Distance between two straight lines

The distance between two straight lines, $$r$$ and $$s$$, is the minimal distance between any point of $$r$$ and any point of $$s$$.

  • If the straight lines are secant or coincidental, their distance is obviously zero. Namely $$d (r, s) = 0$$.
  • If the straight lines are parallel, the distance between $$r$$ and $$s$$ is the distance from a point of any of both straight lines to any other.

To find the analytical expression of the distance from $$r$$ to $$s$$, we will suppose that we have $$r: Ax + By + C = 0$$ and $$s: Ax + By + C' = 0$$. As the straight lines need to have parallel director vectors, we can suppose that they do have the same, which is why $$A = A'$$ and $$B = B'$$.

As the straight lines cannot be coincidental, we will obviously have $$C\neq C'$$.

Let $$P =(p_1,p_2)$$ now be a point belonging to the straight line $$r$$. Then we have: $$$\displaystyle d(r,s)=d(P,s)=\frac{|A\cdot p_1+B\cdot p_2+C'|}{\sqrt{A^2+b^2}}$$$ But since $$P$$ belongs to the straight line $$r$$ we have $$$A\cdot a_1+B\cdot a_2+C=0 \Leftarrow A\cdot a_1+B\cdot a_2=-C$$$ substituting, $$$d(r,s)=d(P,S)=\displaystyle \frac{|C'-C|}{\sqrt{A^2+B^2}}$$$

Calculate the distance between the straight lines $$r: 2x + 3y - 4 = 0$$ and $$s:-4x - 6y + 24 = 0$$.

To begin, we divide the equation of the straight line $$s$$ by $$-2$$: $$$s: 2x + 3y - 12 = 0$$$ Now we are in condition to apply the formula: $$$\displaystyle d (r, s) = d (P, s) =\frac{|C'-C|}{\sqrt{A^2+b^2}}=\frac{|-4-(-12)|}{\sqrt{2^2+3^2}}=\frac{8}{\sqrt{13}}$$$