If area of a triangle is $35$ square units with vertices $\left( {2, - 6} \right),\,\,\left( {5,\,\,4} \right)$ and $({k},\,\,4)$ then ${k}$ is :

Area of triangle $=35$ sq.units

$\begin{array}{c} \frac{1}{2}\left|\begin{array}{ccc} 2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{array}\right|=35 \\ |2(4-4)+6(5-k)+1(20-4 k)|=70 \\ |30-6 k+20-4 k|=70 \\ |50-10 k|=70 \end{array}$

Ou Taking +ve sign,

$\begin{array}{c} 50-10 k=70 \\ 10 k=-20 \\ k=-2 . \end{array}$

On taking - ve sign,

$\begin{array}{c} 50-10 k=-70 \\ 10 k=120 \\ k=12 . \end{array}$

$\therefore K=12,-2$