Taxicab

Recall that one model for our axiom system to this point is the usual Cartesian , with

Points: {(x,y)| x0ú, y0ú} Lines: {(x,y)| ax + by + c = 0, x, y 0ú, a, b, c 0ú, not both a = 0 and b = 0}

Distance between P(x1, y1) and Q(x2, y2) is

This model satisfies: • The incidence axioms and axioms • Plane Separation • Angle measure, angle addition, and linear pair axioms • The Protractor Postulate (how?) • The Ruler Postulate. We’ll take some time to show this. How do we match points with their “ruler coordinates?” We relate them to the (x,y) coordinates of points. Given a y=mx+b, we can write two ponts

P(x1, y1) and Q(x2, y2) on the line as (x1, mx1+b) and (x2, mx2+b). Then

Since PQ is supposed to be the absolute value of the difference of ruler coordinates, we can take these ruler coordinates to be the x-coordinates of the point, multiplied by a constant that depends only on the slope of the line. (Actually, we would need to subtract from each the x-coordinate of our “zero” point, and account for the chosen direction, but these are easily done.) A new model:

Points and Lines are the same as in the previous model:

Points: {(x,y)| x0ú, y0ú} Lines: {(x,y)| ax + by + c = 0, x, y 0ú, a, b, c 0ú, not both a = 0 and b = 0}

But in this model, distance is different:

Distance between P(x1, y1) and Q(x2, y2) is

Since lines, points, slopes, etc. are the same, this model satisfies all the axioms the original model does, with the possible exception of those involving distance. But if we again consider two points P and Q on a line y = mx +b, we get: In other words, the ruler postulate works here, too, with just a different choice of constant.

Thus,

So betweenness is exactly the same for the two models!

For any three points A, B, C in taxicab geometry, A-B-C* iff A-B-C. Two interesting Triangles: