Question

Show that points P(2, –2), Q(7, 3), R(11, –1) and S (6, –6) are vertices of a parallelogram.

Answer 1

The given points are P(2, –2), Q(7, 3), R(11, –1) and S (6, –6).
$$\begin{gathered} \mathrm{PQ}=\sqrt{{\left(3-\left(-2\right)\right)}^2+{\left(7-2\right)}^2}=\sqrt{25+25}=5\sqrt{2} \\ \mathrm{QR}=\sqrt{{\left(7-11\right)}^2+{\left(3-\left(-1\right)\right)}^2}=\sqrt{16+16}=4\sqrt{2} \\ \mathrm{RS}=\sqrt{{\left(11-6\right)}^2+{\left(-1-\left(-6\right)\right)}^2}=\sqrt{25+25}=5\sqrt{2} \\ \mathrm{PS}=\sqrt{{\left(2-6\right)}^2+{\left(-2-\left(-6\right)\right)}^2}=\sqrt{16+16}=4\sqrt{2} \end{gathered}$$
So, PQ = RS and QR = PS
Thus, opposite sides are equal.
Hence, the given points form a parallelogram.