If area of a triangle is $35 s q u n i t s$ with vertices $(2, - 6)$ , $(5, 4)$ and $(k, 4)$ , then find the value of $k$ .

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Solution

$\frac{1}{2} ∣ ∣ ∣ ∣ \begin{matrix} 2 & - 6 & 1 5 & 4 & 1 k & 4 & 1 \end{matrix} ∣ ∣ ∣ ∣ = 35 ∣ ∣ ∣ ∣ \begin{matrix} 2 & - 6 & 1 5 & 4 & 1 k & 4 & 1 \end{matrix} ∣ ∣ ∣ ∣ = 70 | 2 (4 - 4) + 6 (5 - k) + 1 (20 - 4 k) | = 70 | 50 - 10 k | = 70 50 - 10 k = \pm 70 k = - 2, 12$

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Similar questions

If area of a triangle is 35 sq unit with vertices (2,-6), (5,4) and (k,4), then k is
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d) 12, -2

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