ABC is a triangle formed by the vertices (0, 5), (6, 13), and (6, −3). If AD⊥BC, then what is the length of AD?

Using the distance formula, we obtain

$BC=\sqrt{{\left(6-6\right)}^2+{\left(13+3\right)}^2}=16$
Area of ΔABC=$\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]$

$\Rightarrow \frac{1}{2}\times AD\times BC=\frac{1}{2}\left[0\left(13+3\right)+6(-3-5)+6(5-13)\right]$

$\Rightarrow AD\times 16=\left|-48-48\right|$

$\Rightarrow AD=\frac{96}{16}=6$

Thus, the length of AD is 6 units.