Question

A parallelogram has vertices at the points A(4,-3), B(-2,-3), C(-1, 1), and D(5,1). What is the area of this parallelogram

Answer 1

Area of a parallelogram = 6√17

Explanation:

Points of a parallelogram : A(4,-3), B(-2,-3), C(-1, 1), and D(5,1)

Distance of AB = $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

                         $=\sqrt{(-2-4)^2 + (-3+3)^2} \\\\= \sqrt{36} \\\\= 6$

Distance of BC = $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

                         $= \sqrt{(-1+2)^2 + (1+3)^2} \\\\=\sqrt{17}$

Distance of CD = $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

                         $= \sqrt{(5+1)^2 - (1-1)^2} \\\\= \sqrt{36} \\\\= 6$

Distance of AD = $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

                         $=\sqrt{(5-4)^2 + (1+3)^2} \\\\= \sqrt{17}$

Therefore,

area of parallelogram = Length X width

                                    = 6 X √17

                                    = 6√17