Find the points of trisection of the line segment joining the points:

(3, -2) and (-3, -4)

#### Solution

The coordinates of a point which divided two points `(x_1,y_1)` and `(x_2, y_2)` internally in the ratio m:n is given by the formula,

`(x,y) = ((mx_2 + nx_1)/(m + n), (my_2 + ny_1)/(m + n))`

The points of trisection of a line are the points which divide the line into the ratio 1: 2.

Here we are asked to find the points of trisection of the line segment joining the points *A*(3*,*−2) and *B*(−3*,*−4).

So we need to find the points which divide the line joining these two points in the ratio 1: 2 and 2: 1.

Let *P*(*x, y*) be the point which divides the line joining ‘*AB*’ in the ratio 1: 2.

(x,y) = `(((1(3) + 2(-3))/(1 + 2)), ((1(-2) + 2(-4))/(1 + 2))`

`(e, d) = (-1, -10/3)`

Therefore the points of trisection of the line joining the given points are `(1, 8/3) and (-1, -10/3)`