Obtaining normal form of a line from the general form

This is a question relating to SL Loney's coordinate geometry book (article 56).

We have $Ax + By + C = 0$ as the general form of a line.

Want to arrive at $xcos(\alpha)+ ysin(\alpha) - p = 0$ as the 'normal form'

So we have the equations represent the same line thus we have the common ratios of coefficients

$cos(\alpha)\over A$ =$sin(\alpha)\over B$ =$-p\over c$

Now here is where I can't arrive myself.

$p\over c$ = $cos(\alpha)\over -A$ = $sin(\alpha)\over -B$ = $\sqrt{cos^2(\alpha) + sin^2(\alpha) \over A^2 + B^2}$ = $1 \over \sqrt{A^2 + B^2}$

Ultimately, I'm hoping someone can shed some light on how he arrived at the $\sqrt{cos^2(\alpha) + sin^2(\alpha) \over A^2 + B^2}$ part...

• thanks all for the insight, very much appreciated. – saul Nov 19 '14 at 7:13

$$\frac{p}{c}=\frac{cos\alpha}{-A}=\frac{sin\alpha}{-B}=n$$ let's take each term apart:

$$\frac{cos\alpha}{-A}=n ==> -An=cos\alpha$$ $$\frac{sin\alpha}{-B}=n ==> -Bn=sin\alpha$$

Now square both sides :

$$A^2n^2=cos^2\alpha$$ $$B^2n^2=sin^2\alpha$$

$A^2n^2 + B^2n^2 = sin^2\alpha+cos^2\alpha$

$n^2(A^2+B^2)=1$

$n^2=\frac{1}{A^2+B^2}$

$n=\sqrt{\frac{1}{A^2+B^2}}$

So you can say:

$$\frac{p}{c}=\frac{cos\alpha}{-A}=\frac{sin\alpha}{-B}=\sqrt{\frac{1}{A^2+B^2}}$$

If $\dfrac p c = \dfrac{\cos(\alpha)}{-A } = \dfrac{ \sin(\alpha)}{ -B}=k$

$$1=\cos^2\alpha+\sin^2\alpha=k^2(A^2+B^2)\implies k^2=?$$

So, we have a line represented by an equation: $Ax + By + C = 0$. To get the equation of this form we need to multiply the $x \sin \alpha + y \cos \alpha + r = 0$ form by some koefficient $\mu$. Now, lets find the $\mu$:

$\mu \cdot A x + \mu \cdot B y + \mu \cdot C = x \cos \alpha + y \sin \alpha + r;$

from which(because x and y are separate axises):

$\mu \cdot A = \cos \alpha$

and

$\mu \cdot B = \sin \alpha$

from which

$(\mu \cdot A)^2 + (\mu \cdot B)^2 = (\cos \alpha)^2 + (\sin \alpha)^2$ => $\mu^2 (A^2+B^2) = 1$ => $\mu = \sqrt{\frac{1}{A^{2} + B^{2}}}$;

also, $C = \frac{r}{\mu}$ => $\mu = \frac{r}{C}$;