Question

The area of a triangle with vertices (–3, 0), (3, 0) and (0, k) is 9 sq.units. The value of k will be ______.

Options

• 9

• ±3

• – 9

• 6

Answer 1

The area of a triangle with vertices (–3, 0), (3, 0) and (0, k) is 9 sq.units. The value of k will be ±3.

Explanation:

We know that, area of triangle with vertices (x₁, y₁), (x₂, y₂) and (x₃, y₃) is given by

Δ = `1/2|(x_1, y_1, 1),(x_2, y_2, 1),(x_3, y_3, 1)|`

∴ Area of triangle with verticles (–3, 0), (3, 0) and (0, k) is

∴ Δ = `1/2|(-3, 0, 1),(3, 0, 1),(0, "k", 1)|` = 9  ...(Given)

⇒ `[-3(-"k") - 0 + 1(3"k")]` = ±18

⇒ 6k = ±18

∴ k = `+- 18/6` = ±3