A) Prove that every point, P, on the perpendicular bisector of segment AB is the same euclidean distance from A as from B. [br][br]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). [br][br]Using the Euclidean distance formula, we see that: [br][br]1) dist(A, P) = [math]\sqrt[]{\left(a-0\right)^2+\left(0-p\right)^2}\Longrightarrow\sqrt[]{a^2+\left(p\right)^2}[/math].[br][br]2) dist(B, P) = [math]\sqrt[]{\left(b-0\right)^2+\left(0-p\right)^2}\Longrightarrow\sqrt[]{b^2+\left(p\right)^2}[/math].[br][br]Note that since AD = DB, we know that [math]a^2=b^2[/math]. [br][br]Therefore, we know that dist(A, P) = dist(B, P) for any point P such that P is on the perpendicular bisector of AB.[br][br][br][br]B) Using the taxicab metric, it will hold true for line segments for which the slope, m, is either:[br][br][list][*]m = 0[/*][*]m = 1[/*][*]m is undefined[/*][/list][br]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.