Question

Find the coordinates of the points of trisection of the line segment joining the points (3, – 2) and (– 3, – 4).

Answer 1

Let us suppose the given points are A (3, – 2) and B(– 3, – 4).

Let P and Q be the point of trisection. Therefore, we have

$AP=PQ=QB$

Trisection means is to divide a line segment into three equal parts. Hence, we can say that P divides AB in the ratio of 1:2 and Q divides in 2:1 .

Thus, coordinate of P is given by

$P=(\frac{mx_2+nx_1}{m+n} ,\frac{my_2+ny_1}{m+n})\\ \\ P=(\frac{1\times (-3)+2\times 3}{1+2} , \frac{1\times (-4)+2\times (-2)}{1+2} )\\ \\ P=(\frac{-3+6}{3} ,\frac{-8}{3} )\\ \\ P=(1,-\frac{8}{3} )$

Similarly the coordinate of Q is given by

$Q=(\frac{2\times (-3)+1\times 3}{1+2} , \frac{2\times (-4)+1\times (-2)}{2+1} )\\ \\ Q=\left (\frac{-6+3}{3},\frac{-8-2}{3}  \right )\\ \\ Q=\left ( \frac{-3}{3},-\frac{10}{3} \right )\\ \\ Q=\left ( -1,-\frac{10}{3} \right )$

Therefore, the coordinates of the point of trisection are

$(1,-\frac{-8}{3})\text{ and }\left ( -1,-\frac{10}{3} \right )$