[tex]\begin{gathered} In\text{ a parallelogram diagonals bisects each other} \\ \text{hence midpoint of }BD\text{ = midpoint of AC} \\ A=(-4,2) \\ B=(6,-5) \\ C=(3,-11) \end{gathered}[/tex][tex]\begin{gathered} (\frac{-4+3}{2},\frac{2-11}{2})=(\frac{6+x}{2},\frac{-5+y}{2}) \\ \text{hence} \end{gathered}[/tex][tex]\begin{gathered} (\frac{-1}{2},\frac{-9}{2})=(\frac{6+x}{2},\frac{-5+y}{2}) \\ \text{this implies that} \\ 6+x=-1 \\ \text{and} \\ -5+y=-9 \\ \text{therefore} \\ x=-1-6\Rightarrow x=-7 \\ y=-9+5\Rightarrow x=-4 \\ \text{hence the vertex is at (-7,-4)} \end{gathered}[/tex]
