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The value of $' k'$ if area of triangle is 4 sq. units and vertices are $(k, 0), (4, 0), (0, 2)$ is

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0, 8

5, 6

3, 4

1, 2

answer is A.

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Detailed Solution

$Δ = \frac{1}{2} |\begin{matrix} k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1 \end{matrix}| = 4$

$= \frac{1}{2} \times (- 2 (k - 4)) = 4$

$= - k + 4 = 4$

Given, $- k + 4 = 4$ , then $k = 0$

(or)

$- k + 4 = - 4$ , then $k = 8$

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