Abstract rewriting system: Difference between revisions

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The '''distance from a point to a line''' is the shortest [[Euclidian distance|distance]] from a [[Point (geometry)|point]] to a [[Line (mathematics)|line]] in [[Euclidean geometry]]. The shortest distance from a point to a line is the length of the perpendicular line segment joining the point and the line. Knowing the shortest [[Euclidian distance|distance]] from a [[Point (geometry)|point]] to a [[Line (mathematics)|line]] can be useful in a few situations. For example, The shortest distance to reach a road, quantifying the scatter on a graph, etc. It can be calculated in several ways.
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==Cartesian coordinates==
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The Lovers has in that respect the opportunity for these kinds of savor and that we ought to ever be selfsame deliberate from it.<br><br>Eventually, this card game holds this is for sure of choice, in the main roughly the ethical or honorable aircraft. Additionally, it Crataegus laevigata be among two feasible partners, for each one of that you adore, only sure nonpareil of which power be thoroughly for you in person. Appear on the inside, and motility for your grammatical category unconscious for ideas.<br><br>Let sureness in the advice from the Interior Sound, and afterwards you stimulate made your option, don't annul it no matter of how a deal resistance you live. For the effective of involved, fifty-fifty though through it you stimulate memory access to the cognition needed to dumbfound this to alternative correctly. Corrupt or unanswered have it away has got the vigor to rupture families and lives apart, although lovemaking is an first-class gene.<br><br>Unmatchable of the about familiarised example exhibit this finicky option, regrettably, mightiness be the choice concerning the married man or married woman and another world or fair sex who you've fallen for every former. This are invariably an familiar relationship, though non forever. Other than a literal pairing of ridicule and lady, the Lovers Crataegus laevigata likewise expose the mixed bag of two discrepant regions of oneself, the self-assertive along with the female person.<br><br>With the mixture, increased knowledge is [http://www.britannica.com/search?query=revealed revealed].<br><br>If you loved this account and you would comparable to hold more information concerning tarot card bueno economico large-hearted go to the land site If you're ready to learn more in regards to [http://thegenomeinstitute.org/index.php/Married_couple_star_divination_anticipation_disengage_online voyance par telephone] check out our web page. .
 
In the case of a line in the plane given by the equation {{nowrap|1=''ax'' + ''by'' + ''c'' = 0,}} where ''a'', ''b'' and ''c'' are [[real number|real]] constants with ''a'' and ''b'' not both zero, the distance from the line to a point (''x''<sub>0</sub>,''y''<sub>0</sub>) is<ref>{{harvnb|Larson|Hostetler|2007|loc=p. 452}}</ref>
:<math>\operatorname{distance}(ax+by+c=0, (x_0, y_0)) = \frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}. </math>
 
If the point on this line which is closest to (''x''<sub>0</sub>,''y''<sub>0</sub>) has coordinates (''x''<sub>1</sub>,''y''<sub>1</sub>), then:<ref>{{harvnb|Larson|Hostetler|2007|loc=p. 522}}</ref>
:<math>x_1 = \frac{b(bx_0 - ay_0)-ac}{a^2 + b^2} \text{ and } y_1 = \frac{a(-bx_0 + ay_0) - bc}{a^2+b^2}.</math>
 
==Vector formulation==
[[File:Distance from a point to a line.svg|thumb|300px|Illustration of the vector formulation.]]
 
Write the line in [[Euclidean vector|vector]] form:
 
: <math> \mathbf{x} = \mathbf{a} + t\mathbf{n}</math>
 
where {{math|'''n'''}} is a [[unit vector]] in the direction of the line, {{math|'''a'''}} is a point on the line and ''t'' is a scalar. That is, a point {{math|'''x'''}} on the line, is found by moving to a point {{math|'''a'''}} in space, then moving {{math|''t''}} units along the direction 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{a}-\mathbf{p}) - ((\mathbf{a}-\mathbf{p}) \cdot  \mathbf{n})\mathbf{n} \|. </math>
 
This formula is constructed geometrically as follows: <math>\mathbf{a}-\mathbf{p}</math> is a vector from {{math|'''p'''}} to the point {{math|'''a'''}} on the line. Then <math>(\mathbf{a} - \mathbf{p}) \cdot  \mathbf{n}</math> is the projected length onto the line and so
:<math>((\mathbf{a} - \mathbf{p}) \cdot  \mathbf{n})\mathbf{n}</math>
is a vector that is the projection of <math>\mathbf{a}-\mathbf{p}</math> onto the line. Thus
:<math>(\mathbf{a}-\mathbf{p}) - ((\mathbf{a}-\mathbf{p}) \cdot  \mathbf{n})\mathbf{n}</math>
is the component of <math>\mathbf{a}-\mathbf{p}</math> perpendicular to the line. The distance from the point to the line is then just the [[norm (mathematics)|norm]] of that vector.<ref>{{cite web|last=Sunday|first=Dan|title=Lines and Distance of a Point to a Line|url=http://geomalgorithms.com/a02-_lines.html|publisher=softSurfer|accessdate=6 December 2013}}</ref> This more general formula can be used in dimensions other than two.
 
==Another vector projection proof==
[[File:vectorpoint-to-line.svg|250px|left|Diagram for vector projection proof]]
 
Let ''P'' be the point with coordinates (''x''<sub>0</sub>, ''y''<sub>0</sub>) and let the given line have equation ''ax'' + ''by'' + ''c'' = 0. Also, let ''Q'' = (''x''<sub>1</sub>, ''y''<sub>1</sub>) 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 <math>\overrightarrow{QP}</math> on '''n'''. The length of this projection is given by:
:<math>d = \frac{|\overrightarrow{QP} \cdot \mathbf{n}|}{\| \mathbf{n}\|}.</math>
Now,
:<math> \overrightarrow{QP} = (x_0 - x_1, y_0 - y_1),</math> so <math> \overrightarrow{QP} \cdot \mathbf{n} = a(x_0 - x_1) + b(y_0 - y_1)</math> and <math> \| \mathbf{n} \| = \sqrt{a^2 + b^2},</math>
thus
:<math> d = \frac{|a(x_0 - x_1) + b(y_0 - y_1)|}{\sqrt{a^2 + b^2}}.</math>
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>
 
==Two special cases==
In proofs that involve slopes of lines (as the ones below will), the main argument usually fails if the line in question is either vertical or horizontal. However, the result remains valid in these cases. In the general equation of a line, ''ax'' + ''by'' + ''c'' = 0, ''a'' and ''b'' can not both be zero. If ''a'' = 0, the line is horizontal and has equation ''y'' = -''c''/''b''. The distance from (''x''<sub>0</sub>, ''y''<sub>0</sub>) to this line is measured along a vertical line segment of length |''y''<sub>0</sub> - (-''c''/''b'')| = |''by''<sub>0</sub> + ''c''| / |''b''| in accordance with the formula. Similarly, for vertical lines (''b'' = 0) the distance between the same point and the line is |''ax''<sub>0</sub> + ''c''| / |''a''|, as measured along a horizontal line segment.
 
==An algebraic proof==
This proof is only valid 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 (''x''<sub>0</sub>, ''y''<sub>0</sub>). 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,
<math>a(y_0 -n) - b(x_0 - m) = 0,</math>
and by squaring this equation we obtain:
:<math>a^2(y_0 - n)^2 + b^2(x_0 - m)^2 = 2ab(y_0 - n)(x_0 - m).</math>
 
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>
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 (''m'', ''n'') is on ''ax'' + ''by'' + ''c'' = 0.
Thus,
:<math>(a^2 + b^2)((x_0 - m)^2 + (y_0 - n)^2) = (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==
[[File:point-to-line2.svg|thumb|Diagram for geometric proof]]
 
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''<sub>0</sub>, ''y''<sub>0</sub>) 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 -''B''/''A''.
 
∆''SRP'' and ∆''TVU'' are similar triangles since they are both right triangles and ∠''PSR'' ≅ ∠''VUT'' 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''<sub>0</sub>,''m'') then |''PS''| = |''y''<sub>0</sub> - ''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,
:<math>m = \frac{-Ax_0 - C}{B},</math>
and finally obtain:<ref>{{harvnb|Ballantine|Jerbert|1952}}</ref>
:<math> |\overline{PR}| = \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 is not vertical or horizontal.
 
The point P is given with coordinates (<math>x_0, y_0</math>).
The equation of a line is given by <math>y=mx+k</math>. The equation of the normal of that line which passes through the point P is given <math>y=\frac{x_0-x}{m}+y_0</math>.
 
The point at which these two lines intersect is the closest point on the original line to the point P. Hence:
:<math>mx+k=\frac{x_0-x}{m}+y_0.</math>
We can solve this equation for ''x'',
:<math>x=\frac{x_0+my_0-mk}{m^2+1}.</math>
The y coordinate of the point of intersection can be found by substituting this value of ''x'' into the equation of the original line,
:<math>y=m\frac{(x_0+my_0-mk)}{m^2+1}+k.</math>
 
Using the equation for finding the distance between 2 points, <math>d=\sqrt{(X_2-X_1)^2+(Y_2-Y_1)^2}</math>, we can deduce that the formula to find the shortest distance between a line and a point is the following:
 
:<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 }.</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>{{harvnb|Larson|Hosttler|2007|loc=p. 522}}</ref>
 
== See also ==
* [[Line-line intersection]]
 
==Notes==
{{reflist}}
 
==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|pages=242-243}}
 
* {{citation|first1=Ron|last1=Larson|first2=Robert|last2=Hostetler|title=Precalculus: A Concise Course|year=2007|publisher=Houghton Mifflin Co.|isbn=0-618-62719-7}}
 
==Further reading==
{{citation|title=Encyclopedia of Distances|first1=Michel Marie|last1=Deza|author1-link=Michel Deza|first2=Elena|last2=Deza|edition=2nd|publisher=Springer|year=2013|isbn=9783642309588|page=86|url=http://books.google.co.uk/books?id=QxX2CX5OVMsC&pg=PA86}}
 
==External links==
* {{cite web|last=Garner|first=W|title=Shortest Distance from a Point to a Line|url=http://math.ucsd.edu/~wgarner/math4c/derivations/distance/distptline.htm|publisher=University of California San Diego|accessdate=6 December 2013}}
 
 
 
[[Category:Euclidean geometry]]
[[Category:Vectors]]

Latest revision as of 20:42, 7 October 2014

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