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Question
Answer the following question:
Find the value of k:
If area of triangle is 4 square unit and vertices are P(k, 0), Q(4, 0), R(0, 2)
Solution
Here, P(x1, y1) ≡ P(k, 0), Q(x2, y2) ≡ Q(4, 0), R(x3, y3) ≡ R(0, 2)
A(ΔPQR)) = 4 sq. units
Area of a triangle = `1/2|(x_1, y_1, 1),(x_2, y_2, 1),(x_3, y_3, 1)|`
∴ ± 4 = `1/2|("k", 0, 1),(4, 0, 1),(0, 2, 1)|`
∴ ± 4 = `1/2["k"(0 - 2) - 0 + 1(8 - 0)]`
∴ ± 8 = –2k + 8
∴ 8 = –2k + 8 or –8 = –2k + 8
∴ –2k = 0 or 2k = 16
∴ k = 0 or k = 8
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