If the points A(2, –2), B (8,4), C(5,7) are the three vertices of a parallelogram ABCD taken in order, find the fourth vertex D.

Let D(a,b) be the fourth vertex since ABCD is a parallelogram, the diagonals \$\overline{AC}\$ and \$\overline{BD}\$ bisect each other.

That is the mid point of \$\overline{AC}\$ is the same as the mid point of \$\overline{BD}\$.

But the mid point of AC is

\$ ( \frac{2 + 5}{2}, \frac{-2 + 7}{2} ) = ( \frac{7}{2}, \frac{5}{2}) \$

And the mid point of BD is

\$( \frac{8 + a}{2}, \frac{4 + b}{2} ) \$

Equating the co-ordinates, we get

\begin{align} \frac{8 + a}{2} &= \frac{7}{2}, & \frac{4 + b}{2} &= \frac{5}{2} \\ 8 + a &= 7, & 4 + b &= 5 \\ a &= -1, & b &= 1 \\ \end{align}

Answer

The fourth vertex D \$= (−1,1)\$