Question

P and Q be the points of trisection of the line segment joining points A(2,-2) and B(-7,4) such that P is nearer to A. Then find the coordinates of P and Q.
$\left[2Marks\right]$

Answer 1

P and Q are the points of trisection of AB, therefore AP = PQ = QB.
Thus, P divides AB internally in the ratio 1:2 and Q divides AB internally in the ratio 2:1.

$P=(\frac{1\times (-7)+2\times \left(2\right)}{1+2},\frac{1\times \left(4\right)+2\times (-2)}{1+2})$
$=(\frac{-7+4}{3},\frac{4-4}{3})$
$=(\frac{-3}{3},\frac{0}{3})$
$=(-1,0)$
$\left[1Mark\right]$

$Q=(\frac{2\times (-7)+1\times \left(2\right)}{2+1},\frac{2\times \left(4\right)+1\times (-2)}{2+1})$
$=(\frac{-14+2}{3},\frac{8-2}{3})$
$=(\frac{-12}{3},\frac{6}{3})$
$=(-4,2)$
$\left[1Mark\right]$