Find the area of quadrilateral ABCD whose vertices are A(–3, –1), B(–2, – 4), C(4, –1) and D(3, 4).
For triangle ABC
Area of triangle
= 1/2(x1(y2−y3) + x2(y3−y1) + x3(y1−y2))
= 1/2(–3(–4 + 1)–2(–1 + 1) + 4(–1 + 2))
= 1/2(–9 + 0 + 4)
= 1/2(5)
For triangle ACD
Area of triangle
= 1/2(x1(y2−y3) + x2(y3−y1) + x3(y1−y2))
= 1/2(–3(–1–4) + 4(4 + 1) + 3(–1 + 1))
= 1/2(–15 + 20 + 0)
= 1/2(5)
Area of ABCD = Area of ABC + Area of ACD
= 5 sq units