2 x3 the whole minus x into y1 plus y2 minus 2 y3 the whole minus y3 into x1 plus 3 the whole plus x3 into y1 plus y2 minus 2 y3 the whole is equal to 0. And let us explain this as equation number 3. Now we have this as the equation of the median A D that is represented by equation number 1. This is the equation of B E which is given by equation number 2 and this is the equation of C F which is given by equation number 3. Now adding 2 and 3 minus 2 x1 plus x1 plus x3 minus 2 x1 plus 3 the whole into y2 plus y3 minus 2 y1 plus y1 plus y3 minus 2 y2 plus y1 plus y2 minus 2 y3 the whole minus x3 minus 2 y3 the whole is equal to 0. Now we are going to add minus x1 plus y1 plus x1 y1 plus x1 y2 plus x1 y3 minus 2 x1 y1 plus x1 y2 minus x3 y2 plus 2 x2 y2 plus x2 y1 plus x2 y3 minus 2 x2 y2 plus 2 x3 plus 2 x3 y3 minus 2 x1 y2 plus 2 x1 y3 plus x3y1 plus x3y2 minus 2x3y3. Now these all terms will be cancelled with each other equal to y into 0 minus x into 0 plus 0 which is equal to 0 minus 0 plus 0 which is equal to the idea it is given that the three lines p is equal to 0 q is equal to 0 and r is equal to 0 where p, q and r are the linear expression in x and y is equal to 0 then the given lines are concurrent. Now in this case and this was the line this is equal to 0 the lines represented by 1, 2 and 3 sides of the triangle to the vertices opposite to each other solution of the given question and that's all for this session. Hope you all have enjoyed this session.