Distance Between Parallel Lines

Given two parallel lines

$l_1:Ax+By+C_1=0$

$l_2:Ax+By+C_2=0$

Noticed that the coefficient of $x$ and $y$ of both line are respectively the same, this is because the slope of both line, $m_1$ and $m_2$ are the same, with

$$m_1=m_2=-\frac{A}{B}$$

However, for easier understanding, I'll let it be $A_1, B_1$ for $l_1$ and $A_2,B_2$ for $l_2$, but bear in mind that they are actually the same.

Since the distance between parallel lines are the same in any given point, we simply set a point $P(x_0,y_0)$ on $l_1$ and take its distance from $l_2$. Here, we use the formula of distance between a point and a line.

$$d=\left|\frac{A_2x_0+B_2y_0+C_2}{\sqrt{A_2+B_2}}\right|$$

Since $A_1=A_2$ and $B_1=B_2$, the formula can be expressed as

$$d=\left|\frac{A_1x_0+B_1y_0+C_2}{\sqrt{A_2+B_2}}\right|$$

Then, we also have

$l_1:A_1x_0+B_1y_0+C_1=0$

$l_1:A_1x_0+B_1y_0=-C_1$

Substitute to the formula we have

$$d=\left|\frac{-C_1+C_2}{\sqrt{A_2+B_2}}\right|$$

Finally, we can remove everything unnecessary and arrange the formula, we get

$$d=\left|\frac{C_2-C_1}{\sqrt{A^2+B^2}}\right|$$

Suppose that we have $2$ parallel planes: $$P_1: Ax+By+Cz+D_1=0$$ $$P_2: Ax+By+Cz+D_2=0$$ We wish to find the distance between these $2$ planes. Note that we want to find the shortest possible distance between the two planes, i.e. the distance if one traveled from $P_1$ to $P_2$ along a segment parallel to the normal vector.

A normal vector to both of these planes is $\vec{n}=\left<A,B,C\right>$. Note that $|\vec{n}|=\sqrt{A^2+B^2+C^2}$.

Suppose you move along the normal vector such that your displacement is $k\vec{n}=\left<kA,kB,kC\right>$, for some scalar $k$. For all points $P=(x,y,z)\in\mathbb{R}^3$, define $f(P)=Ax+By+Cz$. Then: $$f(P+k\vec{n})=A(x+kA)+B(y+kB)+C(z+kC)$$ $$f(P+k\vec{n})=(Ax+By+Cz)+k(A^2+B^2+C^2)$$ $$f(P+k\vec{n})-f(P)=k|\vec{n}|^2$$ Therefore, a movement of $k|\vec{n}|$ in the normal direction produces a change of $k|\vec{n}|^2$ in $f(P)$. Suppose that we have $2$ points $M_1\in P_1$ and $M_2\in P_2$ such that $M_1M_2\perp P_1,P_2$. Then: $$f(M_2)-f(M_1)=(-D_2)-(-D_1)=D_1-D_2$$ $$D_1-D_2=k|\vec{n}|^2$$ $$k|\vec{n}|=\frac{D_1-D_2}{\sqrt{A^2+B^2+C^2}}$$ Note that $k|\vec{n}|$ is just the displacement in the normal direction! Thus, since geometric distances are always nonnegative, we can say: $$M_1M_2=\frac{|D_1-D_2|}{\sqrt{A^2+B^2+C^2}}$$ This means that the distance $d$ between $2$ parallel planes of the above form is: $$d=\frac{|D_1-D_2|}{\sqrt{A^2+B^2+C^2}}$$ This argument can easily be generalized to any number of dimensions. Suppose we have $2$ parallel $(n-1)$-dimensional hyperplanes in $n$-dimensional hyperspace of the form: $$P_1: A_1x_1+A_2x_2+...+A_nx_n+B_1=0$$ $$P_2: A_1x_1+A_2x_2+...+A_nx_n+B_2=0$$ Then, the distance $d$ between them is: $$d=\frac{|B_1-B_2|}{\sqrt{A_1^2+A_2^2+...+A_n^2}}$$