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Distance from a Point to a Line

Date: 08/20/99 at 11:35:31
From: Carolyn Gallagher
Subject: Deriving the formula for distance from a point to a line

I am a math teacher in Ontario, Canada looking for a straightforward 
way to derive the formula for the distance from a point to a line

     abs(Ax + By + C)/sqrt(A^2+B^2) = D

Sorry about the form, but you probably get the general idea.

This topic is in the new curriculum for tenth grade mathematics as 
part of the analytic geometry strand.

I know that it is easy enough to show specific cases given the point 
and the equation of the line, but how do I explain the equation to 
such a young audience?


Date: 08/20/99 at 14:15:47
From: Doctor Anthony
Subject: Re: Deriving the formula for distance from a point to a line

If the equation of the straight line is ax + by + c = 0, and (h,k) is 
the point from which we require the perpendicular distance, we can 
prove the result as follows.

Assume the line has positive slope and lies in the first quadrant in a 
convenient position.  Let (h,k) be a point in the first quadrant below 
the given line but with k greater than the y coordinate where the 
given line cuts the y axis. This diagram will avoid unnecessary 
complications in explaining the geometry.

Draw the perpendicular from P(h,k) to the line.  Let the foot of this 
perpendicular be point N. Draw a horizontal line from P to meet the 
given line at M. If theta is the angle of slope of the given line then 
the perpendicular PN is given by PM*sin(theta).

The x coordinate of M is found by putting y = k in the equation

     ax + by + c = 0

     ax + bk + c = 0   so   ax = -(bk+c)    x = -(bk+c)/a

Therefore PM =  h - [-(bk+c)/a]

             =  [ah + bk + c]/a

Now tan(theta) = -(a/b) from equation of line.

Then from Pythagoras      

       / |
      /  | a         hypotenuse = sqrt(a^2+b^2)
     /   |
    /    |           sin(theta) = a/sqrt(a^2+b^2)

and so   PM.sin(theta) = [ah+bk+c]/a * a/sqrt(a^2+b^2)

                    PN = [ak+bk+c]/sqrt(a^2+b^2)

So the perpendicular distance from (h,k) to the line ax + by + c = 0 
is given by

          ah + by + c

If this result is positive the point (h,k) is on the same side of the 
given line as the origin. If the answer is negative then (h,k) is on 
the opposite side of the line from the origin. This can be used as a 
check to see on which side of a line a given point lies.

 - Doctor Anthony, The Math Forum   
Associated Topics:
High School Euclidean/Plane Geometry
High School Geometry

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