The Normal Form of a Line

Try this

1. Turn only $p$. How does the line move? Where does it pass when $p=0$?
2. Fix $p$ at $3$ and turn only $\theta$ through a full turn. Which circle do all these lines just touch?
3. Look at the two coefficients of the equation. What is ${\cos }^2\theta +{\sin }^2\theta$, always? And what is the sign of the constant term?
4. Drag $Q$ across the line. Where does $\delta$ change sign? When $Q$ and the origin are on the same side of the line, is $\delta$ positive or negative?

Drop a perpendicular from the origin $O$ to the line $l$, with foot $N$. Let $\theta$ be the angle from the positive $x$-axis to $\overrightarrow{ON}$, and let $|ON|=p$. Every point $M(x,y)$ of the line has projection exactly $p$ onto the direction of $\overrightarrow{ON}$, so

This is called the normal form of the line. It has two features: the squares of the coefficients of $x$ and $y$ add up to $1$, and the constant term is negative or zero. (When the line passes through the origin, $p=0$ and we require $0^{\circ }\le \theta <180^{\circ }$.)

Why: turning the general form into the normal form

$Ax+By+C=0$ and the normal form describe the same line, so their coefficients are proportional: $\cos \theta =\mu A$, $\sin \theta =\mu B$, $-p=\mu C$. From ${\cos }^2\theta +{\sin }^2\theta =1$,

and $-p=\mu C\le 0$ fixes the sign: $\mu$ and $C$ have opposite signs. $\mu$ is called the normalizing factor.

What the normal form is for: signed distance

Substituting a point $Q(x_0,y_0)$ into the left side of the normal form gives

The geometric meaning is clear: $x_0\cos \theta +y_0\sin \theta$ is the projection of $\overrightarrow{OQ}$ onto the normal direction, and subtracting $p$ tells you how far $Q$ “sticks out” past the line. $|\delta |$ is the distance, and the sign tells you the side: positive on the side away from the origin, negative on the origin’s side, zero on the line.

The next stop puts this to full use: the distance from a point to a line, angle bisectors, and the angle between two lines.

Practice