If the area of a triangle with vertices $(-3,0),(3,0)$, and (0, a) is 9 square units, then what is the value of a?

Here, the vertices of the triangle are  $(-3,0),(3,0)$, and $(0, a)$.

Given, Area of the triangle is 9 square units.

We know, area of required triangle $=\frac{1}{2}\left|{\mathrm{x}}_1\left({\mathrm{y}}_2-{\mathrm{y}}_3\right)+{\mathrm{x}}_2\left({\mathrm{y}}_3-{\mathrm{y}}_1\right)+{\mathrm{x}}_3\left({\mathrm{y}}_1-{\mathrm{y}}_2\right)\right|$.

where $\left({\mathrm{x}}_1,{\mathrm{y}}_1\right),\left({\mathrm{x}}_2,{\mathrm{y}}_2\right)$ and $\left({\mathrm{x}}_3,{\mathrm{y}}_3\right)$ are the vertices of the triangle.

$\Rightarrow \frac{1}{2}\left|-3\left(0-\mathrm{a}\right)+3\left(\mathrm{a}-0\right)+0\left(0-0\right)\right|=9$

$\Rightarrow \frac{1}{2}\left|3\mathrm{a}+3\mathrm{a}\right|=9$

$\Rightarrow \frac{1}{2}\times 6\mathrm{a}=9$

$\Rightarrow 3\mathrm{a}=9$

$\therefore \mathrm{a}=3$

Hence, the value of a is 3.