L : Ax+By+C=0

Key concept: This note is about the distance from a point to a line written in standard form. The formula comes directly from projecting the point onto the line's normal vector.

Step by step

For a line in standard form

$L: Ax+By+C=0,$

the perpendicular distance from a point $P(x_1,y_1)$ to the line is

$d=\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}.$

What each symbol means

• $A,B,C$ are the coefficients of the line
• $(x_1,y_1)$ is the given point
• $d$ is the shortest distance, measured perpendicularly to the line

Why this works

The vector $(A,B)$ is normal to the line. The numerator gives the signed projection of the point onto that normal direction, and the absolute value makes the distance nonnegative.

How to use it

1. Write the line as $Ax+By+C=0$.
2. Substitute the point coordinates into $Ax_1+By_1+C$.
3. Take the absolute value.
4. Divide by $\sqrt{A^2+B^2}$.

Pitfall alert

A frequent error is to forget the absolute value, which can make the result negative. Another common mistake is using the distance formula for a point to a point instead of point to a line. Also, the line must be in the form $Ax+By+C=0$ before applying the formula directly.

Try different conditions

If the line is written as $y=mx+b$, first rearrange it to $mx-y+b=0$ before using the formula. If the point is the origin $(0,0)$, the distance becomes $\frac{|C|}{\sqrt{A^2+B^2}}$.

Further reading

perpendicular distance, standard form, normal vector