Question

Show that the points A(3, 1), B(0, −2), C(1, 1) and D(4, 4) are the vertices of a parallelogram ABCD.

Answer 1

The points are A(3, 1), B(0, −2), C(1, 1) and D(4, 4).
Join AC and BD, intersecting at O.


We know that the diagonals of a parallelogram bisect each other.
$$\begin{gathered} \mathrm{Midpoint}\mathrm{of}AC=\left(\frac{3+1}{2},\frac{1+1}{2}\right)=\left(\frac{4}{2},\frac{2}{2}\right)=\left(2,1\right) \\ \mathrm{Midpoint}\mathrm{of}BD=\left(\frac{0+4}{2},\frac{-2+4}{2}\right)=\left(\frac{4}{2},\frac{2}{2}\right)=\left(2,1\right) \end{gathered}$$
Thus, the diagonals AC and BD have the same midpoint.
Therefore, ABCD is a parallelogram.