Find the area of quadrilateral ABCD whose vertices are A(3, -1), B(9, -5), C(14, 0) and D(9, 19).
By joining A and C, we get two triangles ABC and ACD
Let A (3, -1), B (9, -5) and C (14, 0) and D (9, 19)
(x1=3,y1=−1),(x2=9,y2=−5),(x3=14,y3=0),(x4=9,y4=19)
Then
Area of triangle ABC
=12(x1(y2−y3)+x2(y3−y1)+x3(y1−y2))
=12×(3(−5−0)+(9)(0+1)+(14)(−1+5))
=12×(3(−5)+9(1)+14(4))
=12×(−15+9+56)
=12×(50)
= 25 sq.units
Area of triangle ACD
12×(x1(y2−y3)+x2(y3−y1)+x3(y1−y2))
=12×(3(0−19)+(14)(19+(1))+(9)(−1–0)))
=12×(3(−19)+14(20)+9(−1))
=12×(−57+280–9)
=12×(214)
= 107 sq.units
So, the area of the quadrilateral is 25 + 107 = 132 sq. units