Foot of Perpendicular from a Point to a Line

Foot of the perpendicular from (x₁, y₁) to the line ax + by + c = 0.

Given point $P(x_1, y_1)$ and line $L: ax + by + c = 0$.

The perpendicular from $P$ to $L$ has direction $(a, b)$ (the normal to $L$). Parametric equations of this perpendicular:

$$x = x_1 + at, \qquad y = y_1 + bt$$

The foot $N$ is where this meets $L$:

$$a(x_1 + at) + b(y_1 + bt) + c = 0$$ $$(ax_1 + by_1 + c) + t(a^2 + b^2) = 0$$ $$t = -\frac{ax_1 + by_1 + c}{a^2 + b^2}$$

Substituting back gives the foot:

$$\boxed{\frac{x - x_1}{a} = \frac{y - y_1}{b} = -\frac{ax_1 + by_1 + c}{a^2 + b^2}}$$

The coordinates of $N$ explicitly:

$$N = \left(x_1 - \frac{a(ax_1+by_1+c)}{a^2+b^2},\quad y_1 - \frac{b(ax_1+by_1+c)}{a^2+b^2}\right)$$