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▲In [[Euclidean geometry]], the '''distance from a point to a line''' is the shortest [[Euclidean distance|distance]] from a given [[Point (geometry)|point]] to any point on an infinite [[Line (mathematics)|straight line]]. It is the [[perpendicular distance]] of the point to the line, the length of the [[line segment]] which joins the point to nearest point on the line. The [[algebraic expression]] 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
==Cartesian coordinates==
[[File:PointToLineDistance v2.png|thumb|''This '''picture''' is a stub, help improve it.''. It tries to explain the roles of the values in the formula. Note: values are NOT to scale. Cyan line represents the formula to the left.]]
In the case of a line in the plane given by the equation {{
▲== Line defined by an equation ==
▲In the case of a line in the plane given by the equation {{math|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''<sub>0</sub>, ''y''<sub>0</sub>)}} 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 = \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,
== Line defined by two points ==▼
If the line passes through two points {{math|1=''P''<sub>1</sub> = (''x''<sub>1</sub>, ''y''<sub>1</sub>)}} and {{math|1=''P''<sub>2</sub> = (''x<sub>2</sub>'', ''y<sub>2</sub>'')}} then the distance of {{math|(''x''<sub>0</sub>, ''y''<sub>0</sub>)}} from the line is:<ref name=GEO />▼
:<math>\operatorname{distance}(P_1, P_2, (x_0, y_0)) = \frac{|(x_2-x_1)(y_0-y_1)-(x_0-x_1)(y_2-y_1)|}{\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}. </math>▼
The denominator of this expression is the distance between {{math|''P''<sub>1</sub>}} and {{math|''P''<sub>2</sub>}}. The numerator is twice the area of the triangle with its vertices at the three points, {{math|(''x''<sub>0</sub>, ''y''<sub>0</sub>)}}, {{math|''P''<sub>1</sub>}} and {{math|''P''<sub>2</sub>}}. 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 two points
▲:<math>\operatorname{distance}(P_1, P_2, (x_0, y_0)) = \frac{|(
▲The denominator of this expression is the distance between
==Proofs==
===An algebraic proof===
This proof is valid only if the line is neither vertical nor horizontal, that is, we assume that neither
The line with equation
:<math>\frac{y_0 - n}{x_0 - m}=\frac{b}{a}.</math>
Thus,
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Now consider,
:<math> (a(x_0 - m) + b(y_0 - n))^2
▲(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 \\
using the above squared equation. But we also have,
:<math> (a(x_0 - m) + b(y_0 - n))^2 = (ax_0 + by_0 - am -
since
Thus,
:<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}
===A geometric proof===
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and finally obtain:<ref>{{harvnb|Ballantine|Jerbert|1952}}</ref>
:<math> |\overline{PR}| = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}.</math>
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 hypoteneuse 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===
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Since ''Q'' is a point on the line, <math>c = -ax_1 - by_1</math>, and so,<ref>{{harvnb|Anton|1994|loc=pp. 138-9}}</ref>
:<math> d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}.</math>
== Another formula ==
It is possible to produce another expression to find the shortest distance of a point to a line. This derivation also requires that the line
The point P is given with coordinates (<math>x_0, y_0</math>).
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:<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
==Vector formulation==
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: <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{
This formula can be derived as follows: <math>\mathbf{
:<math>
is a vector that is the [[projection (linear algebra)|projection]] of <math>\mathbf{
:<math>(\mathbf{
is the component of <math>\mathbf{
== Another vector formulation ==
If the vector space is [[orthonormality|orthonormal]] and if the line (''l'' ) goes through point
: <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>.
Note that cross products only exist in dimensions 3 and 7.
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==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
* {{citation|first1=Ron|last1=Larson|first2=Robert|last2=Hostetler|title=Precalculus: A Concise Course|year=2007|publisher=Houghton Mifflin Co.|isbn=
* {{citation|first=Barry|last=Spain|title=Analytical Conics|year=2007|
==Further reading==
*{{citation|title=Encyclopedia of Distances|first1=Michel Marie|last1=Deza|author1-link=Michel Deza|first2=Elena|last2=
[[Category:Euclidean geometry]]
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