How to Find the Distance from a Point to a Line

This finds the distance between the line $ax + by + c = 0$ and the point $(x, y)$, meaning the length of the perpendicular dropped from the point to the line.

$a$, $b$ and $c$ are the coefficients of the line, $(x, y)$ is the point and $d$ is the distance. The nearest point on the line, the foot of the perpendicular, is also given.

What the two parts do

The top, $ax + by + c$, is what you get by putting the point into the equation of the line. It is 0 when the point lies on the line and grows as the point moves away. That value scales with the size of the coefficients, so dividing by $\sqrt{a^2 + b^2}$ turns it into a length that does not depend on how the line was written. The absolute value keeps the distance positive on either side of the line.

Example

For the line $3x + 4y - 5 = 0$ and the point $(2, 3)$, the top is $|3 \times 2 + 4 \times 3 - 5| = 13$ and the bottom is $\sqrt{3^2 + 4^2} = 5$, so $d = 13 \div 5 = 2.6$. The foot of the perpendicular is $(0.44, 0.92)$.

Notes

Write the line as $ax + by + c = 0$. A line given as $y = 2x + 1$ becomes $2x - y + 1 = 0$.

With $a$ and $b$ both 0 the equation is not a line at all, so nothing can be calculated.