Question

If the vertices of a parallelogram are $A(-2,3),B(3,-1),C(p,q)$ and $D(-1,9)$, then the value of $p+q$ is

Answer 1

Given : $A(-2,3),B(3,-1),C(p,q)$ and $D(-1,9)$
We know that diagonals of a parallelogram bisect each other.


Then,
$(\frac{p-2}{2},\frac{q+3}{2})=(\frac{3-1}{2},\frac{-1+9}{2})\Rightarrow (\frac{p-2}{2},\frac{q+3}{2})=(1,4)\Rightarrow p=4,q=5\therefore p+q=9$