Deriving the Midpoint Formula

Date: 08/08/99 at 16:04:07
From: Jason
Subject: Midpoint formula (proof)

I'm wondering how the midpoint formula is derived from the distance 
formula. Is there a better way to prove the midpoint formula?

Sincerely,
Jason

Date: 08/09/99 at 04:08:48
From: Doctor Floor
Subject: Re: Midpoint formula (proof)

Hi Jason,

Thanks for your question.

Let us consider the points P1 = (x1,y1) and P2 = (x2,y2). For ease of 
use, we can rewrite P2 = (x2,y2) = (x1+2t,y1+2u). This second way of 
writing is easy, because we immediately see that, following the 
"midpoint formula," the midpoint should be M =(x1+t,y1+u). We can also 
immediately see that the slope of the line P1P2 is equal to the slope 
of the line P1M, so the three points are on one line.

The distance formula gives:

     d(P1,P2) = sqrt((2t)^2 + (2u)^2) 
              = sqrt( 4(t^2 + u^2) )
              = 2 sqrt(t^2 + u^2)

while

     d(P1,M) = sqrt(t^2 + u^2)

so, indeed the distance d(P1,M) = d(P1,P2)/2. As desired.

But I think it is much easier to use geometry for the proof. We can 
use the fact that if we perpendicularly project a segment onto an 
image segment on a given line, then the midpoint of the original 
projects to the midpoint of the image. When we apply this on segment 
P1P2, and perpendicularly project it to the x-axis and y-axis 
respectively, we immediately find the midpoint formula. See the 
following figure:

                           P2
                          /|
                         / |
                        /  |
                       /   |
                      /    |
                     M     |
                    /|     |
                   / |     |
                  /  |     |
                 /   |     |
                /    |     |
              P1     |     |
               |     |     |
               |     |     |
     ---------x1-----*-----x2--------- x-axis
                     |
                  (x1+x2)/2

If you need more help, just write us back.

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/