A triangle has vertices $A(-4, -1)$, $B (2, -1)$ and $C(1, 3)$. What is the area, in square units, of triangle $ABC$?
To find the area of a triangle, I need to know its base and height. I can choose any side of the triangle as the base, but then I need to find the perpendicular distance from the opposite vertex to that side. One possible choice is to use $AB$ as the base, since it is horizontal and has a length of $6$ units. To find the height of the triangle, I need to find the distance from $C$ to $AB$. Since $AB$ is horizontal, the height is just the difference in the $y$-coordinates of $C$ and any point on $AB$. For example, I can use $A$ as the point on $AB$, and then the height is $3 - (-1) = 4$ units. Therefore, the area of the triangle is half the product of the base and the height, which is $\frac{1}{2} \times 6 \times 4 = 12$ square units. 