Strange Points of LocusDate: 6/6/96 at 8:24:33 From: Anonymous Subject: Strange Points of Locus Given two fixed points, A and B, on a plane, if P is a moving point such that PA and PB are perpendicular and the locus of P is a circle, should we exclude points A and B? Date: 6/14/96 at 0:54:4 From: Doctor Brian Subject: Re: Strange Points of Locus Yes, we should. P can not be the same as A. If it were, then we'd have the statement that AA and AB are perpendicular. 1. AA would have to be a line segment, but it's not....line segments have positive length, not a length of zero. Now, you might want to make the argument that this is a special case, that AA is a limiting case of the length of all perpendicular segments to AB through A. The problem with that argument is that an identical argument can be made for all segments forming an 89 degree angle with AB. Or any angle you want! You can't really put a specific degree measure on the angle. 2. Another argument for this comes from the definition of cosine in vector spaces. If you've ever worked with vectors, you may have run across the definition of the cosine of the angle between two vectors: cos theta = (the "dot product" of the vectors)/(the product of the lengths of the vectors) If one of the vectors here is AA, then its length is zero, and so the denominator of the expression above would be zero. So the formula basically doesn't allow us to come up with an angle. Now, you can repeat the whole thing to show why B must be eliminated from the locus as well. -Doctor Brian, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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