• question_answer 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.

 Since, P and Q are the points of trisection of AB then, P divides AB in $1:2$. $\therefore$ Coordinates of P $=\left( \frac{1(-7)+2(2)}{1+2},\frac{1(4)+2(-2)}{1+2} \right)$ $=\left( \frac{-3}{3},\frac{0}{3} \right)=(-1,0)$ And, Q is the mid-point of PB $\therefore$Coordinates of $Q=\left( \frac{-1+(-7)}{2},\frac{0+4}{2} \right)$ $=(-4,2)$ So,                   $P=(-1,0),Q=(-4,2)$