Given:
ABCD is the parallelogram.
vertices are A(8,2), B(6,-4), and C(-5,-4)
We know the diagonals of the parallelogram bisect each other.
Find the midpoint of AC.
[tex]\begin{gathered} m=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}) \\ (x_1,y_1)=(8,2) \\ (x_2,y_2)=(-5,-4) \\ m=(\frac{8-5}{2},\frac{2-4}{2}) \\ m=(\frac{3}{2},-\frac{2}{2}) \\ m=(\frac{3}{2},-1) \end{gathered}[/tex]Now, the midpoint of BD is given as,
[tex]\begin{gathered} m=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}) \\ m=(\frac{3}{2},-1) \\ B\mleft(6,-4\mright),D(x,y) \\ (\frac{3}{2},-1)=(\frac{6+x}{2},\frac{-4+y}{2}) \\ \frac{6+x}{2}=\frac{3}{2},\frac{-4+y}{2}=-1 \\ 6+x=3,-4+y=-2 \\ x=-3,y=2 \end{gathered}[/tex]The coordinate of D is (-3,2).
