Difference between revisions of "Distance formula"
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So | So | ||
− | + | x = x_1 + a {times} t/\sqrt(a^2+b^2) | |
and | and | ||
− | + | y = y_1 + b {times} t/\sqrt(a^2+b^2) | |
This meets the given line ax+by+c = 0 where: | This meets the given line ax+by+c = 0 where: |
Revision as of 12:41, 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:
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---- = ---- =
where t is a parameter. a b
t will be the distance from the point along the perpendicular line to (x,y).
So
x = x_1 + a {times} t/\sqrt(a^2+b^2)
and
y = y_1 + b {times} t/\sqrt(a^2+b^2)
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)