NOTES and Answers: Working with the Ruler Axiom

What is known so far

Recall that the Ruler Axiom states that for any line m, the points on m can be put in correspondence with the real numbers, so that if points A and B correspond to numbers a and b, the the distance |AB| = the absolute value |b-a|.

Also, the triangle inequality has a corollary that points A, B, C are collinear if one of the distances |AB|. |BC|, |CA| is the sum of the other two.

Suppose point C is on line AB, distinct from A and B, and corresponds to number c. Then |AC| + |CB| = |AB| if and only if c is between a and b (i.e., either a < c < b or b < c < a). In this case, we say that C is on segment AB.

Some Exercises

Suppose in each case that A, B, C are points. In each case there is a ruler on line AB and numbers a and b correspond to A and B. Answer these questions: (1) are the points collinear? (2) if a and b are given, what number c corresponds to C?

1. Suppose a = 1, b = 3, |AC| = 6, |BC| = 4.
• |AB| = 3-1 = 2, so 6 = 4+2. Collinear in order A - B - C. Since a < b, c = a + |AC| = 7. Check |BC| = 7 - 3 = 4.
2. Suppose a = -1, b = 4, |AC| = 6, |BC| = 3.
• |AB| = 4 - (-1) = 4, so 6 < 5 + 3 and points form a triangle, are not collinear.
3. Suppose a = 970, b = -23, |AC| = 80, |BC| = 913.
• |AB| = 970 - (-23) = 993. 993 = 913+80. Collinear in order A - C - B. Since b < a, c = b + |BC| = -23 + 913 = 890. Check |AC| = 970 - 890 = 80.
4. Suppose a = 5, b = 11, |AC| = |BC| = (1/2)|AB|.
• |AC| + |CB| = (1/2)|AB| + (1/2)|AB| = |AB| so collinear with C between A and B. |AB| = 11 - 5 = 6, so |AC| = 3. Then C = a + |AC| = 5 + 3 = 8.
5. Suppose a = a, b = b, |AC| = |BC| = (1/2)|AB|.
• |AC| + |CB| = (1/2)|AB| + (1/2)|AB| = |AB| so collinear with C between A and B. If a <c< b, then c = a + |CA| = a + (1/2)(c-a) = (1/2)(a+b). Note that we used |c-a| = c-a because of the order. In the other case, b < c < a, we get the same result, c = (1/2)(a+b).
6. Suppose a = a, b = b, |AC| = (1/3)|AB| and |BC| = (2/3)|AB|.
   • |AC| + |CB| = (1/3)|AB| + (2/3)|AB| = |AB| so collinear with C between A and B. If a <c< b, then c = a + |CA| = a + (1/3)(c-a) = (2/3)a + (1/3)b. Note that we used |c-a| = c-a because of the order. In the other case, b < c < a, we get the same result, c = b + (2/3)(a-b) = (2/3)a + (1/3)b.

Answer this related question.

7. If M is the midpoint of AB, and a ruler on line AB corresponds M = 3 and A to x, what number x' does B correspond to?
   (Note: B is called the point reflection of A with center M.)

Use the formula found in 5 to get 3 = (1/2)(x+x'), or x' = 6 - x. If in general the number corresponding to M is m, then the same formula gives x' = 2m - x.