Distance from a point to a line

The distance from a point to a line is the shortest distance from a point to a line in Euclidean geometry. It can be calculated in the following ways.

1 Cartesian coordinates
2 Vector formulation
3 Proof 1 (algebraic proof)
4 Proof 2 (geometric proof)
5 Sample code
6 See also

Cartesian coordinates

In the case of a line in the plane given by the equation ax + by + c = 0, where a, b and c are real constants with a and b not both zero, the distance from the line to a point (x₀,y₀) is

$\operatorname{distance}(ax%2Bby%2Bc=0, (x_0, y_0)) = \frac{|ax_0%2Bby_0%2Bc|}{\sqrt{a^2%2Bb^2}}.$

Vector formulation

Suppose we express the line in vector form:

$\mathbf{x} = \mathbf{a} %2B t\mathbf{n}$

where $n$ is a unit vector. That is, a point, $x$ , on the line is found by moving to a point $a$ in space, then moving $t$ units along the direction of the line..

The distance of an arbitrary point $p$ to this line is given by

$\operatorname{distance}(\mathbf{x} = \mathbf{a} %2B t\mathbf{n}, \mathbf{p}) = \| (\mathbf{a}-\mathbf{p}) - ((\mathbf{a}-\mathbf{p}) \cdot \mathbf{n})\mathbf{n} \|.$

This more general formula can be used in dimensions other than two. This equation is constructed geometrically as follows: $\mathbf{a}-\mathbf{p}$ is a vector from $p$ to the point $a$ on the line. Then $(\mathbf{a} - \mathbf{p}) \cdot \mathbf{n}$ is the projected length onto the line and so

$((\mathbf{a} - \mathbf{p}) \cdot \mathbf{n})\mathbf{n}$

is a vector that is the projection of $\mathbf{a}-\mathbf{p}$ onto the line and so

$(\mathbf{a}-\mathbf{p}) - ((\mathbf{a}-\mathbf{p}) \cdot \mathbf{n})\mathbf{n}$

is the component of $\mathbf{a}-\mathbf{p}$ perpendicular to the line. The distance from the point to the line is then just the norm of that vector.

Proof 1 (algebraic proof)

Let point (x,y) be the intersection between the line ax + by + c = 0 and its perpendicular which contains (m,n), where point (m,n) is any arbitrary point on the perpendicular line to ax + by + c = 0.

Then it is necessary to show $a^2(n-y)^2 %2B b^2(m-x)^2 = 2ab(m-x)(n-y).$

The above equation can be changed to $(a^2(n-y))/(m-x) %2B (b^2(m-x))/(n-y) = 2ab,$ because the slope of the perpendicular to the ax+by+c which contains (x,y) and (m,n) is b/a.

Then

$(a^2%2Bb^2)((m-x)^2%2B(n-y)^2)=[a(m-x)%2Bb(n-y)]^2=(am%2Bbn%2Bc)^2.$

So the distance is

$d=\sqrt{(m-x)^2%2B(n-y)^2}= |am%2Bbn%2Bc|/\sqrt{a^2%2Bb^2}$

Proof 2 (geometric proof)

Let the point S(m,n) connect to the point G(x,y) which is on the line ax+by+c=0, both lines being perpendicular to each other.

Draw a line am+bn+d=0, containing the point S(m,n), which is parallel to ax+by+c=0.

The absolute value of (c-d)/b, which is the distance of the line connecting the point G and some point F on the line am+bn+d=0 and parallel to the y-axis, is equal to the absolute value of (am+bn+c)/b.

Then the desired distance SG can be derived from the right triangle SGF, which is in the ratio of a:b: $\sqrt{a^2%2Bb^2}$ .

The absolute value of (am+bn+c)/b is the diagonal of the right triangle, so just multiply by the absolute value of b and divide by $\sqrt{a^2%2Bb^2}$ , and the proof is complete.

Sample code

The following Java snippet provide distance from point P to the line A-B:

public double pointToLineDistance(Point A, Point B, Point P)
{
 double normalLength = Math.sqrt((B.x - A.x) * (B.x - A.x) + (B.y - A.y) * (B.y - A.y));
 return Math.abs((P.x - A.x) * (B.y - A.y) - (P.y - A.y) * (B.x - A.x)) / normalLength;
}