1. Show geometrically that given any two points in the hyperbolic plane there is always a unique line passing through them. § 23.1 Given point A(x1, y1)

1. Show geometrically that given any two points in the hyperbolic plane there is always a unique line passing through them.

§ 23.1

Given point A(x1, y1) and B(x2, y2). These points can always be substituted into the general half-circle with center on the x-axis.

x 2 + y 2 + ax = b

The resulting two equations (one from each point) can be solved except in the case where x1 = x2. However, in this case, the line is a vertical ray with the equation x = x1.

2. Given two points A and B on the hyperbolic with point C between them that the betweenness property holds. I.e. verify that if A – B – C then AC + CB = AB. Note there are two cases for the two “different” types of lines in this model

Case 1 – points on a ray: A (a, m), B (b, m) and C (c, m)

Case 2 – points on a semicircle:

a cAC CB ln ln

c b

a c aln ln AB

c b b

AM CN CM BNAC CB ln ln

AN CM CN BM

AM CN CM BNln

AN CM CN BM

AM BNln ln (AB,MN) AB

AN BM

3. Let A = (1, 3) B = (1, 6), and C = (5, 7)

a. Find the equation of the line AB.b. Find the equation of the line AC

a. X coordinates are the same. It is a vertical line with equation x = 1.

b. Substitute (1, 3) and (5, 7) into x2 + y2 + ax = b and solve the resulting two equations

10 + a = b and 74 + 5a = b

for a and b yielding a = - 16 and b = - 6. hence the equation of the line AC is

x2 + y 2 – 16 x = - 6

4. Let A = (3, 1), B = (3, 10), C = (12, 20) and D = (24, 16) a. Find the distance from A to B.b. Find the distance from C to D.

a. It is a ray so distance AB = 2.30

b. Not a ray so distance CD = 0.69

You need to find M and N first. Use the method of problem 3 to find the equation of the line CD. x 2 + y 2 - 24x = 256. this circle has x-intercepts at M = 32 and N = - 8.

10AB ln 2.30

1

2 2 2 2

2 2 2 2

(12 32) 20 (24 ( 8)) 16CM DNCD ln ln

CN DM (12 ( 8)) 20 (24 32) 16

1ln 4 0.69

2

2 2 2 2

2 2 2 2

20 20 32 16 1 800 1280ln ln

2 800 32020 20 8 16

5. Find the angle between the two h-lines x 2 + y 2 = 25 and x 2 + y 2 – 10x = - 9

Solve the two equations simultaneously to get the point of intersection at I(3.4, 3.6661).

I like to use the derivative of the lines to get the slopes at the point of intersection.

First Line: dy/dx = - x/y = - 0.9274 = m 1

Second Line : dy/dx = (5 – x)/y = 0.4364 = m 2

Tan = (m 2 – m 1)/(1 + m 1 m 2) = 1.3638/.5953 = 2.2910

And = 66.4218You may also make an accurate drawing and use a protractor to measure the angle.

6. Show that the two lines x = 5 and x 2 + y 2 – 6x + 5 = 0 are parallel.

a. Substituting x = 5 into the second equations yields a y value of 0. this means that both lines have the point (5, 0) in common. This point is on the x-axis and is an ideal point (at infinity) and the two lines are parallel.

7. Let A = (2, 1), B = (2, 3), C = (2, 14) and D = (2, 16). Segments AB and CD have the same Euclidean length. Find the hyperbolic lengths of the two segments and compare.

AB = 1.10 and CD = 0.13. What is going on with distance in this “hyperbolic” model?

8. Given the hyperbolic triangle defined by the points A = (7, 8) B = (1, 20), and C = (1, 10).

a. Calculate AC.b. Calculate BCc. Verify that ABC is an isosceles right triangle.d. Verify that the point D(-24, 0) is the center of the semicircle AB, and

use this to calculate each of the angles of ABC using the slope formula for tan . What is the angle sum of the triangle?

a. AC = 0.6931

b. BC = 0.6931

c. See parts a and b. I will confirm it is a right triangle in part d.

d. The equation for the three sides are:

AB: x 2 + y 2 + 48x = 449 AC: x 2 + y 2 - 2x = 99 and BC: x = 1

Note that the center AB is (- 24, 0). It is always (-a/2, 0)

The center of AC is (1, 0) and thus AC and BC form a right angle at vertex C.

continued

Consider making a drawing of this triangle.


d. Verify that the point D(-24, 0) is the center of the semicircle AB, and use this to calculate each of the angles of ABC using the slope formula for tan . What is the angle sum of the triangle?



C = 90. I will use the method of problem 5 to find the other two angles.

To find A use sides AB and AC.

First Line, AB: dy/dx = x + 24/- y = - 3.875 = m 1 at point A

Second Line, AC : dy/dx = (1 – x)/y = - 0.75 = m 2 at point A

Tan = (m 2 – m 1)/(1 + m 1 m 2) = - 3.125/3.90625 = - 0.8

And = A = 38.6598

continued





C = 90 and A = 38.6598.

To find B use sides AB and BC.

First Line, AB: dy/dx = x + 24/- y = - 1.25 = m 1 at point B

Second Line, BC: BC is a vertical line with slope undefined. To find B we will first find the angle between line AB and a horizontal line (m 2 = 0) at point B. B will be 90 – the angle we find.

Tan = (m 2 – m 1)/(1 + m 1 m 2) = 1.25

And = 51.3402 B = 90 – 51.3402 = 38.6598.





C = 90 and A = 38.6598. and B = 38.6598.

Notice that the angles opposite equal sides are equal.

The sum of the angles of ABC is 167.3196.

9. Using the information from the previous problem does the Pythagorean Theorem hold in Hyperbolic Geometry?

BC = a = 0.6931, AC = b = 0.6931, and AB = c = 1.0163

a 2 + b 2 = 0.9608 while c 2 = 1.0329. The Pythagorean Theorem does not hold in hyperbolic geometry.

10. Using the information from the previous problem does the following relationship cosh c = cosh a cosh b hold?

BC = a = 0.6931, AC = b = 0.6931, and AB = c = 1.0163

cosh a cosh b = 1.5625 and cosh c = 1.5624 Other than round off error the relationship holds. Strange “Pythagorean Theorem”!!!!

11. Find the two h-lines parallel to x 2 + y 2 = 25 through the point (5, 10).

The line x 2 + y 2 = 25 has center at the origin and radius 5. thus it crosses the x-axis at (-5, 0) and (5, 0). Each of lines we need will go through one of these points and the given point (5, 10).

Consider making a drawing of this triangle.

First line. Through (5, 10) and (-5, 0). Using the method of problem 3 substituting into x2 + y2 + ax = b gives

125 + 5a = b and 25 – 5a = b which solve giving a = - 10 and b = 75

So the lines is x2 + y2 - 10x = 75

Second line. Through (5, 10) and (5, 0). Notice that the x values of these two points are the same so it is a vertical ray with equation x = 5.

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1. Show geometrically that given any two points in the hyperbolic plane there is always a unique line passing through them. § 23.1 Given point A(x1, y1)