If the area of a triangle with vertices (-3, 0), (3, 0) and (0, k) is 9 square units, then what is the value of k ? - Mathematics MCQ Mathematics Mock Test - 2 2026 | ExamTest.live

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If the area of a triangle with vertices (- 3, 0), (3, 0) and (0, k) is 9 square units, then what is the value of k ?

Solution & Step-by-step Explanation

The area of a triangle with vertices (x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3}) is: A re a = \frac{1}{2} ∣ x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) ∣ Substituting the given vertices: 9 = \frac{1}{2} ∣ (- 3) (0 - k) + 3 (k - 0) + 0 (0 - 0) ∣ 9 = \frac{1}{2} ∣3 k + 3 k ∣ 9 = \frac{1}{2} ∣6 k ∣ 9 = 3∣ k ∣ ∣ k ∣ = 3 ⟹ k = 3 or - 3 Based on the options, k = 3 is the correct answer.

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If the area of a triangle with vertices (- 3, 0), (3, 0) and (0, k) is 9 square units, then what is the value of k ?

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