If area of triangle is 35 square units with vertices (2, −6), (5, 4), and (k, 4). Then k is

Solution:

Answer: D

The area of the triangle with vertices (2, −6), (5, 4), and (k, 4) is given by the relation,

$\Delta=\frac{1}{2}\left|\begin{array}{ccc}2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1\end{array}\right|$

$=\frac{1}{2}[2(4-4)+6(5-k)+1(20-4 k)]$

$=\frac{1}{2}[30-6 k+20-4 k]$

$=\frac{1}{2}[50-10 k]$

$=25-5 k$

It is given that the area of the triangle is ±35.

Therefore, we have:

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

$\Rightarrow 5(5-k)=\pm 35$

$\Rightarrow 5-k=\pm 7$

When $5-k=-7, k=5+7=12$.

When $5-k=7, k=5-7=-2$.

Hence, $k=12,-2$.

The correct answer is D.