Distance from a point to a line in the plane

Content updated: October 22, 2026

The distance from a point to a line is the length of its shortest, perpendicular segment to the line. The normal-vector formula provides a quick calculation and explains why an absolute value is needed.

On this page: Concepts · Method · Examples · Practice · Further learning

1. Distance and the shortest segment

Let a point $M$ lie outside a line $d$, and let $H$ be the projection of $M$ onto $d$. For $Q\in d$ with $Q\ne H$, the triangle $MHQ$ is right-angled at $H$, so $MQ^2=MH^2+HQ^2>MH^2$. Thus $MH$ is the minimum length and is called the distance from $M$ to $d$. If $M\in d$, the distance is zero.

2. Formula from the general equation

Let $M(x_0;y_0)$ and $d:ax+by+c=0$, where $a^2+b^2>0$. Then

Set $q=ax_0+by_0+c$. By the projection formula, $\overrightarrow{MH}=-q(a;b)/(a^2+b^2)$. Taking lengths gives $MH=|q|/\sqrt{a^2+b^2}$. The distance denominator is the length of the normal vector; the sum of squares occurs instead in the coefficient used to find the projection.

When $q=0$, the point lies on the line and the distance is zero. Multiplying the equation by $\lambda\ne0$ multiplies both numerator and denominator by $|\lambda|$, leaving the result unchanged. Do not use the formula when $a=b=0$, since the equation then does not define a line.

Procedure: rearrange the equation with zero on one side, read the signed coefficients correctly, substitute the point into the numerator, take the absolute value, and divide by the normal-vector length. For a vertical line $x=k$ and a horizontal line $y=\ell$, the formula reduces to $|x_0-k|$ and $|y_0-\ell|$, respectively.

3. Worked examples

Example 1. Retaining the correct constant term

Find the distance from $M_0(-1;2)$ to $d:4x+3y-1=0$. We have $q=4(-1)+3\cdot2-1=1$ and $\sqrt{4^2+3^2}=5$. Hence

Writing the same line as $-8x-6y+2=0$ gives numerator $|-2|$ and denominator 10, again yielding $1/5$.

Example 2. A negative numerator and a point on the line

For $O(0;0)$ and the same line $4x+3y-1=0$, the numerator before taking its absolute value is $-1$, so the distance is again $1/5$. The point $A(1;-1)$ satisfies $4-3-1=0$ and therefore has distance zero. A negative distance would contradict its definition as a length.

Example 3. Vertical and horizontal lines

Let $M(-2;3)$. Its distance to $x=1$ is $|-2-1|=3$; its distance to $y=-2$ is $|3-(-2)|=5$. No slope conversion is needed, and a vertical line still has a well-defined distance.

4. Practice exercises

1. Find the distance from $M(1;2)$ to $3x+4y-5=0$.
2. Find the distance from $M(-2;3)$ to $2x-y+1=0$.
3. Find the distance from $(0;4)$ to $y=4$.

5. Common mistakes

• Omitting the absolute value when the numerator is negative.
• Using $a^2+b^2$ instead of its square root in the distance denominator.
• Changing the sign of the constant term incorrectly when rearranging the equation.
• Assuming that a vertical line has no point-to-line distance because it has no slope.

6. Related lessons and resources

• Projection and reflection across a line (English lesson)
• The general equation of a line (English lesson)
• Normal and direction vectors (English lesson)

Open the original exercise resource on Scribd. This lesson includes practice exercises with answers.