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

Answer 1

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.