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Question
Find the value of k, if the area of a quadrilateral is 28 sq. units, whose vertices are (– 4, – 2), (– 3, k), (3, – 2) and (2, 3)
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Solution
Let the vertices A(– 4, – 2), B(– 3, k), C(3, – 2) and D(2, 3)
Area of the Quadrilateral = 28 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)]` = 28
`1/2[(-4"k" + 6 + 9 - 4) - (6 + 3"k" - 4 - 12)]` = 28
`1/2[(-4"k" + 11) - (3"k" - 10)]` = 28
– 4k + 11 – 3k + 10 = 56
– 7k + 21 = 56
– 7k = 56 – 21
– 7k = 35
⇒ 7k = – 35
k = `35/7`
= – 5
The value of k = – 5
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