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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
Solution
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
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