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>. |
− | {{ | + | ==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}{b}</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> | |
− | <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> | |
− | |||
− | <cmath> | ||
− | |||
− | |||
− | |||
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− | + | {{stub}} | |
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Latest revision as of 21:37, 1 August 2024
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.