1. 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. 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:

3. 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 x 2 + y 2 + 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 x 2 + y 2 – 16 x = - 6

4. 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.

5. 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  = 66.4218 You may also make an accurate drawing and use a protractor to measure the angle.

6. 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. 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. 8. Given the hyperbolic triangle defined by the points A = (7, 8) B = (1, 20), and C = (1, 10). a. Calculate AC. b. Calculate BC c. 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.

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

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

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

12. 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.

13. 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”!!!!

14. 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 x 2 + y 2 + ax = b gives 125 + 5a = b and 25 – 5a = b which solve giving a = - 10 and b = 75 So the lines is x 2 + y 2 - 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.