Question

Find the coordinates of points of trisection of the line segment joining $A(4,-8)$ and $B(7,4)$

Answer 1

Given points $A(4,-8)$ and $B(7,4)$

Points of Trisection divide the line in ratio $2:1$ and $1:2$

$m=2;n=1x=\frac{m{x}_2+n{x}_1}{m+n};y=\frac{m{y}_2+n{y}_1}{m+n}x=\frac{2\left(7\right)+1\left(4\right)}{2+1};y=\frac{2\left(4\right)+1(-8)}{2+1}x=\frac{14+4}{3};y=\frac{8-8}{3}x=6;y=0$

Again

$m=1;n=2x=\frac{m{x}_2+n{x}_1}{m+n};y=\frac{m{y}_2+n{y}_1}{m+n}x=\frac{1\left(7\right)+2\left(4\right)}{2+1};y=\frac{1\left(4\right)+2(-8)}{2+1}x=\frac{7+8}{3};y=\frac{4-16}{3}x=5;y=-3$
The points of trisection are $(6,0)$ and $(5,-3)$