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

We know that, area of a triangle with vertices $\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right)$

and $\left( {{x_3},{y_3}} \right)$

is given by
$\Delta = \frac{1}{2}\left| {\begin{array}{llllllllllllllllllll}{{x_1}}&{{y_1}}&1\\{{x_2}}&{{y_2}}&1\\{{x_3}}&{{y_3}}&1\end{array}} \right|$

$\therefore \Delta = \frac{1}{2}\left| {\begin{array}{cccccccccccccccccccc}{ - 3}&0&1\\3&0&1\\0&k&1\end{array}} \right|$

Expanding along ${R_1}$,

$9 = \frac{1}{2}[ - 3( - k) - 0 + 1(3k)]$
$\Rightarrow$ $18 = 3k + 3k = 6k$
$\therefore k = \frac{{18}}{6} = 3$