Question

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

• A. $P(-4,2)$
  $Q(-1,0)$
• B. $P(-1,0)$
  $Q(-4,2)$
• C. $P(-2,0)$
  $Q(-3,2)$
• D. $P(-4,-1)$
  $Q(2,0)$

Answer 1

The correct option is B

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

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

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

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