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Question
If (k, 2), (2, 4) and (3, 2) are vertices of the triangle of area 4 square units then determine the value of k
Solution
Given Area of the triangle with vertices (k, 2), (2, 4) and (3, 2) is 4 square units.
The area of the triangle with vertices
(x1, y1), (x2, y2) and (x3, y3) is
Δ = `|1/2||(x_1, y_1, 1),(x_2, y_2, 1),(x_3, y_3, 1)|`
Given Δ = 4
(x1, y1) = (k, 2)
(x2, y2) = (2, 4)
and (x3, y3) = (3, 2)
∴ We have 4 = `|1/2||("k", 2, 1),(2, 4, 1),(3, 2, 1)|`
4 × 2 = `|("k", 2, 1),(2, 4, 1),(3, 2, 1)|`
± 8 = k(4 – 2) – 2(2 – 3) + 1(4 – 12)
± 8 = k × 2 – 2 × – 1 – 8
± 8 = 2k + 2 – 8
± 8 = 2k – 6
2k – 6 = 8 or 2k – 6 = -8
2k = 8 + 6 or 2k = – 8 + 6
2k = 14 or 2k = -2
k = 7 or k = 1
Required values of k are 1, 7.
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