Question

Three consecutive vertices of a parallelogram $ABCD$ are $A(-1,2),B(3,5)$ and $C(2,3)$.Find the fourth vertex $D$

Answer 1

In a parallelogram $ABCD$ are $A(-1,2,4),B(3,5,-2)$ and $C(2,3,4)$.Let the fourth vertex $D$ be $(x,y,z)$

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

Midpoint of $BD=(\frac{3+x}{2},\frac{5+y}{2})$

Since $ABCD$ is a parallelogram and in a parallelogram, the diagonals bisect each other, so the mid-points of $AC$ and $BD$ are same,

$\therefore (\frac{3+x}{2},\frac{5+y}{2})=(\frac{1}{2},\frac{5}{2})$

$\Rightarrow \frac{3+x}{2}=\frac{1}{2},\frac{5+y}{2}=\frac{5}{2}$

$\Rightarrow x+3=1,y+5=5$

$\Rightarrow x=1-3=-2,y=5-5=0$

Hence, the fourth vertex $D$ be $(-2,0)$