Difference between revisions of "Distance formula"
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| − | 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 [[point]]s <math>P_1 = (x_1, y_1)</math> and <math>P_2 = (x_2, y_2)</math> is given by <math>d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}</math>. In the <math>n</math>-dimensional case, the distance between <math>(a_1,a_2,...,a_n)</math> and <math>(b_1,b_2,...,b_n)</math> is <math>\sqrt{(a_1-b_1)^2+(a_2-b_2)^2+\cdots+(a_n-b_n)^2}</math> | + | 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 [[point]]s <math>P_1 = (x_1, y_1)</math> and <math>P_2 = (x_2, y_2)</math> is given by <math>d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}</math>. In the <math>n</math>-dimensional case, the distance between <math>(a_1,a_2,...,a_n)</math> and <math>(b_1,b_2,...,b_n)</math> is <math>\sqrt{(a_1-b_1)^2+(a_2-b_2)^2+\cdots+(a_n-b_n)^2}</math>. |
Revision as of 17:34, 8 September 2018
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 
and 
is given by 
. In the 
-dimensional case, the distance between 
and 
is 
.
Shortest distance from a point to a line
the distance between the line 
and point 
is
![\[\dfrac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}\]](http://latex.artofproblemsolving.com/2/d/5/2d5b5f75564ea734da6afc1f8c0d83378c266fd3.png)
Proof
The equation 
can be written as 
Thus, the perpendicular line through is:
![\[\dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=\dfrac{t}{\sqrt{a^2+b^2}}\]](http://latex.artofproblemsolving.com/5/6/3/5638887bdbc43cbac92922432d41dc3232d947fe.png)
where 
is the parameter.
will be the distance from the point along the perpendicular line to 
.
So ![\[x = x_1 + a \cdot \dfrac{t}{\sqrt{a^2+b^2}}\]](http://latex.artofproblemsolving.com/2/c/c/2ccf383b37c445fd12553d8bfaccd183f5906ab8.png)
and ![\[y = y_1 + b \cdot \dfrac{t}{\sqrt{a^2+b^2}}\]](http://latex.artofproblemsolving.com/4/b/8/4b8f9c3c7b7000415b46e5991cb3e44b18978e97.png)
This meets the given line , where:
![\[a\left(x_1 + a \cdot \dfrac{t}{\sqrt{a^2+b^2}}\right) + b\left(y_1 + b \cdot \dfrac{t}{\sqrt{a^2+b^2}}\right) + c = 0\]](http://latex.artofproblemsolving.com/f/d/6/fd6ba4df459edd33f5fdf86d7825315d84f04f3f.png)
![\[\implies ax_1 + by_1 + c + \dfrac{t(a^2+b^2)}{\sqrt{a^2+b^2}} + c = 0\]](http://latex.artofproblemsolving.com/8/0/7/8077b2738403e402f33929e8c52c36ed245e5fb8.png)
![\[\implies ax_1 + by_1 + c + t \cdot \sqrt{a^2+b^2} = 0\]](http://latex.artofproblemsolving.com/a/c/3/ac3dd3bfe87ee63f416db42389c0024240e91773.png)
, so:
![\[t \cdot \sqrt{a^2+b^2} = -(ax_1+by_1+c)\]](http://latex.artofproblemsolving.com/0/b/5/0b5ce92e2fad16bfb9c26f893a47b2e0bdadfb3f.png)
![\[\implies t = \dfrac{-(ax_1+by_1+c)}{\sqrt{a^2+b^2}}\]](http://latex.artofproblemsolving.com/a/5/4/a546ad78ce2319796a749b795b8251036ecac510.png)
Therefore the perpendicular distance from to the line is:
![\[|t| = \dfrac{|ax_1 + by_1 + c|}{\sqrt{a^2+b^2}}\]](http://latex.artofproblemsolving.com/3/3/e/33e8d2c104ced7933483b4788d5d6be07ef368ad.png)
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