Geometric significance of $\sqrt{A^2 + B^2}$ in general equation of line, if any? To reduce the general equation of first degree $Ax + By + C = 0$ to the so called perpendicular form of equation of line which is $x\cos\theta + y\sin\theta = p$ my book divides the general equation by the term $\sqrt{A^2 + B^2}$. 
And converts the equation into the second form. 
My question is, what is the geometrical significance of the term $\sqrt{A^2 + B^2}$,  if any? And how does all of this work. 
 A: There is a severe limitation on which pairs of numbers can be written as $U=\cos \theta$ and $V = \sin \theta$: this comes from the pythagorean identity $\cos^2 \varphi + \sin^2 \varphi = 1$. Thus, we need $U^2 + V^2 = 1$ in order to be able to do this.
By dividing the equation through by $\sqrt{A^2 + B^2}$, the two coefficients satisfy the needed condition that the sums of their squares is now $1$.
One could give geometric arguments too: e.g. it's expressing your line in terms of a unit normal vector. Thus, the need to normalize by dividing by its length.
A: The vector $\vec{v} = (A, B)$ is a perpendicular vector to your line. This is because the equation $Ax+By+C=0$ can be rewritten as $(A, B) \cdot (x, y) = -C$. This means that all the points in the line must have the same component in the direction of $\vec{v}$, and so the line must be perpendicular to it. Try to make a drawing if this doesn't convince you.
When we put it in the form $x\cos \theta + y\sin \theta = p$, we are using a new vector $\vec{w} = (\cos\theta, \sin\theta)$, which has the property that it is a unit vector, i.e., it has length $1$. So, if we have a vector and we want to make it a unit vector while preserving its direction, what do we do? We divide it by its length, that is, by $\sqrt{A^2+B^2}$. Our new vector $\left(\frac{A}{\sqrt{A^2+B^2}}, \frac{B}{\sqrt{A^2+B^2}}\right)$ has unit length, and so can be put in the form $(\cos\theta, \sin\theta)$.
A: The solution set $g$ of the equation $$Ax+By+C=0\ ,\tag{1}$$ where it is assumed that $(A,B)\ne(0,0)$, remains unchanged under multiplication of $A$, $B$, $C$ with a common factor $\lambda\ne0$. It follows that the quantity $\sqrt{A^2+B^2}>0$ has no geometrical significance at all. 
Among all  equations which are equivalent to a given $(1)$ there is one which really displays geometrical information, namely the equation$A'x+B'y+C'=0$ with
$$A':={A\over\sqrt{A^2+B^2}},\quad B':={B\over\sqrt{A^2+B^2}},\quad C':={C\over\sqrt{A^2+B^2}}\ .$$
Here $A'$ and $B'$ can be interpreted as $\cos\theta$ and $\sin\theta$ for a certain argument $\theta$ related to the direction of the line $g$, and $|C'|$ is the distance of $g$ from the origin – as explained in other answers to this question.
