Equation of Straight Line in Plane/Normal Form

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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

Straight-line-normal-form.png


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
\(\displaystyle \leadsto \ \ \) \(\displaystyle a\) \(=\) \(\displaystyle \dfrac p {\cos \alpha}\)
\(\displaystyle p\) \(=\) \(\displaystyle b \sin \alpha\) Definition of Sine
\(\displaystyle \leadsto \ \ \) \(\displaystyle b\) \(=\) \(\displaystyle \dfrac p {\sin \alpha}\)


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

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

$\blacksquare$


Sources