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 are the points of trisection of AB then, P divides AB in 1 : 2



$\therefore$ Coordinates of P

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

= $(\frac{-3}{3},\frac{0}{3})=(-1,0)$

And, Q is the mid-point of PB

$\therefore$ Coordinates of $Q=(\frac{-1+(-7)}{2},\frac{0+4}{2})$

= $(-4,2)$

So, $P=(-1,0),Q=(-4,2)$