Find the area of quadrilateral ABCD whose vertices are A(–5, 7), B(–4, –5), C(–1, – 6) and D(4, 5).
For triangle ABC
Area of triangle
= 1/2(x1(y2−y3) + x2(y3−y1) + x3(y1−y2))
= 1/2(–5(–5 + 6)–4(–6–7)–1(7 + 5))
= 1/2(–5 + 52–12)
= 1/2(35)
For triangle ACD
Area of triangle
= 1/2(x1(y2−y3) + x2(y3−y1) + x3(y1−y2))
= 1/2(–5(–6–5)–1(5–7) + 4(7 + 6))
= 1/2(–55 + 2 + 52)
= 1/2(1)
Area of ABCD = Area of ABC + Area of ACD
= 18 sq units
5