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.

Answer:

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,                   \[P=(-1,0),Q=(-4,2)\]