 Hello friends welcome to the session I am Alka and its discussed a given question and the coordinates of the points of tri-section of the line segment joining 4 minus 1 and minus 2 minus 3. Before starting with the solution I would like to tell you the basic idea behind the version. The first one is the solution formula and we all know that under section formula for any given line segment a b with the coordinates x1, y1 and x2, y2 there is a point p which divides the line segment in the ratio m1 and m2. So we know the coordinates of the point p which are m1 x2 plus m2 x1 upon m1 plus m2 and m1 y2 plus m2 y1 upon m1 plus m2. So these are the coordinates of the point p which divide the line segment a b in the ratio m1 is to m2. Now let's start with the solution let a b be a line segment a b we are given in the question that coordinates of point a are 4 and minus 1 and coordinates of the point b are minus 2 and minus 3. Now we will divide it now let p and q let p, q be the points of trisection a b such that a p equal to pq equal to qb therefore we can say that p divides a b internally in the ratio 1 is to 2. Therefore coordinates of point p are x equal to we know that m1 x2 plus m2 x1 upon m1 plus m2. Now here m1 and m2 are 1 and 2. Now on substituting the value of m1 and 2 x1 x2 we can see from the figure that m x1 and x2 are 4 and minus 2. So we can write here x1 x2 x1 is 4 and x2 is minus 2. Now m1 is 1 into m2 is minus 2 plus m2 is 2 into x1 is 4 upon m1 plus m2 that is 1 plus 2 this is equal to minus 2 plus a upon 3 equal to 6 upon 3 this is equal to 2. Therefore we can say that x is equal to 2. Now we find the coordinates of y which is equal to m1 y2 plus m2 y1 upon m1 plus m2 here m1 and m2 are 1 and 2 and y1 is minus 1 and y2 is minus 3. This is equal to on substituting the value of m1 m2 y1 y2 we get 1 into minus 3 plus 2 into minus 1 upon m1 plus m2 that is 1 plus 2. This is equal to minus 3 minus 2 upon 3 equal to minus 5 upon 3. Therefore y is equal to minus 5 upon 3. Now we can say that coordinates of the point P are 2 and minus 5 upon 3. Now we can see from the figure that the point Q divides the line segment AB internally in the ratio 2 is to 1. So we can say that point Q divides AB internally in the ratio 2 is to 1. Therefore coordinates coordinates point Q are now here again we will use a section formula to find the coordinates of the point Q which are m1 x2 plus m2 x1 upon m1 plus m2 and m1 y2 plus m2 y1 upon m1 plus m2. Now we will substitute the value of m1 m2 x1 x2 y1 y2 we get now we will substitute the value of m1 m2 x1. Now we will substitute the value of m1 m2 x1 x2 y1 y2 we know that m1 is 2 and m2 is 1. So 2 into x2 is minus 2 plus m2 is 1 into x1 is 4 upon m1 plus 2 that is 2 plus 1 and 4 by m1 is 2 into y2 is minus 3 plus m2 is 1 into y1 is minus 1 upon 2 plus 1. So this is equal to we get minus 4 plus 4 upon 3 and 4 by it is minus 6 minus 1 upon 3 which is equal to 0 and minus 7 upon 3. Therefore the coordinates of the point Q are 0 and minus 7 upon 3. Therefore we can say that the required the required points minus 5 upon 3 and the second point is with the coordinate 0 and minus 7 upon 3. This is the required answer. So hope you understood the solution and enjoyed the session. Goodbye and take care.