Let A(-3, 2), B(-5, -5), C(2, -3) and D(4, 4).
By distance formula,
d = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 ∴ A B = [ − 5 − ( − 3 ) ] 2 + [ − 5 − 2 ] 2 = [ − 5 + 3 ] 2 + [ − 7 ] 2 = [ − 2 ] 2 + [ − 7 ] 2 = 4 + 49 = 53 . ∴ B C = [ 2 − ( − 5 ) ] 2 + [ − 3 − ( − 5 ) ] 2 = [ 2 + 5 ] 2 + [ − 3 + 5 ] 2 = 7 2 + 2 2 = 49 + 4 = 53 . ∴ C D = ( 4 − 2 ) 2 + [ 4 − ( − 3 ) ] 2 = 2 2 + 7 2 = 4 + 49 = 53 . ∴ A D = [ 4 − ( − 3 ) ] 2 + ( 4 − 2 ) 2 = 7 2 + 2 2 = 49 + 4 = 53 . d = \sqrt{(x2 - x 1)^2 + (y2 - y 1)^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}. d = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 ∴ A B = [ − 5 − ( − 3 ) ] 2 + [ − 5 − 2 ] 2 = [ − 5 + 3 ] 2 + [ − 7 ] 2 = [ − 2 ] 2 + [ − 7 ] 2 = 4 + 49 = 53 . ∴ BC = [ 2 − ( − 5 ) ] 2 + [ − 3 − ( − 5 ) ] 2 = [ 2 + 5 ] 2 + [ − 3 + 5 ] 2 = 7 2 + 2 2 = 49 + 4 = 53 . ∴ C D = ( 4 − 2 ) 2 + [ 4 − ( − 3 ) ] 2 = 2 2 + 7 2 = 4 + 49 = 53 . ∴ A D = [ 4 − ( − 3 ) ] 2 + ( 4 − 2 ) 2 = 7 2 + 2 2 = 49 + 4 = 53 .
Calculating diagonals,
A C = [ 2 − ( − 3 ) ] 2 + [ − 3 − 2 ] 2 = [ 2 + 3 ] 2 + [ − 5 ] 2 = 5 2 + [ − 5 ] 2 = 25 + 25 = 50 = 5 2 units . B D = [ 4 − ( − 5 ) ] 2 + [ 4 − ( − 5 ) ] 2 = [ 4 + 5 ] 2 + [ 4 + 5 ] 2 = 9 2 + 9 2 = 81 + 81 = 162 = 9 2 units . 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}. A C = [ 2 − ( − 3 ) ] 2 + [ − 3 − 2 ] 2 = [ 2 + 3 ] 2 + [ − 5 ] 2 = 5 2 + [ − 5 ] 2 = 25 + 25 = 50 = 5 2 units . B D = [ 4 − ( − 5 ) ] 2 + [ 4 − ( − 5 ) ] 2 = [ 4 + 5 ] 2 + [ 4 + 5 ] 2 = 9 2 + 9 2 = 81 + 81 = 162 = 9 2 units .
Since, all sides are equal and diagonals are not equal.
∴ ABCD is a rhombus.
Area of rhombus = 1 2 × d 1 × d 2 \dfrac{1}{2} \times d1 \times d 2 2 1 × d 1 × d 2
= 1 2 × 5 2 × 9 2 = 1 2 × 45 × 2 = 45 sq. units . = \dfrac{1}{2} \times 5\sqrt{2} \times 9\sqrt{2} \\[1em] = \dfrac{1}{2} \times 45 \times 2 \\[1em] = 45 \text{ sq. units}. = 2 1 × 5 2 × 9 2 = 2 1 × 45 × 2 = 45 sq. units .
Hence, ABCD is a rhombus and area = 45 sq. units.
