Show that the points A(2, –1), B(5, 3), C(1, 6) and D(–2, 2) are the vertices of a square.

Answer: All four sides are 5 units and both diagonals are \(\sqrt{50}\) units, so □ABCD is a square.

Sides (by distance formula):

\(AB = \sqrt{(5 - 2)^2 + (3 + 1)^2} = \sqrt{9 + 16} = 5\)

\(BC = \sqrt{(1 - 5)^2 + (6 - 3)^2} = \sqrt{16 + 9} = 5\)

\(CD = \sqrt{(-2 - 1)^2 + (2 - 6)^2} = \sqrt{9 + 16} = 5\)

\(DA = \sqrt{(2 + 2)^2 + (-1 - 2)^2} = \sqrt{16 + 9} = 5\)

So AB = BC = CD = DA, and □ABCD is a rhombus.

Diagonals:

\(AC = \sqrt{(1 - 2)^2 + (6 + 1)^2} = \sqrt{1 + 49} = \sqrt{50}\)

\(BD = \sqrt{(-2 - 5)^2 + (2 - 3)^2} = \sqrt{49 + 1} = \sqrt{50}\)

A rhombus whose diagonals are equal is a square. \(\therefore\) □ABCD is a square.