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Distance from a Line to a Point

Date: 07/10/2003 at 23:09:36
From: Jack
Subject: Find the direct distance from the line to the point

Find the direct distance from the line to the point:

   4x-3y-5 = 0 , (7,6)


Date: 07/17/2003 at 09:53:46
From: Doctor Korsak
Subject: Re: Find the direct distance from the line to the point

Hi Jack,
  
I am sure that a general formula for the answer to your question can 
be found in current school and college textbooks, but you can work 
out the solution yourself like this:

The closest distance to a line is along another line intersecting the
first at a 90-degree angle; the term for that is an orthogonal line. 
The slope m of an orthogonal line is -1/m, where m is the slope of
the given line. (This works only when m is not 0.)    

For the example in your question, the slope of the line 

    4x - 3y - 5 = 0

is 4/3, since you can rewrite the equation as 

    y = (4x - 5)/3 

      = (4/3)x - 5/3.

The orthogonal line you want is therefore 

    y = -(3/4)x + b,

where b is to be determined by the line having to pass through your 
starting point (7,6), which means that

    6 = -(3/4)7 + b, 
    b = 6 + 21/4 = 45/4, 

so then the orthogonal line is

    y = -(3/4)x + 45/4 

or

    3x + 4y = 45.

Now you can compute the shortest distance between the given point and 
line.  First, you can find where the two lines intersect by solving 
the pair of equations of the two lines for x and y.  Then you can 
calculate the distance to the point (7,6) and you will have your 
answer.

There is a more direct way to calculate the distance. You can start at 
the point (7,6) and go out along a line in the orthogonal direction 
toward (or away from) the given line by moving a point (x,y)

    x = 7 + 4s/5, y = 6 - 3s/5

where s is a parameter that moves you along the orthogonal line.

Why did I divide by 5? Well, we want to move along the orthogonal line 
by a distance s and then solve for s when we hit the given line, so we 
prefer to move by multiplying a unit length segment by a factor s. It 
so happens that 3 and 4 are the lengths of the sides of a right 
triangle with a long side of length 5, and 3^2 + 4^2 = 5^2, so the 
point (3/5, -4/5) forms a unit length line segment from the point 
(0,0) in the orthogonal direction we need to travel.

Now we calculate s by requiring our point to lie on the given line:

   4x - 3y - 5 = 0 

substituting our expressions for x and y in terms of s gives

  4(7 + 4s/5) -3(6 - 3s/5) - 5 = 0

and now we can solve for s (the distance) directly. 

I hope this response is of value to you.  Feel free to write back if
you require further assistance. 

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

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