Show that the points (2, 1), (0, 3), (-2, 1) and (0, -1), taken in order, are the vertices of a square. Also find the area of the square.

Let A(2, 1), B(0, 3), C(-2, 1) and D(0, -1) be the four points.

By distance formula,

$d = \sqrt{(x$2 - x1)^2 + (y2 - y1)^2} \\[1em] \therefore AB = \sqrt{(0 - 2)^2 + (3 - 1)^2} \\[1em] = \sqrt{(-2)^2 + (2)^2} \\[1em] = \sqrt{4 + 4} \\[1em] = \sqrt{8} \text{ units}. \\[1em] \therefore BC = \sqrt{(-2 - 0)^2 + (1 - 3)^2} \\[1em] = \sqrt{(-2)^2 + (-2)^2} \\[1em] = \sqrt{4 + 4} \\[1em] = \sqrt{8} \text{ units}.\\[1em] \therefore CD = \sqrt{[0 - (-2)]^2 + [-1 - 1]^2} \\[1em] = \sqrt{[0 + 2]^2 + [-2]^2} \\[1em] = \sqrt{4 + 4} \\[1em] = \sqrt{8} \text{ units}. \\[1em] \therefore AD = \sqrt{(0 - 2)^2 + (-1 - 1)^2} \\[1em] = \sqrt{(-2)^2 + (-2)^2} \\[1em] =\sqrt{4 + 4} \\[1em] = \sqrt{8} \text{ units}.

Since, AB = BC = CD = AD i.e. all sides are equal so, ABCD can be a rhombus or a square.

Calculating diagonals,

$AC = \sqrt{(-2 - 2)^2 + (1 - 1)^2} \\[1em] =\sqrt{(-4)^2 + 0^2} \\[1em] = \sqrt{16} \\[1em] = 4 \text{ units}. \\[1em] BD = \sqrt{(0 - 0)^2 + (-1 - 3)^2} \\[1em] = \sqrt{0^2 + (-4)^2} \\[1em] = \sqrt{16} \\[1em] = \sqrt{4} \text{ units}.$

Since, diagonals are equal.

∴ ABCD is a square.

Area of square = (side)²

= (AB)²

= $(\sqrt{8})^2$

= 8 sq. units.

Hence, proved that (2, 1), (0, 3), (-2, 1) and (0, -1), taken in order, are the vertices of a square and area = 8 sq. units.