Distance between two parallel lines by having linear equations

Question

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 :$$\frac{|c-c_1|}{\sqrt{a^2+b^2}}$$

Answer 1

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.

Answer 2

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: $$d=\sqrt{\left(\frac{a(c-c_1)}{a^2+b^2}\right)^2+\left(\frac{b(c-c_1)}{a^2+b^2}\right)^2}=\frac{|c-c_1|}{\sqrt{a^2+b^2}}.$$

Answer 3

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?)

Answer 4

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.