Question

Find the value of k, if the area of the triangle with vertices at A(k, 3), B(–5, 7), C(–1, 4) is 4 square units.

Answer 1

Here, A(x₁, y₁) ≡ A(k, 3), B(x₂, y₂) ≡ B(– 5, 7), C(x₃, y₃) ≡ C(– 1, 4)
A(ΔABC) = 4 sq. units

Area of a triangle = `1/2|(x_1, y_1, 1),(x_2, y_2, 1),(x_3, y_3, 1)|`

∴ `1/2|("k", 3, 1),(-5, 7, 1),(-1, 4, 1)|` = ± 4

∴ k(7 – 4) – 3(– 5 + 1) + 1(– 20 + 7) = ± 8
∴ 3k + 12 – 13 = ± 8
∴ 3k – 1 = ± 8
∴ 3k – 1 = 8 or 3k – 1 = – 8
∴ 3k = 9 or 3k = – 7

∴ k = 3 or k = `(-7)/3`