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 Exercise [05.01] 
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Joined: 12 Mar 2008, 10:57
Posts: 69
Location: India
Post Exercise [05.01]
The various possibilities are as follows:
Parallelogram Law
1.If w+z=0,then w,0,z are collinear.In this case the parallelogram degenerates into a straight line.
2.If one of the points(w or z)is zero,then we get a straight line joining 0 and the non zero point.
3.If both w and z are zero then we represent just the origin in the plane.That is, the parallelogram has degenerated to a single point.
Similar triangle Law
1.If w=1,then we have two lines;one joining w and the origin and the other joining z and the origin.(z is not zero)
2.If w=0 then we have a straight line through 0 and z and another a part of the real axis.
3.If z=0 or z=1,then we have a single triangle.
4.If |z|=1 then the triangles are congruent.


13 Jul 2008, 13:45
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Joined: 15 May 2009, 20:50
Posts: 1
Post Re: Exercise [05.01]
For the parallelogram, w, 0, z are collinear in any case where z = kw (for real k). Then the assertion is that the points 0, w, z, w+z are on a straight line so that the distance from w to w+z is equal to the distance from 0 to z.
For the triangle, 0, 1, w are collinear in any case where w is real. The assertion is that the two sets of points 0, 1, w and 0, z, wz are both on straight lines so that the ratio of the distances between any two pairs of points in the first set is the same as between any two pairs of points in the second set.


15 May 2009, 22:01
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