Equation of Straight Line in Plane/Normal Form/Proof 1

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Theorem

Let $\LL$ be a straight line such that:

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


Then $\LL$ 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 $\LL$.

Let $B$ be the $y$-intercept of $\LL$.


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

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

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


Then:

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


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

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

$\blacksquare$


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