Distance formula: Difference between revisions

Revision as of 19:25, 7 November 2013

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 $\sqrt{(a_1-b_1)^2+(a_2-b_2)^2+\cdots+(a_n-b_n)^2}$

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

---Proof--- The equation $ax + by + c = 0$ can be written as $y = -(a/b)x - (c/a)$ So the perpendicular line through $(x_1,y_1)$ is:

where $t$ is a parameter.

$t$ will be the distance from the point $(x_1,y_1)$ along the perpendicular line to $(x,y)$ . So $x = x_1 + a * t/\sqrt(a^2+b^2)$ and $y = y_1 + b* t/\sqrt(a^2+b^2)$

This meets the given line $ax+by+c = 0$ where: $a(x_1 + a * t/\sqrt(a^2+b^2)) + b(y1 + b* t/\sqrt(a^2+b^2)) + c = 0$ $ax_1 + by_1 + c + t(a^2+b^2)/\sqrt(a^2+b^2) + c = 0$ $ax_1 + by_1 + c + t * \sqrt(a^2+b^2) = 0$

so $t * sqrt(a^2+b^2) = -(ax_1+by_1+c)$ $t = -(ax_1+by_1+c)/\sqrt(a^2+b^2)$

Therefore the perpendicular distance from $(x_1,y_1)$ to the line ax+by+c = 0 is: $|t| = (ax_1 + by_1 + c)/\sqrt(a^2+b^2)$

@@ Line 11: / Line 11: @@
 The equation <math>ax + by + c = 0</math> can be written as <math>y = -(a/b)x - (c/a)</math>
 So the perpendicular line through <math>(x_1,y_1)</math> is:
-     <math>x-x_1</math>   <math>y-y_1</math>
+     <math>(x-x_1)/a=(y-y_1)/b=t/\sqrt(a^2+b^2)</math>
-     ----   = ---- =  <math>t/\sqrt(a^2+b^2)</math>     where t is a parameter.
+where <math>t</math> is a parameter.
-      a         b
-t 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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Distance formula: Difference between revisions

Revision as of 19:25, 7 November 2013