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

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 trianglesΔABC and ΔACD

Area of the triangle (ABC) = = [-4 (-5 + 2) -3 (-2+ 2) + 3 (-2+ 5)]

= (12 + 0 + 9)

= Square units

Area of the triangle (ACD) = [-4 (-2 – 3) + 3(3 + 2) + 2 (-2+ 2)]

= (20 + 15 + 0)

= Square units

Area of Quadrilateral ABCD = Area of the triangle (ABC) + Area of the triangle (ACD)

= +

= 28 Square units