Difference between revisions of "Distance formula"
Line 25: | Line 25: | ||
So | So | ||
− | <math>x = x_1 + a | + | <math>x = x_1 + a times t/\sqrt(a^2+b^2)</math> |
and | and | ||
− | <math>y = y_1 + b | + | <math>y = y_1 + b times t/\sqrt(a^2+b^2)</math> |
This meets the given line ax+by+c = 0 where: | This meets the given line ax+by+c = 0 where: | ||
− | a(x1+a | + | 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(a^2+b^2)/sqrt(a^2+b^2) + c = 0 | ||
− | ax1 + by1 + c + t | + | ax1 + by1 + c + t times sqrt(a^2+b^2) = 0 |
so | so | ||
− | t | + | t times sqrt(a^2+b^2) = -(ax1+by1+c) |
t = -(ax1+by1+c)/sqrt(a^2+b^2) | t = -(ax1+by1+c)/sqrt(a^2+b^2) |
Revision as of 11:38, 3 April 2011
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:
So the perpendicular line through (x1,y1) is:
---- = ---- = where t is a parameter. a b
t will be the distance from the point along the perpendicular line to (x,y).
So
and
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)