Question

The three vertices of a parallelogram taken in order are $(-1,0)$, $(3,1)$ and $(2,2)$ respectively. Find the coordinates of the fourth vertex.

Answer 1

Let $A(-1,0)$, $B(3,1),C(2,2)$ and $D(x,y)$ be the vertices of a parallelogram $ABCD$ taken in order.

Since the diagonals of a parallelogram bisect each other. Coordinates of the midpoint of $AC$ will be the same as coordinates of the midpoint of $BD$.

If $\left(x_1,y_1\right)$ and $\left(x_1,y_1\right)$ are two points then their midpoint is defined as $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right).$

By using the above conclusion,

$(\frac{-1+2}{2},\frac{0+2}{2})=(\frac{3+x}{2},\frac{1+y}{2})$

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

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

$\Rightarrow x=-2$ and $y=1$

Hence the fourth, vertex of the parallelogram is $(-2,1).$