Difference between revisions of "Distance formula"
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The equation <math>ax + by + c = 0</math> can be written as <math>y = -(a/b)x - (c/a)</math> | 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: | So the perpendicular line through <math>(x_1,y_1)</math> is: | ||
− | <math>x-x_1 | + | <math>(x-x_1)/a=(y-y_1)/b=t/\sqrt(a^2+b^2)</math> |
− | + | where <math>t</math> is a parameter. | |
− | |||
− | 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> | 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> | ||
Revision as of 20:25, 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 and is given by . In the -dimensional case, the distance between and is
This article is a stub. Help us out by expanding it.
--Shortest distance from a point to a line-- the distance between the line and point is
---Proof--- The equation can be written as So the perpendicular line through is:
where is a 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 ax+by+c = 0 is: