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/
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