Show that the points A(1, 1), B(5, 4), C(2, 8) and D(−2, 5) 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.

By the distance formula:

\(AB = \sqrt{(5 - 1)^2 + (4 - 1)^2} = \sqrt{16 + 9} = 5\)
\(BC = \sqrt{(2 - 5)^2 + (8 - 4)^2} = \sqrt{9 + 16} = 5\)
\(CD = \sqrt{(-2 - 2)^2 + (5 - 8)^2} = \sqrt{16 + 9} = 5\)
\(DA = \sqrt{(1 + 2)^2 + (1 - 5)^2} = \sqrt{9 + 16} = 5\)

Diagonals: \(AC = \sqrt{(2 - 1)^2 + (8 - 1)^2} = \sqrt{1 + 49} = \sqrt{50}\) and \(BD = \sqrt{(-2 - 5)^2 + (5 - 4)^2} = \sqrt{49 + 1} = \sqrt{50}\).

All sides are equal, so □ ABCD is a rhombus; its diagonals are also equal, so it is a square. (Also \(AB^2 + BC^2 = 25 + 25 = 50 = AC^2\), so ∠B = 90°.)