The area of the triangle formed by the points $A(-1.5,3), B(6,-2)$, and $C(-3,4)$ is  ____ sq. units.

Concept: The area of a triangle is $\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]$ where $\left(x_1,y_1\right),\left(x_2,y_2\right)$ and $\left(x_3,y_3\right)$ are the vertices of the triangle.

Here, the vertices of the triangle are  $A(-1.5,3), B(6,-2)$, and $C(-3,4)$.

We know, area of $\Delta ABC=\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]$.

$$$\Rightarrow \frac{1}{2}[-1.5(-2-4)+6(4-3)+(-3\left)\right(3+2\left)\right]$

$$\begin{gathered} \Rightarrow \frac{1}{2}[9+6-15] \\ \Rightarrow 0 \end{gathered}$$