Question

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

Answer 1

In a parallelogram $ABCD$ co-ordinates are $A(5,2),B(3,5,)$

and $C(2,3)$.Let the fourth vertex $D$ be $(x,y)$

Midpoint of $AC=(\frac{5+2}{2},\frac{2+3}{2})=(\frac{7}{2},\frac{5}{2})$

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,

$\therefore (\frac{3+x}{2},\frac{5+y}{2})=(\frac{7}{2},\frac{5}{2})$

$\Rightarrow \frac{3+x}{2}=\frac{7}{2},\frac{5+y}{2}=\frac{5}{2}$

$\Rightarrow x+3=7,y+5=5$

$\Rightarrow x=7-3=4,y=5-5=0$

Hence, the fourth vertex $D$ be $(4,0)$