Question

If the three vertices of a parallelogram are $(1,3),(4,2)$ and $(3,5)$, then find the fourth vertex.

Answer 1

Let $A(1,3),B(4,2),C(3,5)$ and $D(x,y)$ be the vertices of the parallelogram $ABCD$.

Midpoint of $AC=(\frac{1+3}{2},\frac{3+5}{2})=(2,4)$

Also midpoint of $BD=(\frac{4+x}{2},\frac{2+y}{2})=(\frac{x+4}{2},\frac{y+2}{2})$

Since the diagonals of a parallelogram bisect each other, the mid-points of $AC$ and $BD$ are the same.

$\Rightarrow \frac{x+4}{2}=2$ and $\frac{y+2}{2}=4$

$\Rightarrow x+4=4$ and $y+2=8$

$\therefore x=4-4=0,y=8-2=6$

Hence the fourth vertex is $(0,6)$