# Difference between revisions of "Distance formula"

m |
|||

(16 intermediate revisions by 5 users not shown) | |||

Line 1: | Line 1: | ||

− | 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>. |

+ | |||

+ | |||

+ | ==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 | ||

+ | <cmath>\dfrac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}</cmath> | ||

+ | |||

+ | ===Proof=== | ||

+ | The equation <math>ax + by + c = 0</math> can be written as <math>y = -\dfrac{a}{b}x - \dfrac{c}{a}</math> | ||

+ | Thus, the perpendicular line through <math>(x_1,y_1)</math> is: | ||

+ | <cmath>\dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=\dfrac{t}{\sqrt{a^2+b^2}}</cmath> | ||

+ | 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>. | ||

+ | 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: | ||

+ | <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>\implies ax_1 + by_1 + c + \dfrac{t(a^2+b^2)}{\sqrt{a^2+b^2}} + c = 0</cmath> | ||

+ | <cmath>\implies ax_1 + by_1 + c + t \cdot \sqrt{a^2+b^2} = 0</cmath> | ||

+ | |||

+ | , so: | ||

+ | <cmath> t \cdot \sqrt{a^2+b^2} = -(ax_1+by_1+c)</cmath> | ||

+ | <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 <math>ax+by+c = 0</math> is: | ||

+ | |||

+ | <cmath>|t| = \dfrac{|ax_1 + by_1 + c|}{\sqrt{a^2+b^2}}</cmath> | ||

{{stub}} | {{stub}} |

## Latest 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

### Proof

The equation can be written as Thus, the perpendicular line through is: where is the parameter.

will be the distance from the point along the perpendicular line to . So and

This meets the given line , where:

, so:

Therefore the perpendicular distance from to the line is:

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