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Let A (1, 3), B (0, 0) and C (k, 0) be three points such that area of Δ(ABC)=3\Delta(ABC) = 3 sq. units, then the value of k is:

Solution

✅ Correct Option: 2

Using the determinant formula, area =12∣1(0−0)−3(0−k)+1(0−0)∣=3∣k∣2= \frac{1}{2}|1(0-0) - 3(0-k) + 1(0-0)| = \frac{3|k|}{2}. Setting this equal to 3 gives ∣k∣=2|k| = 2, so k=±2k = \pm 2.

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