Proof: Given 𝐿1 and 𝐿2 with 𝐿1 ≠ 𝐿2 . If both are vertical, then they do not intersect at all. If one is vertical and the other is not, then the graphs of
𝑥 = 𝑎, 𝑦 = 𝑚𝑥 + 𝑏
Intersect at the unique point (𝑎, 𝑚𝑎 + 𝑏). Suppose finally, that 𝐿1 and 𝐿2 are the graphs of
𝑦 = 𝑚1𝑥 + 𝑏1, 𝑦 = 𝑚2𝑥 + 𝑏2.
If 𝑚1 ≠ 𝑚2 , very elementary algebra gives us exactly one common solution and hence exactly one intersection point. If
𝑚1 = 𝑚2 , then 𝑏1 ≠ 𝑏2 , and the graphs do not intersect at all.
We have already observed that if 𝐿 is the graph of 𝑦 = 𝑚𝑥 + 𝑏, then for every two points 𝑥1, 𝑦1 , (𝑥2, 𝑦2) of 𝐿, we have
𝑦2− 𝑦1
𝑥2− 𝑥1= 𝑚.
Thus, 𝑚 is determined by the non-vertical line 𝐿. As usual, we call 𝑚 the slope of 𝐿. Theorem 5.3.3: Every vertical line intersects every non-vertical line.
Proof: Let 𝐿1 be a vertical line 𝑥 = 𝑎 and 𝐿2 be non-vertical line = 𝑚𝑥 + 𝑏, then by theorem 2, 𝐿1 and 𝐿2 intersect at the point (𝑎, 𝑚𝑎 + 𝑏2).
Theorem 5.3.4: Two lines are parallel if and only if (1) both are vertical, or (2) neither is vertical, and they have the same slope.
Proof: Given 𝐿1 ≠ 𝐿2 . If both are vertical, then 𝐿1 ∥ 𝐿2 . If neither is vertical, and they have the same slope, then the equations
𝑦 = 𝑚𝑥 + 𝑏1, 𝑦 = 𝑚𝑥 + 𝑏2. (𝑏1≠ 𝑏2) have no common solution, and 𝐿1 ∥ 𝐿2 .
Suppose, conversely, that 𝐿1 ∥ 𝐿2 . If both are vertical, then (1) holds. It remains only to show that if neither line is vertical, they have the same slope.
Suppose not. Then 𝐿1 : 𝑦 = 𝑚1𝑥 + 𝑏1, 𝐿2 : 𝑦 = 𝑚2𝑥 + 𝑏2 (𝑚1≠𝑚2) We can now solve for 𝑥 and 𝑦:
0 = 𝑚1− 𝑚2 𝑥 + 𝑏1− 𝑏2 , Solve for 𝑥 we obtain
𝑥 = − 𝑏1− 𝑏2 𝑚1− 𝑚2,
𝑦 = −𝑚1
𝑏1− 𝑏2
𝑚1− 𝑚2 + 𝑏1
We got this value of y by substituting in the equation of 𝐿1 . But, our x and y also satisfy the equation of 𝐿2 . This contradicts the hypothesis 𝐿1 ∥ 𝐿2 .
Theorem 5.3.5: Given a point 𝑃 = (𝑥1, 𝑦1) and a number 𝑚, there is exactly one line which passes through 𝑃 and has slope = 𝑚.
Proof: The lines L with slope 𝑚 are the graphs of equations 𝑦 = 𝑚𝑥 + 𝑏.
If L contains (𝑥1, 𝑦1), then 𝑏 = 𝑦1− 𝑚𝑥1, and conversely. Therefore, our line exists and is unique. Theorem 5.3.6: In the Cartesian model, the Euclidean parallel postulate holds.
Proof: Given a line L and a point 𝑃 = (𝑥1, 𝑦1) not on L.
1. If L is the graph of 𝑥 = 𝑎, then the line 𝐿′: 𝑥 = 𝑥1 is the only vertical line through P, and by theorem 5.3.3, no non-vertical line is parallel to L. Thus, the parallel line L through P is unique.
2. If L is the graph of 𝑦 = 𝑚𝑥 + 𝑏, then the only line parallel to L through P is the line through P with slope = 𝑚. This is unique.
5.4. Translations and Rotations
By a translation of the Cartesian model, we mean a one-to-one correspondence
𝑓: 𝐸 ↔ 𝐸: (𝑥, 𝑦) ↔ (𝑥 + 𝑎, 𝑦 + 𝑏).
Merely by substitution in the distance formula, and observing that 𝑎 and 𝑏 cancel out, we have: Theorem 5.4.1: Translations are isometries.
If L is the graph of the equation
𝐴𝑥 + 𝐵𝑦 + 𝐶 = 0, then the points 𝑥′, 𝑦′ = (𝑥 + 𝑎, 𝑦 + 𝑏) of 𝑓(𝐿) satisfy the equation 𝐴 𝑥′− 𝑎 + 𝐵 𝑦′− 𝑏 + 𝐶 = 0,
Or 𝐴𝑥′+ 𝐵𝑦′+ −𝑎𝐴 − 𝑏𝐵 + 𝐶 = 0 This is linear. Thus, we have proven the theorem. Theorem 5.4.2: Translation preserves lines.
Since translations preserve lines and distance, they preserve everything defined in terms of lines and distance. Theorem 5.4.3: Translations preserve betweenness, segments, rays, angles, triangles, and angle congruences.
Rotations are harder to describe, because at this stage we have no trigonometry to work with. Let us first try using trigonometry, wistfully, to find out what we ought to be doing, and then find a way to do something equivalent, using only the primitive apparatus that we now have at our disposal in our study of the Cartesian model.
Figure 5.4.1
We want to rotate the Cartesian model through and angle of measure ∅. (Fig. 5.4.1) Trigonometrically, this can be done by a one-to-one correspondence,
𝑓: 𝐸 ↔ 𝐸,
defined as the labels in the figure suggest. Now cos(𝜃+𝜙) = 𝑐𝑜𝑠𝜃𝑐𝑜𝑠𝜙 − 𝑠𝑖𝑛𝜃𝑠𝑖𝑛𝜙 Let 𝑎 = 𝑐𝑜𝑠𝜙, 𝑏 = 𝑠𝑖𝑛𝜙
Now 𝑟 = 𝑥2+ 𝑦2, 𝑐𝑜𝑠𝜃 = 𝑥2𝑥+𝑦2, 𝑠𝑖𝑛𝜃 = 𝑥2𝑦+𝑦2 We can therefore rewrite our formulas in the form
𝑓: (𝑥, 𝑦) ↔ (𝑥′, 𝑦′) Where 𝑥′= 𝑟𝑐𝑜𝑠(𝜃+𝜙)
r Y
= 𝑥2+ 𝑦2 𝑥2𝑥 +𝑦2𝑎 − 𝑦 𝑥2+𝑦2𝑏 = 𝑎𝑥 − 𝑏𝑦 And 𝑦′ = 𝑥2+ 𝑦2 𝑥2𝑦+𝑦2𝑎 + 𝑥2𝑥+𝑦2𝑏 = 𝑎𝑦 + 𝑏𝑥.
Any correspondence of this form, with 𝑎2+ 𝑏2= 1, is called a rotation of the Cartesian model. Theorem 5.4.4: Rotations preserve distance.
Proof: We have 𝑃 = 𝑥1, 𝑦1 𝑄 = 𝑥2, 𝑦2 𝑃′= 𝑓 𝑃 = (𝑎𝑥 1− 𝑏𝑦1, 𝑎𝑦1+ 𝑏𝑥1) 𝑄′= 𝑓 𝑄 = (𝑎𝑥 2− 𝑏𝑦2, 𝑎𝑦2+ 𝑏𝑥2)
It is merely an exercise in patience to substitute in the distance formula, calculate
𝑃′𝑄′, simplify with the aid of the equation 𝑎2+ 𝑏2= 1, and observe that 𝑃′𝑄′= 𝑃𝑄. Solving for 𝑥 𝑎𝑛𝑑 𝑦, we get
𝑥 = 𝑎𝑥′+ 𝑏𝑦′, 𝑦 = 𝑎𝑦′− 𝑏𝑥′ Comparing the formulas
𝑥′= 𝑎𝑥 − 𝑏𝑦
, 𝑦′ = 𝑏𝑥 + 𝑎𝑦
For 𝑓 and the corresponding formulas for 𝑓−1, we see that these have the same form:
𝑥 = 𝑎′𝑥′− 𝑏′𝑦′, 𝑦 = 𝑎′𝑦′+ 𝑏′𝑥′, where 𝑎′= 𝑎 and 𝑏′= −𝑏. Therefore, we have the following theorem.
Theorem 5.4.5: The inverse of a rotation is a rotation.