Difference between revisions of "Distance formula"

Revision as of 19:16, 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:

So the perpendicular line through (x1,y1) 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 {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)

@@ Line 7: / Line 7: @@
 the distance
 between the line <math>ax+by+c = 0</math> and point <math>(x_1,y_1)</math> is
+<cmath>|ax_1+by_1+c|/\sqrt(a^2+b^2)</cmath>
-     <math>|ax_1+by_1+c|/\sqrt(a^2+b^2)</math>
               Proof:

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Difference between revisions of "Distance formula"

Revision as of 19:16, 7 November 2013