(-4, -2), (-3, -5), (3, -2) and (2, 3).

Solution:

Let the quadrilateral’s vertices be A (- 4, – 2), B (– 3, – 5), C (3, – 2), and D (2, 3). 

Divide the quadrilateral into two triangles by joining AC.

ABC and ACD are the two triangles we have.

The area of a triangle = 1/2 × [x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]

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

= 1/2 (12 + 0 + 9)

= 21/2 square units

The 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 ΔABC + Area of ΔACD = Area of quadrilateral ABCD

= (21/2 + 35/2)square units = 28 square units