A triangle has vertices (2, 1), (8, 9) and (10, 3), with coordinates in cm…

The distance of $(x_1, y_1)$ from $Ax + By + C = 0$ is $\dfrac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$. Parallel lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$ are $\dfrac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$ apart. The sign of $Ax_1 + By_1 + C$ tells the side of the line: the same sign means the same side. The foot of the perpendicular is $(x_1, y_1) - \dfrac{Ax_1 + By_1 + C}{A^2 + B^2}(A, B)$.