$\triangle A B C$ has vertices $A(-1,2), B(5,2)$ and $C(-4,-3)$. What is the area of $\triangle A B C$ ?
Since $A$ and $B$ have the same $y$-coordinate, then $A B$ is horizontal.

We treat $A B$ as the base of $\triangle A B C$. It has length $5-(-1)=6$.

The height of $\triangle A B C$ is the vertical distance from $C(-4,-3)$ to the line through $A B$ (which is $y=2$ ). This distance is $2-(-3)=5$.

Therefore, the area of $\triangle A B C$ is $\frac{1}{2}(6)(5)=15$.