Question

Let P and Q be the points of trisection of the line segment joining the points A$(2,-2)$ and B$(-7,4)$ such that P is nearer to A. Find the coordinates of P and Q.

Answer 1

Since $P$ and $Q$ be the points of trisection of the line segment joining the points A$(2,-2)$ and B$(-7,4)$ such that $P$ is nearer to $A$.

Therefore, $P$ divides line segment in the ratio $1:2$ and $Q$ divides in $2:1$ as shown in the figure.

Using section formula,

Coordinates of $P$ = $[\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n}]$

= $[\frac{1(-7)+2\left(2\right)}{1+2},\frac{1\left(4\right)+2(-2)}{1+2}]$

= $[\frac{-3}{3},\frac{0}{3}]$

= $[-1,0]$

Coordinates of Q =$[\frac{2(-7)+1\left(2\right)}{2+1},\frac{2\left(4\right)-1(-2)}{2+1}]$

= $[\frac{-12}{3},\frac{6}{3}]$

= $[-4,2]$