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 $r : A x + B y + C = 0$ and $s : A x + B y + C^{'} = 0$ . 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: $d (r, s) = d (P, s) = \frac{| A \cdot p_{1} + B \cdot p_{2} + C^{'} |}{\sqrt{A^{2} + b^{2}}}$ 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, $d (r, s) = d (P, S) = \frac{| C^{'} - C |}{\sqrt{A^{2} + B^{2}}}$

Example

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

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