What is the area, in square units, of a triangle whose vertices are at $(4, -1)$, $(10, 3)$ and $(4, 5)$?
I need to find the area of a triangle given its vertices. One way to do that is to use the formula $\frac{1}{2}bh$, where $b$ is the base and $h$ is the height. I can choose any side of the triangle as the base, but I need to find the height perpendicular to that base. I notice that the side joining $(4, -1)$ and $(4, 5)$ is vertical and has a length of $6$ units, so I can use that as the base. To find the height, I need to find the horizontal distance from the third vertex $(10, 3)$ to the base. I can do that by drawing a horizontal line from $(10, 3)$ until it meets the base at $(4, 3)$. The length of that horizontal line is the height, and it is $6$ units as well. Therefore, the area of the triangle is $\frac{1}{2}(6)(6) = 18$ square units.

