Taxicab Geometry
Recall that one model for our axiom system to
this point is the usual Cartesian plane, 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 distance 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 line 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: