It is possible to prove that the Cartesian coordinate plane is a Euclidean plane with practically no axioms. We define a point to be an ordered pair of real numbers, and the plane to be the set of all ordered pairs of real numbers. This enables us to use what we know about real numbers. The axioms are the axioms for the development of the real number system.

An important definition is the distance between two points.

**Definition**: Let (*x*_{1},
*y*_{1}) and (*x*_{2},
*y*_{2}) be two points. The distance between them
is

What we are doing with this definition is we are actually assuming the Pythagorean Theorem, which is well known to be equivalent to EuclidÕs Parallel Postulate. However, we are not assuming the whole Pythagorean Theorem, just the special case where the perpendicular sides of the triangle are horizontal and vertical. It is possible to prove the entire Pythagorean Theorem from this special case.

Let
*ABC* be a right triangle
with right angle at *C*.

If we let *A*
= (*x*_{1}, *y*_{1}), *B*
= (*x*_{2}, *y*_{2}), and * C* = (*x*_{3}, *y*_{3}), then to say that *C* is the right angle is to say that the slopes of *AC* and
*BC* are negative reciprocals of each other.

Note that If either if these denominators is 0, then the lines are horizontal or vertical, and we have dealt with that case with the definition of distance, so we do not have to worry about zero denominators.

Cross multiply.

(*y*_{1} – *y*_{3})(*y*_{2} – *y*_{3}) =
-(*x*_{1} – *x*_{3})(*x*_{2} – *x*_{3})

Transpose to get rid of the negative

(*y*_{1} – *y*_{3})(*y*_{2} – *y*_{3}) +
(*x*_{1} – *x*_{3})(*x*_{2} – *x*_{3}) =
0

Remove parentheses

*y _{1}y*

Transpose all the terms that do
not involve *x*_{3} or
*y*_{3} to the other side of the equation.

– *y*_{1}*y*_{3} – *y*_{2}*y _{3}* +

Double both sides.

– 2*y*_{1}*y*_{3}
– 2*y*_{2}*y _{3}* + 2

Add *x*_{1}^{2}
+ *x*_{2}^{2} + *y*_{1}^{2} + *y*_{2}^{2} to both sides and split up the 2*x*_{3}^{2} and
2*y*_{3}^{2}.

*x*_{1}^{2} –
2*x*_{1}*x*_{3} + *x*_{3}^{2}
+ *y*_{1}^{2} – 2*y*_{1}*y*_{3} + *y*_{3}^{2}
+ *x*_{2}^{2} – 2x_{2}x_{3}
+ *x*_{3}^{2} + *y*_{2}^{2} - 2*y _{2}y*

(*x*_{1} – *x*_{3})^{2} + (*y*_{1} – *y*_{3})^{2} + (*x*_{2} – *x*_{3})^{2} + (*y*_{2} – *y*_{3})^{2} = (*x*_{1} – *x*_{2})^{2} + (*y*_{1} – *y*_{2})^{2} ^{}

*a ^{2}*
+

This proves that if the triangle
is a right triangle, the sides are related by the Pythagorean Theorem. For the converse, simply reverse the steps. This
direction is actually more straightforward than the way we did it above. The
things that we had to pull out of the air to put into both sides of the
equation will automatically cancel out. Where we multiplied the binomials in
the fourth step to remove the parentheses, we will need to factor when we
reverse our steps, and that may require a bit of cleverness, but the steps are
reversible, and we have the implication both ways. If it is a right triangle, then *a*^{2} + *b*^{2} = *c*^{2}, and if *a*^{2} + *b*^{2} = *c*^{2}, then the triangle is a right triangle.