Find the area of the quadrilateral whose vertices, taken in order, are (– 4, – 2), (– 3, – 5), (3, – 2) and (2, 3).

Let ABC be any triangle whose vertices are A(x₁, y₁), B(x₂, y₂), and C(x₃, y₃).

Then, Area of a triangle is given by 1/2 [x₁ (y₂ - y₃) + x₂ (y₃- y₁) + x₃ (y₁ - y₂)] ------ (1)

Let the vertices of the quadrilateral be A (- 4, - 2), B (- 3, - 5), C (3, - 2), and D (2, 3)

Join AC to form two triangles ∆ABC and ∆ACD.

We know that, area of a triangle = 1/2 [x₁ (y₂ - y₃) + x₂ (y₃- y₁) + x₃ (y₁ - y₂)]

By substituting the values of vertices, A, B, C in the formula.

Area of ΔABC = 1/2 [(- 4){(- 5) - (- 2)} + (- 3){(- 2) - (- 2)} + 3{(- 2) - (- 5)}]

= 1/2 (12 + 0 + 9)

= 21/2 square units

By substituting the values of vertices, A, C, D in the Equation (1),

Area of ΔACD = 1/2 [(- 4){(- 2) - 3} + 3{(3) - (- 2)} + 2{(- 2) - (- 2)}]

= 1/2 (20 + 15 + 0)

= 35/2 square units

Area of ABCD = Area of ΔABC + Area of ΔACD

= (21/2 + 35/2) square units

= 28 square units