Distance from a Point to a Line

Why the shortest segment is perpendicular

Let $P$ be the point and let $H$ be its perpendicular foot on the line. For any other point $Q$ on the line, triangle $PHQ$ is right-angled at $H$, so

Equality occurs at $Q=H$. Therefore $PH$ is the shortest distance. If $P$ already lies on the line, the distance is zero.

Worked example by finding the foot

Find the distance from $P=(3,2)$ to the line $y=x+2$. The line has slope 1, so the perpendicular through $P$ has slope $-1$:

At the intersection,

Then $y=\tfrac32+2=\tfrac72$, giving $H=(\tfrac32,\tfrac72)$.

Apply the distance formula:

The point $(3,5)$ is also on the line, but its vertical distance of 3 from $P$ is longer. The perpendicular distance is approximately 2.121 units.

Direct point-to-line formula

Write the line as

For $P=(x_0,y_0)$, the distance is

The absolute value makes the result non-negative. The denominator accounts for the size of the normal vector $(A,B)$. A derivation using the normal direction is given in Wolfram MathWorld’s point-line distance reference.

For $y=x+2$, use $x-y+2=0$. Hence $A=1$, $B=-1$ and $C=2$:

This agrees with the foot method. Be careful about signs: when a line is initially written $Ax+By=D$, move $D$ to the left, so the formula’s constant is $C=-D$.

Why rescaling the line equation changes nothing

Multiplying the entire line equation by a non-zero constant $\lambda$ gives the same line. In the distance formula, the numerator is multiplied by $|\lambda|$, and so is the denominator:

The factors cancel. This is why $x-y+2=0$ and $2x-2y+4=0$ give the same distance. Multiplying only some terms would not preserve the line.

Another example and the perpendicular foot formula

Find the distance from the origin to $3x+4y-20=0$:

The foot can also be obtained without solving a second line equation. Define

Then the foot is

In this example $q=-20/25=-4/5$, so $H=(12/5,16/5)$. It lies on the line because $3(12/5)+4(16/5)=20$, and its distance from the origin is 4.

Vertical lines, horizontal lines and parallel lines

For a vertical line $x=k$, the distance from $(x_0,y_0)$ is $|x_0-k|$. For a horizontal line $y=k$, it is $|y_0-k|$. Both follow directly from the general formula.

For parallel lines written with identical $A,B$ coefficients,

the distance between them is $|C_2-C_1|/\sqrt{A^2+B^2}$. Normalize the coefficients first; subtracting the constants of differently scaled equations is not valid.

A segment is not an infinite line

If the perpendicular foot lies outside a specified segment, the nearest point on that segment is an endpoint, not the foot on its infinite extension.

For example, consider $P=(4,2)$ and the horizontal segment from $(0,0)$ to $(1,0)$. The distance to the infinite line $y=0$ is 2, but the foot $(4,0)$ is outside the segment. The nearest endpoint is $(1,0)$, at distance $\sqrt{3^2+2^2}=\sqrt{13}$.

Always identify whether the target is a line, ray or segment before choosing a distance method.

Related topics

Circles in Coordinate Geometry compares the distance of a line from a centre with a radius. Parabolas uses equality of point-to-point and point-to-line distances to define a curve.