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

\[PQ = \sqrt{\left( - 2 - 2 \right)^2 + \left( 2 - 2 \right)^2}\]

\[ = \sqrt{\left( - 4 \right)^2 + 0}\]

\[ = \sqrt{\left( - 4 \right)^2 + 0}\]

\[ = \sqrt{16} = 4\]

\[QR = \sqrt{\left( 2 - 2 \right)^2 + \left( 2 - 7 \right)^2}\]

\[ = \sqrt{0 + \left( - 5 \right)^2}\]

\[ = \sqrt{25}\]

\[ = 5\]

\[PR = \sqrt{\left( - 2 - 2 \right)^2 + \left( 2 - 7 \right)^2}\]

\[ = \sqrt{\left( - 4 \right)^2 + \left( - 5 \right)^2}\]

\[ = \sqrt{16 + 25}\]

\[ = \sqrt{41}\]

\[{PQ}^2 + {QR}^2 = {PR}^2 \]

⇒4² + 5² = 16 + 25 = 41 = PR 

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.