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Distance Between a Line and a Point Using Vectors


Date: 09/25/1999 at 17:08:46
From: Ian Bui
Subject: Distance between line and point using vectors

How do you use scalar vector projections to prove that the distance 
between the line ax+by+c = 0 and point (x1,y1) is

     |ax1+by1+c|/sqrt(a^2+b^2)

I tried choosing an arbitrary point on the line, but with each step, 
things got more complex.


Date: 09/25/1999 at 19:46:50
From: Doctor Anthony
Subject: Re: Distance between line and point using vectors

The equation ax + by + c = 0 can be written:

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

So the gradient of line is -(a/b). This means that the cosine of the 
angle between the line and the x axis is -b/sqrt(a^2+b^2) and the 
cosine of the angle between the line and the y axis is 
a/sqrt(a^2+b^2).

So the direction ratios of the line are (-b,a).

If we have a line perpendicular to this, its direction ratios will be 
(a,b) since (-ab) + (ab) = 0.

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

     x-x1   y-y1
     ---- = ---- = t/sqrt(a^2+b^2)     where t is a parameter.
      a      b

t will be the distance from the point (x1,y1) along the perpendicular.
 
So

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

and

     y = y1 + b.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)

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Linear Algebra

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