Distance of a Point from a Line

Given a line $L: Ax+By+C=0$ and a point $P(x_1,y_1)$ not on $L$, this section finds the length of the perpendicular dropped from $P$ to $L$ — the shortest possible distance from the point to the line.

Setting up. Let $M$ be the foot of the perpendicular from $P$ to $L$ (Figure), so the required distance is $d=PM$. Assume, for the derivation, that neither $A$ nor $B$ is zero (the two axis-parallel cases were already covered directly in Section 4). Let $L$ meet the x-axis at $Q\left(-\dfrac{C}{A},0\right)$ and the y-axis at $R\left(0,-\dfrac{C}{B}\right)$ — its intercepts, from Section 8.

Derivation (area method). Consider the triangle $PQR$. Its area can be computed in two different ways, and equating them isolates $d$.

First way — the coordinate area formula, applied to vertices $P(x_1,y_1)$, $Q\left(-\frac{C}{A},0\right)$, $R\left(0,-\frac{C}{B}\right)$; after simplifying the determinant expansion, this reduces to

Second way — as $\frac12\times\text{base}\times\text{height}$, using $QR$ as the base and $PM=d$ as the corresponding height (since $PM\perp QR$ by construction, $QR$ being part of $L$):

(by the distance formula of Section 1, applied to $Q$ and $R$).

Equating the two areas and cancelling the common factor $\dfrac{|C|}{|AB|}$ from both sides gives

— the distance formula. Although derived here assuming $A,B,C$ all non-zero, the formula is checked to hold for every line (including $x=a$ and $y=b$) by direct substitution, and so may be used unconditionally.

Distance between two parallel lines. Two parallel lines share the same $A,B$ (up to a common scalar), differing only in the constant term: $L_1: Ax+By+C_1=0$ and $L_2: Ax+By+C_2=0$. Picking any point $(x_1,y_1)$ on $L_1$ (so $Ax_1+By_1=-C_1$) and applying the distance formula to find its distance from $L_2$:

$d = \frac{|Ax_1+By_1+C_2|}{\sqrt{A^2+B^2}} = \frac{|-C_1+C_2|}{\sqrt{A^2+B^2}} = \boxed{\frac{|C_1-C_2|}{\sqrt{A^2+B^2}}}.$ …