**Theorem
4.1**:
(The distance axioms) Let *A*, *B*, and *C *be any three points in the
plane. Then

1. |*AB*| __>__ 0 with equality if and only if *A* = *B*.

2. |*AB*| = |*BA*|

3. |*AB*| __<__ |*AC*| + |*CB*| with equality if and only if *C *is on the line segment between *A *and *B*.

**Proof**: 1. Let *A* = (*x*_{1}, *y*_{1}), *B* = (*x*_{2}, *y*_{2}), and C = (*x*_{3},
*y*_{3}). Then by the definition of distance,

Since, by convention, we take the nonnegative square root, the distance will be nonnegative. It will be zero if and only if the number inside the square root is zero. But the number inside the square root is a sum of squares, and since squares are never negative, a sum of squares is zero if and only if each term is zero. Thus

if and only if

if and only if

if and only if

2. If we compute |*B**A*| instead of |*A**B*|, we will switch
the order of the subtractions inside the radical in the distance
fromula. If we switch the order of a subtraction problem, we get the
negative of what we would have gotten if we had subtracted in the
other order, but the differences are squared and squares wipe out the
differenct between positives and negatives.

** **

3. This is Theorem 3.5 restated for convenience.