If area of triangle is 35 sq. units with vertices (2,– 6), (5, 4) and (k, 4), then k is

(A) 12

(B) – 2

(C) – 12, – 2

(D) 12, – 2

Option d is correct

Area of triangle $= \cfrac{1}{2}\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr}2&{ - 6}&1\\5&4&1\\k&4&1\end{array}} \right| = \pm 35$

$\Rightarrow$ $\cfrac{1}{2}[2(4 - 4) + 6(5 - k) + 1(20 - 4k)] = \pm 35$

$\Rightarrow$ $\cfrac{1}{2}[30 - 6k + 20 - 4k] = \pm 35$

$\Rightarrow$ $\cfrac{1}{2}[50 - 10k] = \pm 35 \Rightarrow 25 - 5k = \pm 35$

$\Rightarrow$ $25 - 5k = 35$ or $25 - 5k = - 35$

$\Rightarrow$ $= 5k = 10$ or $5k = 60$ $\Rightarrow$ $k = - 2$ or $k = 12$