Distance formula: Difference between revisions

Revision as of 11:35, 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 $P_1 = (x_1, y_1)$ 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

            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

and

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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Distance formula: Difference between revisions

Revision as of 11:35, 3 April 2011

@@ Line 3: / Line 3: @@
 {{stub}}
+Shortest distance from a point to a line:
+the distance
+between the line <math>ax+by+c = 0</math> and point <math>(x_1,y_1)</math> is
+     <math>|ax_1+by_1+c|/\sqrt(a^2+b^2)</math>
+             Proof:
+The equation <math>ax + by + c = 0</math> can be written:
+     <math>y = -(a/b)x - (c/a)</math>
+So the perpendicular line through (x1,y1) is:
+    <math>x-x_1</math>   <math>y-y_1</math>
+     ----   = ---- =  <math>t/\sqrt(a^2+b^2)</math>     where t is a parameter.
+      a         b
+t will be the distance from the point <math>(x_1,y_1)</math> along the perpendicular line to (x,y).
+So
+     <math>x = x_1 + a \dot t/\sqrt(a^2+b^2)</math>
+and
+     <math>y = y_1 + b \dot t/\sqrt(a^2+b^2)</math>
+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)