Distance between a point and a straight line

The distance between a point $P$ and a straight line $r$ is the minimum of the distances between $P$ and any point on the straight line.

We can distinguish two cases:

If $P$ belongs to the straight line $r$ , $d (P, r) = 0$ .
If $P$ does not belong to the straight line $r$ , $d (P, r)$ , it is the module of the vector $\vec{Q P}$ , where $Q$ is the intersection point between the straight line $r$ and the perpendicular to $r$ that crosses $P$ .

Let $A x + B y + C = 0$ be the general equation of the straight line $r$ , and $P = (p_{1}, p_{2})$ the given point and $A = (a_{1}, a_{2})$ any point of the straight line.

If we take a perpendicular vector to $r$ , for example $\vec{n} = (A, B)$ for the properties of the scalar product in the vectors projection we have: $d (P, r) = \frac{| \vec{A P} \cdot \vec{n} |}{\vec{n}} = \frac{| A \cdot p_{1} + B \cdot p_{2} - (A \cdot a_{1} + B \cdot a_{2}) |}{\sqrt{A^{2} + B^{2}}}$ But since $A = (a_{1}, a_{2})$ is a point of the straight line $r$ , it verifies its equation: $A \cdot a_{1} + B \cdot a_{2} + C = 0 \leftarrow A \cdot a_{1} + B \cdot a_{2} = C$ Therefore we obtain the following formula: $d (P, r) = \frac{| A \cdot p_{1} + B \cdot p_{2} + C |}{\sqrt{A^{2} + B^{2}}}$

Example

Let $P = (- 1, 2)$ be a point and $r : 4 x - 3 y + 1 = 0$ a straight line. Calculate the distance between the point and the straight line.

Applying the previous formula we have: $d (P, r) = \frac{A \cdot p_{1} + B \cdot p_{2} + C}{\sqrt{A^{2} + B^{2}}} = \frac{| 4 \cdot (- 1) + (- 3) \cdot 2 + 1 |}{\sqrt{4^{2} + (- 3)^{2}}} = \frac{9}{5}$