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 $\overrightarrow{QP}$, where $Q$ is the intersection point between the straight line $r$ and the perpendicular to $r$ that crosses $P$.

Let $Ax + By + C = 0$ 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 $\overrightarrow{n} = (A, B)$ for the properties of the scalar product in the vectors projection we have: $$\displaystyle d(P,r)=\frac{|\overrightarrow{AP} \cdot \overrightarrow{n}|}{\overrightarrow{n}}=\frac{|A\cdot p_1+B\cdot p_2-(A\cdot a_1+B\cdot a_2)|}{\sqrt{A^2+B^2}}$$ 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: $$d(P,r)=\displaystyle \frac{|A\cdot p_1+B\cdot p_2+C|}{\sqrt{A^2+B^2}}$$

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

Applying the previous formula we have: $$\displaystyle d (P, r) =\frac{A\cdot p_1+B\cdot p_2+C}{\sqrt{A^2+B^2}}=\frac{|4\cdot (-1)+(-3)\cdot 2+1|}{\sqrt{4^2+(-3)^2}}=\frac{9}{5} $$