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Question
Find the value of k, if the area of triangle whose vertices are P(k, 0), Q(2, 2), R(4, 3) is `3/2 "sq.unit"`
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Solution
Here, P(x1, y1) ≡ P(k, 0), Q(x2, y2) ≡ Q(2, 2), R(x3, y3) ≡ R(4, 3)
A(ΔPQR) = `3/2 "sq.units"`
Area of a triangle = `1/2|(x_1, y_1, 1),(x_2, y_2, 1),(x_3, y_3, 1)|`
∴ `± 3/2 = 1/2|("k", 0, 1),(2, 2, 1),(4, 3, 1)|`
∴ `± 3/2 = 1/2["k"(2 - 3) - 0 + 1(6 - 8)]`
∴ `± 3/2 = 1/2(-"k" - 2)`
∴ ± 3 = –k – 2
∴ 3 = –k – 2 or –3 = –k – 2
∴ k = – 5 or k = 1
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