Distance between two parallel lines by having linear equations I wonder where does this formula is coming from?
It is for finding the distance between two parallel lines when we have their linear equation:
First line is:$ax+by+c=0$
Second line is:$ax+by+c_1=0$
Their distance :$$\frac{|c-c_1|}{\sqrt{a^2+b^2}}$$
 A: HINT
Calculate the distance of each line from the origin that is
$$d=\frac{|a\cdot0+b\cdot0+c|}{\sqrt{a^2+b^2}}=\frac{|c|}{\sqrt{a^2+b^2}}$$
than take


*

*$|d_1-d_2|$ for the line on the same side ($c$ and $c_1$ have same sign)

*$|d_1+d_2|$ for the line on diffent sides ($c$ and $c_1$ have different sign)


from which the given formula is obtained.
A: Alternatively: the perpendicular line that passes through origin is: $bx-ay=0$. It crosses the two parallel lines at:
$$\left(-\frac{ac}{a^2+b^2},-\frac{bc}{a^2+b^2}\right) \ \ \text{and} \ \ \left(-\frac{ac_1}{a^2+b^2},-\frac{bc_1}{a^2+b^2}\right).$$
The distance between these points is:
$$d=\sqrt{\left(\frac{a(c-c_1)}{a^2+b^2}\right)^2+\left(\frac{b(c-c_1)}{a^2+b^2}\right)^2}=\frac{|c-c_1|}{\sqrt{a^2+b^2}}.$$
A: Consider $ax+by+c=0$, assume $a,b,c \not =0.$
$y=0$: $X-$intercept: $x=-c/a;$
$x=0:$ $Y-$intercept: $y =-c/b.$
$A(-c/a,0); B(0,-c/b);$ $O(0,0);$ 
form a right $\triangle ABO$ with
lengths of legs $|c/a|$ and $|c/b|.$
Lenght of hypotenuse : $\sqrt{(c/a)^2+(c/b)^2}.$
Height, $h$,  on $AB$ is the desired distance to the origin:
Area of $\triangle ABO$ :
Area $= (1/2)|c/a||c/b| = (1/2)h\sqrt{(c/a)^2+(c/b)^2}.$
Solve for $h:$
$h= \dfrac{c^2}{|ab|}\dfrac {|ab|}{|c|\sqrt{(a^2+b^2)}}$.
$h =\dfrac{|c|}{a^2+b^2}$.
Left to do : 
The above gives you the  distance of one line from the origin, regardless of the sign of $c.$
Now you have 2 lines , with $c,c _1.$
Find the distance between them.
(Does gimusi's answer help?)
A: As an alternative suppose wlog that $b\neq 0$ then translate vertically with $by\to by-c_1$ and obtain


*

*$ax+by+c=0\to ax+by+c-c_1=0$

*$ax+by+c_1\to ax+by=0$


then the distance between the two lines is equal to the distance of the translated line $ax+by+c-c_1=0$ from the origin that is indeed the given expression.
