Find the area of quadrilateral ABCD whose vertices are A(3, –1), B(9, –5), C(14, 0) and D(9, 19).
For triangle ABC
Area of triangle
= 1/2(x1(y2−y3) + x2(y3−y1) + x3(y1−y2))
= 1/2(3(–5–0) + 9(0 + 1) + 14(–1 + 5))
= 1/2(–15 + 9 + 56)
= 1/2(50)
= 25
For triangle ACD
Area of triangle
= 1/2(x1(y2−y3) + x2(y3−y1) + x3(y1−y2))
= 1/2(3(0–19) + 14(19 + 1) + 9(–1–0))
= 1/2(–57 + 280–9)
= 1/2(214)
= 107
Area of ABCD = Area of ABC + Area of ACD
= 25 + 107
= 132 sq units
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