Difference between revisions of "Distance formula"

Latest revision as of 20:37, 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 $P_1 = (x_1, y_1)$ 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 $\sqrt{(a_1-b_1)^2+(a_2-b_2)^2+\cdots+(a_n-b_n)^2}$ .

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 $y = -\dfrac{a}{b}x - \dfrac{c}{b}$ 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:

$|t| = \dfrac{|ax_1 + by_1 + c|}{\sqrt{a^2+b^2}}$

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

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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>.
-{{stub}}
+==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>
-Shortest distance from a point to a line:
+===Proof===
-the distance
+The equation <math>ax + by + c = 0</math> can be written as <math>y = -\dfrac{a}{b}x - \dfrac{c}{b}</math>
-between the line <math>ax+by+c = 0</math> and point <math>(x_1,y_1)</math> is
+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>|ax_1+by_1+c|/\sqrt(a^2+b^2)</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 \cdot \dfrac{t}{\sqrt{a^2+b^2}}</cmath> and <cmath>y = y_1 + b \cdot \dfrac{t}{\sqrt{a^2+b^2}}</cmath>
-             Proof:
+This meets the given line <math>ax+by+c = 0</math>, where:
-The equation <math>ax + by + c = 0</math> can be written:
+<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>
-     <math>y = -(a/b)x - (c/a)</math>
+, 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>
-So the perpendicular line through (x1,y1) is:
+Therefore the perpendicular distance from <math>(x_1,y_1)</math> to the line <math>ax+by+c = 0</math> is:
-    <math>x-x_1</math>   <math>y-y_1</math>
+<cmath>|t| = \dfrac{|ax_1 + by_1 + c|}{\sqrt{a^2+b^2}}</cmath>
-     ----   = ---- =  <math>t/\sqrt(a^2+b^2)</math>     where t is a parameter.
-      a         b
-t will be the distance from the point <math>(x_1,y_1)</math> along the perpendicular line to (x,y).
-So
-     <math>x = x_1 + a {times} t/\sqrt(a^2+b^2)</math>
+{{stub}}
-and
-     <math>y = y_1 + b {times} t/\sqrt(a^2+b^2)</math>
-This meets the given line ax+by+c = 0 where:
-     a(x1 + a {times} t/sqrt(a^2+b^2)) + b(y1 + b {times} t/sqrt(a^2+b^2)) + c = 0
-              ax1 + by1 + c + t(a^2+b^2)/sqrt(a^2+b^2) + c = 0
-                           ax1 + by1 + c + t {times} sqrt(a^2+b^2) = 0
-so
-     t {times} sqrt(a^2+b^2) = -(ax1+by1+c)
-                   t = -(ax1+by1+c)/sqrt(a^2+b^2)
-Therefore the perpendicular distance from (x1,y1) to the line
-ax+by+c = 0 is:
-            ax1 + by1 + c
-     |t| =  -------------
-            sqrt(a^2+b^2)

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Difference between revisions of "Distance formula"

Latest revision as of 20:37, 1 August 2024

Shortest distance from a point to a line

Proof