Minimum Distance between a Line and a Point

Date: 07/07/2003 at 13:33:08
From: Magen
Subject: An Algebra 2 example math problem

Find the distance between point (2,2) and the line y=-2x-6.

I cannot figure out the coordinates that correspond to the line.

Date: 07/07/2003 at 17:28:07
From: Doctor Dotty
Subject: Re: An Algebra 2 example math problem

Hi Magen,

Thanks for the question.

Take the line 'y = 3x + 4' and the point (-5, 4) as an example:

                        |  / 3x + 4
  (-5, 4)               |,'
         x              /
                      ,'|
                     /  |
                   ,'   |
                  /     |
                ,'      |
               /        |
     '''''''',''''''''''|'''''''''''''
            /           |
          ,'            |
         /              |
       ,'               |
      /                 |
     '                  |

We want the shortest distance between the point and the line. This is 
measured along the line that is perpendicular (at 90-degrees) to the 
initial line. This is called the 'normal' and is shown here:

                        |  / y = 3x + 4
  (-5, 4)               |,'
         x_             /
           `.         ,'|
             `-.     /  |
                `._,'   |
                  /`.   |
                ,'   `-.|
               /        `._
     '''''''',''''''''''|''':'''''''''
            /           |    `-.
          ,'            |       `._normal
         /              |          `.
       ,'               |
      /                 |
     '                  |

The equation of a linear line can be written as y = mx + c, where m is 
the gradient (slope) and c is the y-intercept. Our original line was 
y = 3x + 4; therefore its gradient is 3.

To find the gradient of the normal to this, you use:

                                   -1
        gradient of normal = ----------------
                             gradient of line

Therefore the gradient of the normal is -1/3.

The other general equation of a line is:

           y - y1 = m(x - x1)

Where m is the gradient and two points it passes through are (x, y) 
and (x1, y1).

We know that the gradient of the normal is -1/3, and it passes through 
(-5, 4), so:

           y - y1 = m(x - x1)

                   -1
           y - 5 = --(x - -4)
                    3
         3y - 15 = -x - 4

              3y = -x + 11

              3y = 11 - x

Which is the equation of the normal.

We need to find the distance between (-5, 4) and the intersection of 
the two lines, marked 'A'.

                        |  / y = 3x + 4
  (-5, 4)               |,'
         x_             /
           `.         ,'|
             `-.     /  |
                `.A,'   |
                  /`.   |
                ,'   `-.|
               /        `._
     '''''''',''''''''''|''':'''''''''
            /           |    `-.
          ,'            |       `._3y = 11 - x
         /              |          `.
       ,'               |
      /                 |
     '                  |

You can find point A by solving the equations simultaneously. Can 
you see how you can use Pythagoras' theorem to find the distance?

Can you use this method to find the minimum distance between your 
line and your point?

Write back if I can be of any more help - on this or anything else.

- Doctor Dotty, The Math Forum
  http://mathforum.org/dr.math/