Distance formula: Difference between revisions

Revision as of 19:24, 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.
     a         b

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 16: / Line 16: @@
 t will be the distance from the point <math>(x_1,y_1)</math> along the perpendicular line to <math>(x,y)</math>.
-So
+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>
-     <cmath>x = x_1 + a * t/\sqrt(a^2+b^2)</cmath>
-and
-     <cmath>y = y_1 + b {times} t/\sqrt(a^2+b^2)</cmath>
 This meets the given line <math>ax+by+c = 0</math> where:
-<cmath>a(x_1 + a * t/sqrt(a^2+b^2)) + b(y1 + b* t/sqrt(a^2+b^2)) + c = 0</cmath>
+<cmath>a(x_1 + a * t/\sqrt(a^2+b^2)) + b(y1 + b* t/\sqrt(a^2+b^2)) + c = 0</cmath>
-<cmath>ax_1 + by_1 + c + t(a^2+b^2)/sqrt(a^2+b^2) + c = 0</cmath>
+<cmath>ax_1 + by_1 + c + t(a^2+b^2)/\sqrt(a^2+b^2) + c = 0</cmath>
-<cmath>ax_1 + by_1 + c + t * sqrt(a^2+b^2) = 0</cmath>
+<cmath>ax_1 + by_1 + c + t * \sqrt(a^2+b^2) = 0</cmath>
 so
-<math> t * sqrt(a^2+b^2) = -(ax_1+by_1+c)</math>
+<cmath> t * sqrt(a^2+b^2) = -(ax_1+by_1+c)</cmath>
-<math>t = -(ax_1+by_1+c)/\sqrt(a^2+b^2)</math>
+<cmath>t = -(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
 ax+by+c = 0 is:
 <cmath>|t| = (ax_1 + by_1 + c)/\sqrt(a^2+b^2)</cmath>

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Distance formula: Difference between revisions

Revision as of 19:24, 7 November 2013