Verify that points P(–2, 2), Q(2, 2) and R(2, 7) are vertices of a right angled triangle.

$$\begin{gathered} \mathrm{PQ}=\sqrt{{\left(-2-2\right)}^2+{\left(2-2\right)}^2} \\ =\sqrt{{\left(-4\right)}^2+0} \\ =\sqrt{16}=4 \\ \mathrm{QR}=\sqrt{{\left(2-2\right)}^2+{\left(2-7\right)}^2} \\ =\sqrt{0+{\left(-5\right)}^2} \\ =\sqrt{25} \\ =5 \\ \mathrm{PR}=\sqrt{{\left(-2-2\right)}^2+{\left(2-7\right)}^2} \\ =\sqrt{{\left(-4\right)}^2+{\left(-5\right)}^2} \\ =\sqrt{16+25} \\ =\sqrt{41} \end{gathered}$$
$$\begin{gathered} {\mathrm{PQ}}^2+{\mathrm{QR}}^2={\mathrm{PR}}^2 \\ \Rightarrow 4^2+5^2=16+25=41=\mathrm{PR} \end{gathered}$$
Thus, the square of the third is equal to the sum of the squares of the other two sides.
Thus, they are the vertices of the right angled triangle.