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

$$AP:PB=1:2$$
Coordinates of $P$ are: $(\frac{1(-7)+2\left(2\right)}{1+2},\frac{1(4+2(-2)}{1+2})$ by using section formula
$$\Rightarrow (\frac{-7+4}{3},\frac{4-4}{3})=(-1,0)$$
we know $Q$ is the mid point of PB
therefore coordinate of $Q(\frac{-1+(-7)}{3},\frac{0+4}{3})$ using midpoint formula
$$(\frac{-8}{2},\frac{4}{2})=(-4,2)$$