Help in understanding a simple proof.

Question

Let $Ax + By + C = 0$ be a general equation of a line and $x\cos \alpha + y\sin \alpha - p = 0$ be the normal form of the equation.

Then,

$${-p\over C } = { \cos \alpha\over A} = { \sin\alpha\over B}\tag{1}$$ $${p\over C } = { \cos \alpha\over -A} = { \sin\alpha\over -B} \tag{2}$$ $${p\over C } = { \cos \alpha\over -A} = { \sin\alpha\over -B} = {\sqrt{\sin^2 \alpha + \cos^2 \alpha}\over \sqrt{A^2 + B^2}} = {1\over \sqrt{A^2 + B^2}} \tag{3}$$

$$\therefore \bbox[ #FFFDD0, 10px, Border:2px solid #DC143C]{p = {C\over \sqrt{A^2 + B^2}}, \cos \alpha = {-A\over \sqrt{A^2 + B^2}},\sin\alpha = {-B\over \sqrt{A^2 + B^2}}} $$

I did not get the $(3)$ part. Where does $\displaystyle{\sqrt{\sin^2 \alpha + \cos^2 \alpha}\over \sqrt{A^2 + B^2}}$ come from ?

Answer 1

Let $\frac{p}{C}=\frac{\cos\alpha}{A}=\frac{\sin\alpha}{B} = k$

Then we have,

$\cos\alpha = kA$

On squaring we have,

$\cos^2\alpha = k^2A^2$

Also $\sin\alpha = kB$

On squaring $\sin^2\alpha = k^2B^2$

Adding these two squared equations,

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

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

$\frac{\sin^2\alpha + \cos^2\alpha}{(A^2 + B^2)} = k^2$

$\frac{\sqrt{\sin^2\alpha + \cos^2\alpha}}{\sqrt{(A^2 + B^2)}} = k$

Answer 2

$$ {p\over C } = { \cos \alpha\over -A} = { \sin\alpha\over -B} \\ \implies \frac{p^2}{C^2}=\frac{\cos^2(\alpha)}{A^2}=\frac{\sin^2(\alpha)}{B^2} $$

If $$ \frac{a}{b}=\frac{c}{d}=k \\ \implies a=bk, c=dk $$ So the value of : $$ \frac{a+c}{b+d}=\frac{bk+dk}{b+d}=\frac{k(b+d)}{b+d}=k=\frac{a}{b}=\frac{c}{d} $$ Simply put, in a proportion of two or more equal ratios, the sum of the antecedents divided by the sum of the consequents is equal to any one of the ratios.

In the same way : $$ \frac{p^2}{C^2}=\frac{\cos^2(\alpha)}{A^2}=\frac{\sin^2(\alpha)}{B^2}={{\sin^2 \alpha + \cos^2 \alpha}\over {A^2 + B^2}} $$ Hence : $$ \frac{p}{C}={\sqrt{\sin^2 \alpha + \cos^2 \alpha}\over \sqrt{A^2 + B^2}} $$

Answer 3

I'd do it differently: since $A$ and $B$ are not both zero, we can divide by $-\sqrt{A^2+B^2}$, getting $$ \frac{-A}{\sqrt{A^2+B^2}}x+\frac{-B}{\sqrt{A^2+B^2}}-\frac{C}{\sqrt{A^2+B^2}}=0 $$ and we can choose $\alpha$ so that $$ \begin{cases} \cos\alpha=-\dfrac{A}{\sqrt{A^2+B^2}}\\[6px] \sin\alpha=-\dfrac{B}{\sqrt{A^2+B^2}} \end{cases} $$ and set $$ p=\frac{C}{\sqrt{A^2+B^2}} $$

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For the mysterious formula, observe that, if $$ k=\frac{\cos\alpha}{-A}=\frac{\sin\alpha}{-B} $$ then $$ A^2k^2=\cos^2\alpha,\quad B^2k^2=\sin^2\alpha $$ so also $$ (A^2+B^2)k^2=1 $$ and $$ |k|=\frac{1}{\sqrt{A^2+B^2}} $$ Of course one needs to be very precise about the choice of $p$, using this method. It is implied that $p<0$ if $C<0$ and $p>0$ if $C>0$; a special case should be made when $C=0$.

The method above is independent of these considerations.

Answer 4

Let's modify ${p\over C } = { \cos \alpha\over -A} = { \sin\alpha\over -B} $

to ${p\over C } = { \cos \alpha\over -A} = { \sin\alpha\over -B} =K$

And ${p\over C } = { \cos \alpha\over -A} = { \sin\alpha\over -B} = {\sqrt{\sin^2 \alpha + \cos^2 \alpha}\over \sqrt{A^2 + B^2}} = {1\over \sqrt{A^2 + B^2}} \tag{3}$

to $K = {p\over C } = { \cos \alpha\over -A} = { \sin\alpha\over -B} = {K\sqrt{A^2 +B^2}\over \sqrt{A^2 + B^2}}={\sqrt{(-AK)^2 +(-BK)^2}\over \sqrt{A^2 + B^2}}={\sqrt{\sin^2 \alpha + \cos^2 \alpha}\over \sqrt{A^2 + B^2}} = {1\over \sqrt{A^2 + B^2}} \tag{3}$