The area of a triangle with vertices $(-3, 0), (3, 0)$ and $(0, k)$ is $9$ sq. units. The value of $k$ will be

$3$

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

We know that, area of a triangle with vertices $(x_1, y_1), (x_2, y_2)$ and $(x_3, y_3)$ is given by

$\Delta = \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}$

$∴\Delta = \frac{1}{2} \begin{vmatrix} -3 & 0 & 1 \\ 3 & 0 & 1 \\ 0 & k & 1 \end{vmatrix}$

On expanding along $R_1$, we get

$\Delta = \frac{1}{2} [-3(-k) - 0 + 1(3k)] = \frac{1}{2} [6k] = 3k$

Since, area of triangle = $9$ sq. units.

$\pm 9 = 3k$

$∴k = \pm \frac{9}{3} = \pm 3$