Question

If three connective vertices of a parallelogram $ABCD$ are $A(1,-2),B(3,6)$ and $C(5,10)$, find its fourth vertex $D$.

Answer 1

The vertices of a parallelogram $ABCD$ are

$A=(1,-2)$

$B=(3,6)$

$C=(5,10)$

Let, the fourth vertex be $D=(x,y)$ then,

Mid-point of diagonal $AC=$ Mid-point of diagonal $BD$

$⟹(\frac{1+5}{2},\frac{-2+10}{2})=(\frac{3+x}{2},\frac{6+y}{2})$

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

$⟹\frac{3+x}{2}=3$ and $\frac{6+y}{2}=4$

$⟹3+x=6$ and $6+y=8$

$⟹x=3$ and $y=2$

$\therefore$ The fourth vertex $D=(3,2)$