# Equation of Straight Line in Plane/Normal Form

## Theorem

Let $\mathcal L$ be a straight line such that:

the perpendicular distance from $\mathcal L$ to the origin is $p$
the angle made between that perpendicular and the $x$-axis is $\alpha$.

Then $\mathcal L$ can be defined by the equation:

$x \cos \alpha + y \sin \alpha = p$

## Proof Let $A$ be the $x$-intercept of $\mathcal L$.

Let $B$ be the $y$-intercept of $\mathcal L$.

Let $A = \tuple {a, 0}$ and $B = \tuple {0, b}$.

From the Equation of Straight Line in Plane: Two-Intercept Form, $\mathcal L$ can be expressed in the form:

$(1): \quad \dfrac x a + \dfrac y a = 1$

Then:

 $\displaystyle p$ $=$ $\displaystyle a \cos \alpha$ Definition of Cosine of Angle $\displaystyle \leadsto \ \$ $\displaystyle a$ $=$ $\displaystyle \dfrac p {\cos \alpha}$
 $\displaystyle p$ $=$ $\displaystyle b \sin \alpha$ Definition of Sine of Angle $\displaystyle \leadsto \ \$ $\displaystyle b$ $=$ $\displaystyle \dfrac p {\sin \alpha}$

Substituting for $a$ and $b$ in $(1)$:

 $\displaystyle \dfrac x a + \dfrac y a$ $=$ $\displaystyle 1$ $\displaystyle \leadsto \ \$ $\displaystyle \dfrac {x \cos \alpha} p + \dfrac {y \sin \alpha} p$ $=$ $\displaystyle 1$ $\displaystyle \leadsto \ \$ $\displaystyle x \cos \alpha + y \sin \alpha$ $=$ $\displaystyle p$

$\blacksquare$