Question

Find the fourth vertex $D$ of parallelogram $ABCD$ whose three vertices are $A(-2,3),B(6,7)$ and $C(8,3)$.

Answer 1

Diagonals of parallelogram bisect each other

$\Rightarrow BD$ and $AC$ have same mid-point,

$M_x=\frac{6+\alpha }{2}=\frac{-2+8}{2}\Rightarrow \alpha =0$

$M_y=\frac{3+3}{2}=\frac{7+\beta }{2}\Rightarrow \beta =-i\Rightarrow D(0,-1)$