Question

The three vertices of a parallelogram $ABCD$ are $A(3,-4),B(-1,-3)$ and $C(-6,2)$. Find the coordinate of vertex $D$.

Answer 1

Let coordinates of vertex $D$ be $(x,y)$

In parallelogram diagonals are bisect each other.

$\therefore$ Mid-point of $AC=$ Mid-point of $BD$

$\Rightarrow$ $(\frac{3+(-6)}{2},\frac{-4+2}{2})=(\frac{-1+x}{2},\frac{-3+y}{2})$

$\Rightarrow$ $(\frac{-3}{2},\frac{-2}{2})=(\frac{-1+x}{2},\frac{-3+y}{2})$

$\Rightarrow$ $(\frac{-3}{2},-1)=(\frac{-1+x}{2},\frac{-3+y}{2})$

Now,

$\Rightarrow$ $\frac{-3}{2}=\frac{-1+x}{2}$

$\Rightarrow$ $-6=-2+2x$

$\Rightarrow$ $-4=2x$

$\therefore$ $x=-2$

$\Rightarrow$ $-1=\frac{-3+y}{2}$

$\Rightarrow$ $-2=-3+y$

$\Rightarrow$ $1=y$

$\therefore$ $y=1$

$\therefore$ Coordinates of vertex $D$ is $(-2,1)$