Distance formula

Revision as of 11:35, 3 April 2011 by Luie1168e (talk | contribs)

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:

    $y = -(a/b)x - (c/a)$

So the perpendicular line through (x1,y1) is:

   $x-x_1$   $y-y_1$
    ----   = ---- =  $t/\sqrt(a^2+b^2)$     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

    $x = x_1 + a \dot t/\sqrt(a^2+b^2)$

and

    $y = y_1 + b \dot t/\sqrt(a^2+b^2)$

This meets the given line ax+by+c = 0 where:

    a(x1+a.t/sqrt(a^2+b^2)) + b(y1+b.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.sqrt(a^2+b^2) = 0

so

    t.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)
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