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Question
If vertices of a quadrilateral are at A(– 5, 7), B(– 4, k), C(– 1, – 6) and D(4, 5) and its area is 72 sq. units. Find the value of k.
Solution
Area of the quadrilateral ABCD = 72 sq. units.
`1/2 [(x_1y_2 + x_2y_3 + x_3y_4 + x_4y_1) - (x_2y_1 + x_3y_2 + x_4y_3 + x_1y_4)]` = 72
– 5k + 24 – 5 + 28 – (– 28 – K – 24 – 25) = 144
– 5k + 47 – k – 77 = 144
– 5k + 47 + k + 77 = 144
– 4k + 124 = 144
– 4k = 144 – 124
– 4k = 20
k = – 5
The value of k = – 5
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