If the area of the triangle with vertices (-3, 0), (3, 0) and $(0, k)$ is $9$ sq. units, then the values of $k$ will be

$\pm 3$

The correct answer is Option (2) → $\pm 3$ ##

$\text{Area} = \left| \frac{1}{2} \begin{vmatrix} -3 & 0 & 1 \\ 3 & 0 & 1 \\ 0 & k & 1 \end{vmatrix} \right|$

Given that the area $= 9$ sq. units

$\Rightarrow \pm 9 = \frac{1}{2} \begin{vmatrix} -3 & 0 & 1 \\ 3 & 0 & 1 \\ 0 & k & 1 \end{vmatrix}$

Expanding along $C_2$, we get:

$\Rightarrow \pm 9 = \frac{1}{2} \left[ -k \begin{vmatrix} -3 & 1 \\ 3 & 1 \end{vmatrix} \right]$

$\Rightarrow \pm 9 = \frac{1}{2} [ -k(-3 - 3) ]$

$\Rightarrow \pm 9 = \frac{1}{2} [ -k(-6) ]$

$\Rightarrow \pm 9 = 3k$

$\Rightarrow k = \pm 3$