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Difference between revisions of "Distance formula"

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(Fixed LaTeX, minor grammar mistakes, and cleaned up formatting.)
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--Shortest distance from a point to a line--
 
--Shortest distance from a point to a line--
 
the distance between the line <math>ax+by+c = 0</math> and point <math>(x_1,y_1)</math> is
 
the distance between the line <math>ax+by+c = 0</math> and point <math>(x_1,y_1)</math> is
−
<cmath>\dfrac{\|ax_1+by_1+c\|}{\sqrt{a^2+b^2}}</cmath>
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<cmath>\dfrac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}</cmath>
  
 
---Proof---
 
---Proof---
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The equation <math>ax + by + c = 0</math> can be written as <math>y = -(a/b)x - (c/a)</math>
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The equation <math>ax + by + c = 0</math> can be written as <math>y = -\dfrac{a}{b}x - \dfrac{c}{a}</math>
−
So the perpendicular line through <math>(x_1,y_1)</math> is:
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Thus, the perpendicular line through <math>(x_1,y_1)</math> is:
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<cmath>(x-x_1)/a=(y-y_1)/b=t/\sqrt(a^2+b^2)</cmath>
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<cmath>\dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=\dfrac{t}{\sqrt{a^2+b^2}}</cmath>
−
where <math>t</math> is a parameter.
+
where <math>t</math> is the parameter.
  
 
<math>t</math> will be the distance from the point <math>(x_1,y_1)</math> along the perpendicular line to <math>(x,y)</math>.
 
<math>t</math> will be the distance from the point <math>(x_1,y_1)</math> along the perpendicular line to <math>(x,y)</math>.
−
So <cmath>x = x_1 + a * t/\sqrt(a^2+b^2)</cmath> and <cmath>y = y_1 + b* t/\sqrt(a^2+b^2)</cmath>
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So <cmath>x = x_1 + a \cdot \dfrac{t}{\sqrt{a^2+b^2}}</cmath> and <cmath>y = y_1 + b \cdot \dfrac{t}{\sqrt{a^2+b^2}}</cmath>
  
−
This meets the given line <math>ax+by+c = 0</math> where:
+
This meets the given line <math>ax+by+c = 0</math>, where:
−
<cmath>a(x_1 + a * t/\sqrt(a^2+b^2)) + b(y1 + b* t/\sqrt(a^2+b^2)) + c = 0</cmath>
+
<cmath>a\left(x_1 + a \cdot \dfrac{t}{\sqrt{a^2+b^2}}\right) + b\left(y_1 + b \cdot  \dfrac{t}{\sqrt{a^2+b^2}}\right) + c = 0</cmath>
−
<cmath>ax_1 + by_1 + c + t(a^2+b^2)/\sqrt(a^2+b^2) + c = 0</cmath>
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<cmath>\implies ax_1 + by_1 + c + \dfrac{t(a^2+b^2)}{\sqrt{a^2+b^2}} + c = 0</cmath>
−
<cmath>ax_1 + by_1 + c + t * \sqrt(a^2+b^2) = 0</cmath>
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<cmath>\implies ax_1 + by_1 + c + t \cdot \sqrt{a^2+b^2} = 0</cmath>
  
−
so
+
, so:
−
<cmath> t * sqrt(a^2+b^2) = -(ax_1+by_1+c)</cmath>
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<cmath> t \cdot \sqrt{a^2+b^2} = -(ax_1+by_1+c)</cmath>
−
<cmath>t = -(ax_1+by_1+c)/\sqrt(a^2+b^2)</cmath>
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<cmath>\implies t = \dfrac{-(ax_1+by_1+c)}{\sqrt{a^2+b^2}}</cmath>
  
−
Therefore the perpendicular distance from <math>(x_1,y_1)</math> to the line  
+
Therefore the perpendicular distance from <math>(x_1,y_1)</math> to the line <math>ax+by+c = 0</math> is:
−
ax+by+c = 0 is:
+
 
−
<cmath>|t| = (ax_1 + by_1 + c)/\sqrt(a^2+b^2)</cmath>
+
<cmath>|t| = \dfrac{|ax_1 + by_1 + c|}{\sqrt(a^2+b^2)}</cmath>

Revision as of 13:34, 30 July 2014

The distance formula is a direct application of the Pythagorean Theorem in the setting of a Cartesian coordinate system. In the two-dimensional case, it says that the distance between two points $P_1 = (x_1, y_1)$ and $P_2 = (x_2, y_2)$ is given by $d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$. In the $n$-dimensional case, the distance between $(a_1,a_2,...,a_n)$ and $(b_1,b_2,...,b_n)$ is $\sqrt{(a_1-b_1)^2+(a_2-b_2)^2+\cdots+(a_n-b_n)^2}$


This article is a stub. Help us out by expanding it.

--Shortest distance from a point to a line-- the distance between the line $ax+by+c = 0$ and point $(x_1,y_1)$ is \[\dfrac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}\]

---Proof--- The equation $ax + by + c = 0$ can be written as $y = -\dfrac{a}{b}x - \dfrac{c}{a}$ Thus, the perpendicular line through $(x_1,y_1)$ is: \[\dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=\dfrac{t}{\sqrt{a^2+b^2}}\] where $t$ is the parameter.

$t$ will be the distance from the point $(x_1,y_1)$ along the perpendicular line to $(x,y)$. So \[x = x_1 + a \cdot \dfrac{t}{\sqrt{a^2+b^2}}\] and \[y = y_1 + b \cdot \dfrac{t}{\sqrt{a^2+b^2}}\]

This meets the given line $ax+by+c = 0$, where: \[a\left(x_1 + a \cdot \dfrac{t}{\sqrt{a^2+b^2}}\right) + b\left(y_1 + b \cdot \dfrac{t}{\sqrt{a^2+b^2}}\right) + c = 0\] \[\implies ax_1 + by_1 + c + \dfrac{t(a^2+b^2)}{\sqrt{a^2+b^2}} + c = 0\] \[\implies ax_1 + by_1 + c + t \cdot \sqrt{a^2+b^2} = 0\]

, so: \[t \cdot \sqrt{a^2+b^2} = -(ax_1+by_1+c)\] \[\implies t = \dfrac{-(ax_1+by_1+c)}{\sqrt{a^2+b^2}}\]

Therefore the perpendicular distance from $(x_1,y_1)$ to the line $ax+by+c = 0$ is:

\[|t| = \dfrac{|ax_1 + by_1 + c|}{\sqrt(a^2+b^2)}\]

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