Distance from a point to a line: Difference between revisions

Revision as of 00:21, 1 May 2021

Latest revision as of 20:17, 6 July 2025

(37 intermediate revisions by 27 users not shown)

Line 1:

{{Short description|Geometry problem}}

InThe [[Euclidean'''distance''' geometry]], the(or '''perpendicular distance''') '''from a point to a line''' is the shortest [[Euclidean distance|distance]] from a givenfixed [[Point (geometry)|point]] to any point on ana fixed infinite [[Line (mathematics)|straight line]]. It is thein [[perpendicularEuclidean geometry]]. distanceIt of the point to the line,is the length of the [[line segment]] which joins the point to nearestthe pointline onand is [[perpendicular]] to the line. The formula for calculating it can be derived and expressed in several ways.

Knowing the shortest distance from a point to a line can be useful in various situations—for example, finding the shortest distance to reach a road, quantifying the scatter on a graph, etc. In [[Deming regression]], a type of linear curve fitting, if the dependent and independent variables have equal variance this results in [[orthogonal regression]] in which the degree of imperfection of the fit is measured for each data point as the perpendicular distance of the point from the regression line.

==Cartesian coordinates==

<!--''This '''picture''' is a stub. You can help Wikipedia by improving it.''

[[File:PointToLineDistance.png|PointToLineDistance]] -->

===Line defined by an equation===

In the case of a line in the plane given by the equation {{mathnowrap|1=''ax'' + ''by'' + ''c'' = 0,}}, where {{mvar|''a}}'', {{mvar|''b}}'' and {{mvar|''c}}'' are [[real number|real]] constants with {{mvar|''a}}'' and {{mvar|''b}}'' not both zero, the distance from the line to a point {{math|(''x''₀, ''y''₀)}} is<ref>{{harvnb|Larson|Hostetler|2007|loc=p. 452}}</ref><ref>{{harvnb|Spain|2007}}</ref>{{rp|p.14}}

:<math>\operatorname{distance}(ax+by+c=0, (x_0, y_0)) = \frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}. </math>

The point on this line which is closest to {{math|(''x''₀, ''y''₀)}} has coordinates:<ref name="Larson 2007 loc=p. 522">{{harvnb|Larson|Hostetler|2007|loc=p. 522}}</ref>

:<math>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}.</math>

'''Horizontal and vertical lines'''

In the general equation of a line, {{math|1=''ax'' + ''by'' + ''c'' = 0}}, {{mvar|''a}}'' and {{mvar|''b}}'' cannot both be zero unless {{mvar|''c}}'' is also zero, in which case the equation does not define a line. If {{math|1=''a'' &nbsp;= &nbsp;0}} and {{math|''b'' &nbsp;{{math|≠ 0}}&nbsp;0, the line is horizontal and has equation {{math|1=''y'' = −{{sfrac|''c''|/''b''}}}}. The distance from {{math|(''x''₀, ''y''₀)}} to this line is measured along a vertical line segment of length {{math|1={{!}}''y''₀ − (−{{sfrac|''c''|/''b''}}){{!}}| = {{sfrac|{{!}}''by''₀ + ''c''{{!}}|{{!}} / |''b''{{!}}}}}}| in accordance with the formula. Similarly, for vertical lines (''b'' = 0) the distance between the same point and the line is {{math|1={{sfrac|{{!}}''ax''₀ + ''c''{{!}}|{{!}} / |''a''{{!}}}} }}|, as measured along a horizontal line segment.

===Line defined by two points===

If the line passes through two points {{math|1=''P''₁ = (''x''₁, ''y''₁)}} and {{math|1=''P''₂ = (''x₂'', ''y₂'')}} then the distance of {{math|(''x''₀, ''y''₀)}} from the line is:<ref name=GEO />

:<math>\operatorname{distance}(P_1, P_2, (x_0, y_0)) = \frac{|(x_2y_2-x_1)(y_1-y_0)x_0-(x_1x_2-x_0x_1)(y_2-y_0+x_2 y_1)-y_2 x_1|}{\sqrt{(x_2y_2-x_1y_1)^2+(y_2x_2-y_1x_1)^2}}. </math>

The denominator of this expression is the distance between {{math|''P''₁}}'' and {{math|''P''₂}}''. The numerator is twice the area of the triangle with its vertices at the three points, {{math|(''x''₀, ''y''₀)}}, {{math|''P''₁}}'' and {{math|''P''₂}}''. See: {{slink|Area of a triangle|Using coordinates}}. The expression is equivalent to {{<math|1 display=''"inline">h'' = \frac{2A}{sfrac|2''A''|''b''}}}}</math>, which can be obtained by rearranging the standard formula for the area of a triangle: {{<math|1 display=''"inline">A'' = \frac{1}{sfrac|1|2}} ''bh''}}</math>, where {{mvar|''b}}'' is the length of a side, and {{mvar|''h}}'' is the perpendicular height from the opposite vertex.

== Line defined by point and angle ==

The denominator of this expression is the distance between {{math|''P''₁}} and {{math|''P''₂}}. The numerator is twice the area of the triangle with its vertices at the three points, {{math|(''x''₀, ''y''₀)}}, {{math|''P''₁}} and {{math|''P''₂}}. See: {{slink|Area of a triangle|Using coordinates}}. The expression is equivalent to {{math|1=''h'' = {{sfrac|2''A''|''b''}}}}, which can be obtained by rearranging the standard formula for the area of a triangle: {{math|1=''A'' = {{sfrac|1|2}} ''bh''}}, where {{mvar|b}} is the length of a side, and {{mvar|h}} is the perpendicular height from the opposite vertex.

If the line passes through the point {{math|1=''P'' = (''P''_x, ''P''_y)}} with angle {{math|''θ''}}, then the distance of some point {{math|(''x''₀, ''y''₀)}} to the line is

:<math>\operatorname{distance}(P, \theta, (x_0, y_0)) = |\cos(\theta)(P_y-y_0) -\sin(\theta)(P_x-x_0)| </math>

==Proofs==

===An algebraic proof===

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

The line with equation {{math|1=''ax'' + ''by'' + ''c'' = 0}} has slope {{math|−''a''/''b''}}, so any line perpendicular to it will have slope {{math|''b''/''a''}} (the negative reciprocal). Let {{math|(''m'', ''n'')}} be the point of intersection of the line {{math|1=''ax'' + ''by'' + ''c'' = 0}} and the line perpendicular to it which passes through the point ({{math|''x''₀}}, {{math|''y''₀}}). The line through these two points is perpendicular to the original line, so

:<math>\frac{y_0 - n}{x_0 - m}=\frac{b}{a}.</math>

Thus,

Line 39 ⟶ 40:

Now consider,

:<math> (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)</math>

:<math>

\begin{align}

(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{align}

</math>

using the above squared equation. But we also have,

:<math> (a(x_0 - m) + b(y_0 - n))^2 = (ax_0 + by_0 - am - bn )^2 = (ax_0 + by_0 + c)^2</math>

since {{math|(''m'', ''n'')}} is on {{math|1=''ax'' + ''by'' + ''c'' = 0}}.

Thus,

:<math>\left(a^2 + b^2\right) \left((x_0 - m)^2 + (y_0 - n)^2\right) = (ax_0 + by_0 + c)^2 </math>

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

:<math>d=\sqrt{(x_0 - m)^2+(y_0 - n)^2} = \frac{|ax_0+ by_0 +c|}{\sqrt{a^2+b^2}}.</math><ref>Between Certainty and Uncertainty: Statistics and Probability in Five Units With Notes on Historical Origins and Illustrative Numerical Examples</ref>

===A geometric proof===

Line 58 ⟶ 54:

This proof is valid only if the line is not horizontal or vertical.<ref>{{harvnb|Ballantine|Jerbert|1952}} do not mention this restriction in their article</ref>

Drop a perpendicular from the point ''P'' with coordinates (''x''₀, ''y''₀) 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).<ref>If the two triangles are on opposite sides of the line, these angles are congruent because they are alternate interior angles.</ref> Corresponding sides of these triangles are in the same ratio, so:

:<math>\frac{|\overline{PR}|}{|\overline{PS}|} = \frac{|\overline{TV}|}{|\overline{TU}|}.</math>

If point ''S'' has coordinates (''x''₀,''m'') then |''PS''| = |''y''₀ -− ''m''| and the distance from ''P'' to the line is:

:<math> |\overline{PR} | = \frac{|y_0 - m||B|}{\sqrt{A^2 + B^2}}.</math>

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

Line 70 ⟶ 66:

A variation of this proof is to place V at P and compute the area of the triangle ∆''UVT'' two ways to obtain that <math>D|\overline{TU}| = |\overline{VU}||\overline{VT}|</math>

where D is the altitude of ∆''UVT'' drawn to the hypotenusehypoteneuse of ∆''UVT'' from ''P''. The distance formula can then used to express <math>|\overline{TU}|</math>, <math>|\overline{VU}|</math>, and <math>|\overline{VT}|</math>in terms of the coordinates of P and the coefficients of the equation of the line to get the indicated formula.{{citation needed|date=April 2015}}

===A vector projection proof===

Line 101 ⟶ 97:

:<math>d=\sqrt{ \left( {\frac{x_0 + m y_0-mk}{m^2+1}-x_0 } \right) ^2 + \left( {m\frac{x_0+m y_0-mk}{m^2+1}+k-y_0 }\right) ^2 } = \frac{|k + m x_0 - y_0|}\sqrt{1 + m^2} .</math>

Recalling that ''m'' = -−''a''/''b'' and ''k'' = -− ''c''/''b'' for the line with equation ''ax'' + ''by'' + c = 0, a little algebraic simplification reduces this to the standard expression.<ref name=">{{harvnb|Larson |Hostetler|2007 |loc=p. 522"}}</ref>

==Vector formulation==

Line 110 ⟶ 106:

: <math> \mathbf{x} = \mathbf{a} + t\mathbf{n}</math>

Here {{math|'''a'''}} is the position of a point on the line, and {{math|'''n'''}} is a [[unit vector]] in the direction of the line. Then as scalar ''t'' varies, {{math|'''x'''}} gives the [[locus (mathematics)|locus]] of the line.

The distance of an arbitrary point {{math|'''p'''}} to this line is given by

: <math>\operatorname{distance}(\mathbf{x} = \mathbf{a} + t\mathbf{n}, \mathbf{p}) = \| (\mathbf{pa}-\mathbf{ap}) - ((\mathbf{pa}-\mathbf{ap}) \cdot \mathbf{n})\mathbf{n} \|. </math>

This formula can be derived as follows: <math>\mathbf{pa}-\mathbf{ap}</math> is a vector from {{math|'''ap'''}} to the point {{math|'''pa'''}} on the line. Then <math>(\mathbf{pa} - \mathbf{ap}) \cdot \mathbf{n}</math> is the projected length onto the line and so

:<math>\mathbf{a} + ((\mathbf{pa} - \mathbf{ap}) \cdot \mathbf{n})\mathbf{n}</math>

is a vector that is the [[projection (linear algebra)|projection]] of <math>\mathbf{pa}-\mathbf{ap}</math> onto the line and represents the point on the line closest to <math>\mathbf{p}</math>. Thus

:<math>(\mathbf{pa}-\mathbf{ap}) - ((\mathbf{pa}-\mathbf{ap}) \cdot \mathbf{n})\mathbf{n}</math>

is the component of <math>\mathbf{pa}-\mathbf{ap}</math> perpendicular to the line. The distance from the point to the line is then just the [[norm (mathematics)|norm]] of that vector.<ref name=GEO>{{cite web|last=Sunday|first=Dan|title=Lines and Distance of a Point to a Line|url=http://geomalgorithms.com/a02-_lines.html|archive-url=https://archive.today/20131206184414/http://geomalgorithms.com/a02-_lines.html|publisherurl-status=softSurferusurped|accessarchive-date=December 6, 2013|publisher=softSurfer|accessdate=6 December 2013}}</ref> This more general formula is not restricted to two dimensions.

== Another vector formulation ==

If the vectorline space is [[orthonormality|orthonormal]] and if the line(''l'' ) goes through point {{math|'''a'''}}A and has a [[Euclidean vector|direction vector]] {{<math|'''n'''}}>\vec u</math>, the distance between point {{math|P and line (''l'p'''}} and the line) is

: <math>d(\mathrm{P}, (l))= \frac{\left\|\overrightarrow{\mathrm{AP}} \times\vec u\right\|}{\|\vec u\|}</math>

where <math>\overrightarrow{\mathrm{AP}} \times\vec u</math> is the [[cross product]] of the vectors <math>\overrightarrow{\mathrm{AP}}</math> and <math>\vec u</math> and where <math>\|\vec u\|</math> is the vector norm of <math>\vec u</math>.

<math>\overrightarrow{\mathrm{AP}}=P-A</math>

: <math>\operatorname{distance}(\mathbf{x} = \mathbf{a} + t\mathbf{n}, \mathbf{p}) = \frac{\left\|(\mathbf{a}-\mathbf{p}) \times \mathbf{n}\right\|}{\|\mathbf{n}\|}.</math>

Note that cross products only exist in dimensions 3 and 7 and trivially in dimensions 0 and 1 (where the cross product is constant 0).

== See also ==

* [[Hesse normal form]]

* [[Line-lineLine–line intersection]]

* [[Distance between two parallel lines]]

* [[Distance from a point to a plane]]

* [[{{slink|Skew lines#|Distance]]}}

==Notes==

Line 142 ⟶ 140:

==References==

* {{citation|first=Howard|last=Anton|title=Elementary Linear Algebra|edition=7th|year=1994|publisher=John Wiley & Sons|isbn=0-471-58742-7}}

* {{citation|first1=J.P.|last1=Ballantine|first2=A.R.|last2=Jerbert|year=1952|volume=59|title=Distance from a line or plane to a point|journal=American Mathematical Monthly|issue=4 |pages=242–243|doi=10.2307/2306514|jstor=2306514 }}

* {{citation|first1=Ron|last1=Larson|first2=Robert|last2=Hostetler|title=Precalculus: A Concise Course|year=2007|publisher=Houghton Mifflin Co.|isbn=978-0-618-62719-6|url-access=registration|url=https://archive.org/details/precalculusconci00lars}}

* {{citation|first=Barry|last=Spain|title=Analytical Conics|year=2007|orig-yeardate=1957|publisher=Dover Publications|isbn=978-0-486-45773-4}}

==Further reading==