Show that the points (-3, 2), (-5, -5), (2, -3) and (4, 4), taken in order, are the vertices of rhombus. Also, find its area. Do the given points form a square?

Let A(-3, 2), B(-5, -5), C(2, -3) and D(4, 4).

By distance formula,

$d = \sqrt{(x$2 - x1)^2 + (y2 - y1)^2} \\[1em] \therefore AB = \sqrt{[-5 - (-3)]^2 + [-5 - 2]^2} \\[1em] = \sqrt{[-5 + 3]^2 + [-7]^2} \\[1em] = \sqrt{[-2]^2 + [-7]^2} \\[1em] = \sqrt{4 + 49} \\[1em] = \sqrt{53}. \\[1em] \therefore BC = \sqrt{[2 - (-5)]^2 + [-3 - (-5)]^2} \\[1em] = \sqrt{[2 + 5]^2 + [-3 + 5]^2} \\[1em] = \sqrt{7^2 + 2^2} \\[1em] = \sqrt{49 + 4} \\[1em] = \sqrt{53}. \\[1em] \therefore CD = \sqrt{(4 - 2)^2 + [4 - (-3)]^2} \\[1em] = \sqrt{2^2 + 7^2} \\[1em] = \sqrt{4 + 49} \\[1em] = \sqrt{53}. \\[1em] \therefore AD = \sqrt{[4 - (-3)]^2 + (4 - 2)^2} \\[1em] = \sqrt{7^2 + 2^2} \\[1em] = \sqrt{49 + 4} \\[1em] = \sqrt{53}.

Calculating diagonals,

$AC = \sqrt{[2 - (-3)]^2 + [-3 - 2]^2} \\[1em] = \sqrt{[2 + 3]^2 + [-5]^2} \\[1em] = \sqrt{5^2 + [-5]^2} \\[1em] = \sqrt{25 + 25} \\[1em] = \sqrt{50} \\[1em] = 5\sqrt{2} \text{ units}. \\[1em] BD = \sqrt{[4 - (-5)]^2 + [4 - (-5)]^2} \\[1em] = \sqrt{[4 + 5]^2 + [4 + 5]^2} \\[1em] = \sqrt{9^2 + 9^2} \\[1em] = \sqrt{81 + 81} \\[1em] = \sqrt{162} \\[1em] = 9\sqrt{2} \text{ units}.$

Since, all sides are equal and diagonals are not equal.

∴ ABCD is a rhombus.

Area of rhombus = $\dfrac{1}{2} \times d$1 \times d2

$= \dfrac{1}{2} \times 5\sqrt{2} \times 9\sqrt{2} \\[1em] = \dfrac{1}{2} \times 45 \times 2 \\[1em] = 45 \text{ sq. units}.$

Hence, ABCD is a rhombus and area = 45 sq. units.