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Exercise 6.3.13

A) Prove that every point, P, on the perpendicular bisector of segment AB is the same euclidean distance from A as from B. Proof: Let AB be a line segment with A = (a, 0) and B = (b, 0). Let D be the midpoint of AB such that D = (0, 0). Note that the y-axis is the perpendicular bisector of AB by definition. Furthermore, note that AD = DB. Pick some point on the perpendicular bisector of AB, say P = (0, p). Using the Euclidean distance formula, we see that: 1) dist(A, P) = . 2) dist(B, P) = . Note that since AD = DB, we know that . Therefore, we know that dist(A, P) = dist(B, P) for any point P such that P is on the perpendicular bisector of AB. B) Using the taxicab metric, it will hold true for line segments for which the slope, m, is either:
  • m = 0
  • m = 1
  • m is undefined
C) In the Taxi-cab metric, it is obvious that the perpendicular bisectors of a line segment AB will be the set of points equidistant from AB if m is one of the slopes listed above. However, if the slope of AB differs from one of those slopes, you know that m must either be greater than 1 or less than 1. See picture below. The dashed line segments represent the points equidistant from A and B.

AB with m < 1

AB with m > 1

AB with m = 1