Question

If three consecutive vertices of a parallelogram ABCD are A(1, −2), B(3, 6) and C(5, 10), find its fourth vertex D.

Answer 1

Let A(1, −2), B(3, 6) and C(5, 10) be the three vertices of a parallelogram ABCD and the fourth vertex be D(a, b).
Join AC and BD intersecting at O.

We know that the diagonals of a parallelogram bisect each other.
Therefore, O is the midpoint of AC as well as BD.
$$\begin{gathered} \mathrm{Midpoint}\mathrm{of}AC=\left(\frac{1+5}{2},\frac{-2+10}{2}\right)=\left(\frac{6}{2},\frac{8}{2}\right)=\left(3,4\right) \\ \mathrm{Midpoint}\mathrm{of}BD=\left(\frac{3+a}{2},\frac{6+b}{2}\right) \\ \mathrm{Therefore},\frac{3+a}{2}=3\text{and}\frac{6+b}{2}=4 \\ \Rightarrow 3+a=6\text{and}6+b=8 \\ \Rightarrow a=6-3\text{and}b=8-6 \\ \Rightarrow a=3\text{and}b=2 \end{gathered}$$
Therefore, the fourth vertex is D(3, 2).