Question

The line joining the points $(1,-2)$ and $(-3,4)$ is trisected. Find the coordinates of the points of trisection.

Answer 1

Let $A(1,-2)$ and $B(-3,4)$ is trisected by point $P$ and $Q$

$Q$ divides $AB$ in $2:1$

The coordinates of point when ${(x}_1,y_1)$ and ${(x}_2,y_2)$ are divided in $m:n$ internally is $(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n})$

Let the coordinates of $Q$ be $(h,k)$

$h=\frac{2(-3)+1\left(1\right)}{2+1}=\frac{-6+1}{3}=-\frac{5}{3}k=\frac{2\left(4\right)+1(-2)}{2+1}=\frac{8-2}{3}=2\Rightarrow Q(-\frac{5}{3},2)$

Now $P$ divides $AQ$ in $1:1$

Let the coordinates of $P$ be $(a,b)$

$a=\frac{1(-\frac{5}{3})+1\left(1\right)}{1+1}=\frac{-\frac{5}{3}+1}{2}=\frac{-\frac{2}{3}}{2}=-\frac{1}{3}b=\frac{1\left(2\right)+1(-2)}{1+1}=\frac{2-2}{2}=0\Rightarrow P(-\frac{1}{3},0)$

So, the coordinates of point of trisection are $(-\frac{1}{3},0)$ and $(-\frac{5}{3},2)$