Distance from a point to a line

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[1][2]^:p.14

${\displaystyle \operatorname {distance} (ax+by+c=0,(x_{0},y_{0}))={\frac {|ax_{0}+by_{0}+c|}{\sqrt {a^{2}+b^{2}}}}.}$

The point on this line which is closest to (x₀, y₀) has coordinates:[3]

${\displaystyle x={\frac {b(bx_{0}-ay_{0})-ac}{a^{2}+b^{2}}}{\text{ and }}y={\frac {a(-bx_{0}+ay_{0})-bc}{a^{2}+b^{2}}}.}$

Horizontal and vertical lines

In the general equation of a line, ax + by + c = 0, a and b cannot both be zero unless c is also zero, in which case the equation does not define a line. If a = 0 and b ≠ 0, the line is horizontal and has equation y = −c/b. The distance from (x₀, y₀) to this line is measured along a vertical line segment of length |y₀ − (−c/b)| = |by₀ + c|/|b| in accordance with the formula. Similarly, for vertical lines (b = 0) the distance between the same point and the line is |ax₀ + c|/|a|, as measured along a horizontal line segment.

If the line passes through two points P₁ = (x₁, y₁) and P₂ = (x₂, y₂) then the distance of (x₀, y₀) from the line is:[4]

${\displaystyle \operatorname {distance} (P_{1},P_{2},(x_{0},y_{0}))={\frac {|(x_{2}-x_{1})(y_{1}-y_{0})-(x_{1}-x_{0})(y_{2}-y_{1})|}{\sqrt {(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}}}.}$

The denominator of this expression is the distance between P₁ and P₂. The numerator is twice the area of the triangle with its vertices at the three points, (x₀, y₀), P₁ and P₂. See: Area of a triangle § Using coordinates. The expression is equivalent to h = 2A/b, which can be obtained by rearranging the standard formula for the area of a triangle: A = 1/2 bh, where b is the length of a side, and h is the perpendicular height from the opposite vertex.

This proof is valid only if the line is neither vertical nor horizontal, that is, we assume that neither a nor b in the equation of the line is zero.

The line with equation ax + by + c = 0 has slope −a/b, so any line perpendicular to it will have slope b/a (the negative reciprocal). Let (m, n) be the point of intersection of the line ax + by + c = 0 and the line perpendicular to it which passes through the point (x₀, y₀). The line through these two points is perpendicular to the original line, so

${\displaystyle {\frac {y_{0}-n}{x_{0}-m}}={\frac {b}{a}}.}$

Thus, ${\displaystyle a(y_{0}-n)-b(x_{0}-m)=0,}$ and by squaring this equation we obtain:

${\displaystyle a^{2}(y_{0}-n)^{2}+b^{2}(x_{0}-m)^{2}=2ab(y_{0}-n)(x_{0}-m).}$

Now consider,

${\displaystyle {\begin{aligned}(a(x_{0}-m)+b(y_{0}-n))^{2}&=a^{2}(x_{0}-m)^{2}+2ab(y_{0}-n)(x_{0}-m)+b^{2}(y_{0}-n)^{2}\\&=\left(a^{2}+b^{2}\right)\left((x_{0}-m)^{2}+(y_{0}-n)^{2}\right)\end{aligned}}}$

using the above squared equation. But we also have,

${\displaystyle (a(x_{0}-m)+b(y_{0}-n))^{2}=(ax_{0}+by_{0}-am-bn)^{2}=(ax_{0}+by_{0}+c)^{2}}$

since (m, n) is on ax + by + c = 0. Thus,

${\displaystyle \left(a^{2}+b^{2}\right)\left((x_{0}-m)^{2}+(y_{0}-n)^{2}\right)=(ax_{0}+by_{0}+c)^{2}}$

and we obtain the length of the line segment determined by these two points,

${\displaystyle d={\sqrt {(x_{0}-m)^{2}+(y_{0}-n)^{2}}}={\frac {|ax_{0}+by_{0}+c|}{\sqrt {a^{2}+b^{2}}}}.}$[5]

Diagram for geometric proof

This proof is valid only if the line is not horizontal or vertical.[6]

Drop a perpendicular from the point P with coordinates (x₀, y₀) to the line with equation Ax + By + C = 0. Label the foot of the perpendicular R. Draw the vertical line through P and label its intersection with the given line S. At any point T on the line, draw a right triangle TVU whose sides are horizontal and vertical line segments with hypotenuse TU on the given line and horizontal side of length |B| (see diagram). The vertical side of ∆TVU will have length |A| since the line has slope -A/B.

∆PRS and ∆TVU are similar triangles, since they are both right triangles and ∠PSR ≅ ∠TUV since they are corresponding angles of a transversal to the parallel lines PS and UV (both are vertical lines).[7] Corresponding sides of these triangles are in the same ratio, so:

${\displaystyle {\frac {|{\overline {PR}}|}{|{\overline {PS}}|}}={\frac {|{\overline {TV}}|}{|{\overline {TU}}|}}.}$

If point S has coordinates (x₀,m) then |PS| = |y₀ - m| and the distance from P to the line is:

${\displaystyle |{\overline {PR}}|={\frac {|y_{0}-m||B|}{\sqrt {A^{2}+B^{2}}}}.}$

Since S is on the line, we can find the value of m,

${\displaystyle m={\frac {-Ax_{0}-C}{B}},}$

and finally obtain:[8]

${\displaystyle |{\overline {PR}}|={\frac {|Ax_{0}+By_{0}+C|}{\sqrt {A^{2}+B^{2}}}}.}$

A variation of this proof is to place V at P and compute the area of the triangle ∆UVT two ways to obtain that ${\displaystyle D|{\overline {TU}}|=|{\overline {VU}}||{\overline {VT}}|}$ where D is the altitude of ∆UVT drawn to the hypotenuse of ∆UVT from P. The distance formula can then used to express ${\displaystyle |{\overline {TU}}|}$, ${\displaystyle |{\overline {VU}}|}$, and ${\displaystyle |{\overline {VT}}|}$in terms of the coordinates of P and the coefficients of the equation of the line to get the indicated formula.

Diagram for vector projection proof

Let P be the point with coordinates (x₀, y₀) and let the given line have equation ax + by + c = 0. Also, let Q = (x₁, y₁) be any point on this line and n the vector (a, b) starting at point Q. The vector n is perpendicular to the line, and the distance d from point P to the line is equal to the length of the orthogonal projection of ${\displaystyle {\overrightarrow {QP}}}$ on n. The length of this projection is given by:

${\displaystyle d={\frac {|{\overrightarrow {QP}}\cdot \mathbf {n} |}{\|\mathbf {n} \|}}.}$

Now,

${\displaystyle {\overrightarrow {QP}}=(x_{0}-x_{1},y_{0}-y_{1}),}$ so ${\displaystyle {\overrightarrow {QP}}\cdot \mathbf {n} =a(x_{0}-x_{1})+b(y_{0}-y_{1})}$ and ${\displaystyle \|\mathbf {n} \|={\sqrt {a^{2}+b^{2}}},}$

thus

${\displaystyle d={\frac {|a(x_{0}-x_{1})+b(y_{0}-y_{1})|}{\sqrt {a^{2}+b^{2}}}}.}$

Since Q is a point on the line, ${\displaystyle c=-ax_{1}-by_{1}}$, and so,[9]

${\displaystyle d={\frac {|ax_{0}+by_{0}+c|}{\sqrt {a^{2}+b^{2}}}}.}$