Distance from a point to a line

Line defined by an equation

In the case of a line in the plane given by the equation ${\displaystyle ax+by+c=0}$ where ${\displaystyle a}$, ${\displaystyle b}$ and ${\displaystyle c}$ are real constants with ${\displaystyle a}$ and ${\displaystyle b}$ not both zero, the distance from the line to a point ${\displaystyle (x_{0},y_{0})}$ 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 that is closest to ${\displaystyle (x_{0},y_{0})}$ has coordinates:^[3]

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

or equivalently

${\displaystyle x=x_{0}-a{\frac {ax_{0}+by_{0}+c}{a^{2}+b^{2}}}}$ and ${\displaystyle y=y_{0}-b{\frac {ax_{0}+by_{0}+c}{a^{2}+b^{2}}}}$

Horizontal and vertical lines

In the general equation of a line, ${\displaystyle ax+by+c=0}$, ${\displaystyle a}$ and ${\displaystyle b}$ cannot both be zero unless ${\displaystyle c}$ is also zero; in that case, the equation does not define a line. If ${\displaystyle a=0}$ and ${\displaystyle b\neq 0}$, the line is horizontal and has equation ${\displaystyle y=-c/b}$. The distance from ${\displaystyle (x_{0},y_{0})}$ to this line is measured along a vertical line segment of length ${\displaystyle |y_{0}-(-c/b)|=|by_{0}+c|/|b|}$ in accordance with the formula. Similarly, for vertical lines ${\displaystyle (b=0)}$ the distance between the same point and the line is ${\displaystyle |ax_{0}+c|/|a|}$, as measured along a horizontal line segment.

Line defined by two points

If the line passes through two points ${\displaystyle P_{1}=(x_{1},y_{1})}$ and ${\displaystyle P_{1}=(x_{2},y_{2})}$ then the distance of ${\displaystyle (x_{0},y_{0})}$ from the line is:

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

The denominator of this expression is the distance between ${\displaystyle P_{1}}$ and ${\displaystyle P_{2}}$. The numerator is twice the area of the triangle with its vertices at the three points, ${\displaystyle (x_{0},y_{0})}$, ${\displaystyle P_{1}}$ and ${\displaystyle P_{2}}$. See: Area of a triangle § Using coordinates. The expression is equivalent to ${\textstyle h={\frac {2A}{b}}}$, which can be obtained by rearranging the standard formula for the area of a triangle: ${\textstyle A={\frac {1}{2}}bh}$, where ${\displaystyle b}$ is the length of a side, and ${\displaystyle h}$ is the perpendicular height from the opposite vertex.

An algebraic proof

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

The line with equation ${\displaystyle ax+by+c=0}$ has slope ${\displaystyle -a/b}$, so any line perpendicular to it will have slope ${\displaystyle b/a}$ (the negative reciprocal). Let ${\displaystyle (m,n)}$ be the point of intersection of the line ${\displaystyle ax+by+c=0}$ and the line perpendicular to it which passes through the point ${\displaystyle (x_{0},y_{0})}$. 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,

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

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,

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

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

${\displaystyle d={\sqrt {(x_{0}-m)^{2}+(y_{0}-n)^{2}}}={\frac {|ax_{0}+by_{0}+c|}{\sqrt {a^{2}+b^{2}}}}.}$^[4]

A geometric proof

This proof is valid only if the line is not horizontal or vertical.^[5]

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

${\displaystyle \triangle PRS}$ and ${\displaystyle \triangle TVU}$ are similar triangles, since they are both right triangles and ${\displaystyle \angle PSR\cong \angle TUV}$ since they are corresponding angles of a transversal to the parallel lines ${\displaystyle PS}$ and </math>UV</math> (both are vertical lines).^[6] Corresponding sides of these triangles are in the same ratio, so:

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

If point ${\displaystyle S}$ has coordinates ${\displaystyle (x_{0},m)}$ then ${\displaystyle |PS|=|y_{0}-m|}$ and the distance from ${\displaystyle P}$ to the line is:

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

Since ${\displaystyle S}$ is on the line, we can find the value of ${\displaystyle m}$,

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

and finally obtain:^[7]

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

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

A vector projection proof

Let ${\displaystyle P}$ be the point with coordinates ${\displaystyle (x_{0},y_{0})}$ and let the given line have equation ${\displaystyle ax+by+c=0}$. Also, let ${\displaystyle Q=(x_{1},y_{1})}$ be any point on this line and ${\displaystyle n}$ the vector ${\displaystyle (a,b)}$ starting at point ${\displaystyle Q}$. The vector n is perpendicular to the line, and the distance ${\displaystyle d}$ from point ${\displaystyle P}$ 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 ${\displaystyle Q}$ is a point on the line, ${\displaystyle c=-ax_{1}-by_{1}}$, and so,^[8]

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