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Solution Manual For Trigonometry: A Unit Circle Approach, 10th Edition

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    ’ S S OLUTIONS M ANUAL T IM B RITT Jackson State Community College T RIGONOMETRY : A U NIT C IRCLE A PPROACH T ENTH E DITION Michael Sullivan Chicago State University

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    Table of Contents Preface Chapter 1 Graphs and Functions 1.1 The Distance and Midpoint Formulas ......................................................................................... 1 1.2 Graphs of Equations in Two Variables; Circles........................................................................ 12 1.3 Functions and Their Graphs ...................................................................................................... 33 1.4 Properties of Functions ............................................................................................................. 50 1.5 Library of Functions; Piecewise-defined Functions ................................................................. 66 1.6 Graphing Techniques: Transformations ................................................................................... 77 1.7 One-to-One Functions; Inverse Functions ................................................................................ 93 Chapter Review.............................................................................................................................. 112 Chapter Test ................................................................................................................................... 121 Chapter Projects ............................................................................................................................. 125 Chapter 2 Trigonometric Functions 2.1 Angles and Their Measure ...................................................................................................... 126 2.2 Trigonometric Functions: Unit Circle Approach .................................................................... 134 2.3 Properties of the Trigonometric Functions ............................................................................. 151 2.4 Graphs of the Sine and Cosine Functions ............................................................................... 162 2.5 Graphs of the Tangent, Cotangent, Cosecant, and Secant Functions...................................... 182 2.6 Phase Shift; Sinusoidal Curve Fitting ..................................................................................... 191 Chapter Review.............................................................................................................................. 203 Chapter Test ................................................................................................................................... 211 Cumulative Review........................................................................................................................ 214 Chapter Projects ............................................................................................................................. 216 Chapter 3 Analytic Trigonometry 3.1 The Inverse Sine, Cosine, and Tangent Functions .................................................................. 220 3.2 The Inverse Trigonometric Functions (Continued) ................................................................ 231 3.3 Trigonometric Equations ........................................................................................................ 243 3.4 Trigonometric Identities ......................................................................................................... 262 3.5 Sum and Difference Formulas ................................................................................................ 274 3.6 Double-angle and Half-angle Formulas .................................................................................. 297 3.7 Product-to-Sum and Sum-to-Product Formulas...................................................................... 321 Chapter Review.............................................................................................................................. 331 Chapter Test ................................................................................................................................... 345 Cumulative Review........................................................................................................................ 350 Chapter Projects ............................................................................................................................. 353 Chapter 4 Applications of Trigonometric Functions 4.1 Right Triangle Trigonometry; Applications ........................................................................... 357 4.2 The Law of Sines .................................................................................................................... 369 4.3 The Law of Cosines ................................................................................................................ 383 4.4 Area of a Triangle ................................................................................................................... 394 4.5 Simple Harmonic Motion; Damped Motion; Combining Waves ........................................... 402 Chapter Review.............................................................................................................................. 411 Chapter Test ................................................................................................................................... 418 Cumulative Review........................................................................................................................ 422 Chapter Projects ............................................................................................................................. 425

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    Chapter 5 Polar Coordinates; Vectors 5.1 Polar Coordinates.................................................................................................................... 429 5.2 Polar Equations and Graphs .................................................................................................... 436 5.3 The Complex Plane; De Moivre’s Theorem ........................................................................... 465 5.4 Vectors .................................................................................................................................... 476 5.5 The Dot Product ...................................................................................................................... 488 5.6 Vectors in Space ..................................................................................................................... 494 5.7 The Cross Product................................................................................................................... 500 Chapter Review.............................................................................................................................. 510 Chapter Test ................................................................................................................................... 519 Cumulative Review........................................................................................................................ 523 Chapter Projects ............................................................................................................................. 525 Chapter 6 Analytic Geometry 6.2 The Parabola ........................................................................................................................... 529 6.3 The Ellipse .............................................................................................................................. 543 6.4 The Hyperbola ........................................................................................................................ 559 6.5 Rotation of Axes; General Form of a Conic ........................................................................... 578 6.6 Polar Equations of Conics ....................................................................................................... 590 6.7 Plane Curves and Parametric Equations ................................................................................. 597 Chapter Review.............................................................................................................................. 610 Chapter Test ................................................................................................................................... 619 Cumulative Review........................................................................................................................ 624 Chapter Projects ............................................................................................................................. 625 Chapter 7 Exponential and Logarithmic Functions 7.1 Exponential Functions ............................................................................................................ 629 7.2 Logarithmic Functions ............................................................................................................ 648 7.3 Properties of Logarithms ........................................................................................................ 668 7.4 Logarithmic and Exponential Equations ................................................................................. 677 7.5 Financial Models .................................................................................................................... 696 7.6 Exponential Growth and Decay Models; Newton’s Law; Logistic Growth and Decay Models .................................................................................................................... 703 7.7 Building Exponential, Logarithmic, and Logistic Models from Data..................................... 713 Chapter Review.............................................................................................................................. 717 Chapter Test ................................................................................................................................... 726 Cumulative Review........................................................................................................................ 729 Chapter Projects ............................................................................................................................. 731 Appendix A Review A.1 Algebra Essentials .................................................................................................................. 734 A.2 Geometry Essentials ............................................................................................................... 739 A.3 Factoring Polynomials; Completing the Square..................................................................... 745 A.4 Solving Equations .................................................................................................................. 748 A.5 Complex Numbers; Quadratic Equations in the Complex Number System .......................... 762 A.6 Interval Notation; Solving Inequalities .................................................................................. 768 A.7 n th Roots; Rational Exponents ............................................................................................... 779 A.8 Lines....................................................................................................................................... 788 A.9 Building Linear Models from Data ........................................................................................ 804

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    Appendix B Graphing Utilities B.1 The Viewing Rectangle .......................................................................................................... 809 B.2 Using a Graphing Utility to Graph Equations ........................................................................ 810 B.3 Using a Graphing Utility to Locate Intercepts and Check for Symmetry .............................. 814 B.5 Square Screens ....................................................................................................................... 816

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    1 Chapter 1 Graphs and Functions Section 1.1 1. 0 2. ( ) 5 3 8 8 − − = = 3. 2 2 3 4 25 5 + = = 4. 2 2 2 11 60 121 3600 3721 61 + = + = = Since the sum of the squares of two of the sides of the triangle equals the square of the third side, the triangle is a right triangle. 5. 1 2 bh 6. true 7. x- coordinate or abscissa; y -coordinate or ordinate 8. quadrants 9. midpoint 10. False; the distance between two points is never negative. 11. False; points that lie in quadrant IV will have a positive x -coordinate and a negative y -coordinate. The point ( ) 1, 4 − lies in quadrant II. 12. True; 1 2 1 2 , 2 2 x x y y M + +   =     13. b 14. a 15. (a) quadrant II (b) x -axis (c) quadrant III (d) quadrant I (e) y -axis (f) Quadrant IV 16. (a) quadrant I (b) quadrant III (c) quadrant II (d) quadrant I (e) y -axis (f) x -axis 17. The points will be on a vertical line that is two units to the right of the y -axis.

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    Chapter 1: Graphs and Functions 2 18. The points will be on a horizontal line that is three units above the x -axis. 19. 2 2 1 2 2 2 ( , ) (2 0) (1 0) 2 1 4 1 5 d P P = − + − = + = + = 20. 2 2 1 2 2 2 ( , ) ( 2 0) (1 0) ( 2) 1 4 1 5 d P P = − − + − = − + = + = 21. 2 2 1 2 2 2 ( , ) ( 2 1) (2 1) ( 3) 1 9 1 10 d P P = − − + − = − + = + = 22. ( ) 2 2 1 2 2 2 ( , ) 2 ( 1) (2 1) 3 1 9 1 10 d P P = − − + − = + = + = 23. ( ) ( ) ( ) 2 2 1 2 2 2 ( , ) (5 3) 4 4 2 8 4 64 68 2 17 d P P = − + − − = + = + = = 24. ( ) ( ) ( ) ( ) 2 2 1 2 2 2 ( , ) 2 1 4 0 3 4 9 16 25 5 d P P = − − + − = + = + = = 25. ( ) 2 2 1 2 2 2 ( , ) 6 ( 3) (0 2) 9 ( 2) 81 4 85 d P P = − − + − = + − = + = 26. ( ) ( ) 2 2 1 2 2 2 ( , ) 4 2 2 ( 3) 2 5 4 25 29 d P P = − + − − = + = + = 27. ( ) 2 2 1 2 2 2 ( , ) (6 4) 4 ( 3) 2 7 4 49 53 d P P = − + − − = + = + = 28. ( ) ( ) 2 2 1 2 2 2 ( , ) 6 ( 4) 2 ( 3) 10 5 100 25 125 5 5 d P P = − − + − − = + = + = = 29. 2 2 1 2 2 2 2 2 ( , ) (0 ) (0 ) ( ) ( ) d P P a b a b a b = − + − = − + − = + 30. 2 2 1 2 2 2 2 2 2 ( , ) (0 ) (0 ) ( ) ( ) 2 2 d P P a a a a a a a a = − + − = − + − = + = = 31. ( 2,5), (1,3), ( 1, 0) A B C = − = = − ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 2 ( , ) 1 ( 2) (3 5) 3 ( 2) 9 4 13 ( , ) 1 1 (0 3) ( 2) ( 3) 4 9 13 ( , ) 1 ( 2) (0 5) 1 ( 5) 1 25 26 d A B d B C d A C = − − + − = + − = + = = − − + − = − + − = + = = − − − + − = + − = + = Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem: [ ] [ ] [ ] ( ) ( ) ( ) 2 2 2 2 2 2 ( , ) ( , ) ( , ) 13 13 26 13 13 26 26 26 d A B d B C d A C + = + = + = = The area of a triangle is 1 2 A bh = ⋅ . In this problem,

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    Section 1.1: The Distance and Midpoint Formulas 3 [ ] [ ] 1 ( , ) ( , ) 2 1 1 13 13 13 2 2 13 square units 2 A d A B d B C = ⋅ ⋅ = ⋅ = ⋅ ⋅ = 32. ( 2, 5), (12, 3), (10, 11) A B C = − = = − ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 2 ( , ) 12 ( 2) (3 5) 14 ( 2) 196 4 200 10 2 ( , ) 10 12 ( 11 3) ( 2) ( 14) 4 196 200 10 2 ( , ) 10 ( 2) ( 11 5) 12 ( 16) 144 256 400 20 d A B d B C d A C = − − + − = + − = + = = = − + − − = − + − = + = = = − − + − − = + − = + = = Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem: [ ] [ ] [ ] ( ) ( ) ( ) 2 2 2 2 2 2 ( , ) ( , ) ( , ) 10 2 10 2 20 200 200 400 400 400 d A B d B C d A C + = + = + = = The area of a triangle is 1 2 A bh = . In this problem, [ ] [ ] 1 ( , ) ( , ) 2 1 10 2 10 2 2 1 100 2 100 square units 2 A d A B d B C = ⋅ ⋅ = ⋅ ⋅ = ⋅ ⋅ = 33. ( 5,3), (6, 0), (5,5) A B C = − = = ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 2 ( , ) 6 ( 5) (0 3) 11 ( 3) 121 9 130 ( , ) 5 6 (5 0) ( 1) 5 1 25 26 ( , ) 5 ( 5) (5 3) 10 2 100 4 104 2 26 d A B d B C d A C = − − + − = + − = + = = − + − = − + = + = = − − + − = + = + = = Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem: [ ] [ ] [ ] ( ) ( ) ( ) 2 2 2 2 2 2 ( , ) ( , ) ( , ) 104 26 130 104 26 130 130 130 d A C d B C d A B + = + = + = = The area of a triangle is 1 2 A bh = . In this

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    Chapter 1: Graphs and Functions 4 problem, [ ] [ ] 1 ( , ) ( , ) 2 1 104 26 2 1 2 26 26 2 1 2 26 2 26 square units A d A C d B C = ⋅ ⋅ = ⋅ ⋅ = ⋅ ⋅ = ⋅ ⋅ = 34. ( 6, 3), (3, 5), ( 1, 5) A B C = − = − = − ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 2 ( , ) 3 ( 6) ( 5 3) 9 ( 8) 81 64 145 ( , ) 1 3 (5 ( 5)) ( 4) 10 16 100 116 2 29 ( , ) 1 ( 6) (5 3) 5 2 25 4 29 d A B d B C d A C = − − + − − = + − = + = = − − + − − = − + = + = = = − − − + − = + = + = Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem: [ ] [ ] [ ] ( ) ( ) ( ) 2 2 2 2 2 2 ( , ) ( , ) ( , ) 29 2 29 145 29 4 29 145 29 116 145 145 145 d A C d B C d A B + = + = + ⋅ = + = = The area of a triangle is 1 2 A bh = . In this problem, [ ] [ ] 1 ( , ) ( , ) 2 1 29 2 29 2 1 2 29 2 29 square units A d A C d B C = ⋅ ⋅ = ⋅ ⋅ = ⋅ ⋅ = 35. (4, 3), (0, 3), (4, 2) A B C = − = − = ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 2 ( , ) (0 4) 3 ( 3) ( 4) 0 16 0 16 4 ( , ) 4 0 2 ( 3) 4 5 16 25 41 ( , ) (4 4) 2 ( 3) 0 5 0 25 25 5 d A B d B C d A C = − + − − − = − + = + = = = − + − − = + = + = = − + − − = + = + = = Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem: [ ] [ ] [ ] ( ) 2 2 2 2 2 2 ( , ) ( , ) ( , ) 4 5 41 16 25 41 41 41 d A B d A C d B C + = + = + = = The area of a triangle is 1 2 A bh = . In this

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    Section 1.1: The Distance and Midpoint Formulas 5 problem, [ ] [ ] 1 ( , ) ( , ) 2 1 4 5 2 10 square units A d A B d A C = ⋅ ⋅ = ⋅ ⋅ = 36. (4, 3), (4, 1), (2, 1) A B C = − = = ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 2 ( , ) (4 4) 1 ( 3) 0 4 0 16 16 4 ( , ) 2 4 1 1 ( 2) 0 4 0 4 2 ( , ) (2 4) 1 ( 3) ( 2) 4 4 16 20 2 5 d A B d B C d A C = − + − − = + = + = = = − + − = − + = + = = = − + − − = − + = + = = Verifying that ∆ ABC is a right triangle by the Pythagorean Theorem: [ ] [ ] [ ] ( ) 2 2 2 2 2 2 ( , ) ( , ) ( , ) 4 2 2 5 16 4 20 20 20 d A B d B C d A C + = + = + = = The area of a triangle is 1 2 A bh = . In this problem, [ ] [ ] 1 ( , ) ( , ) 2 1 4 2 2 4 square units A d A B d B C = ⋅ ⋅ = ⋅ ⋅ = 37. The coordinates of the midpoint are: 1 2 1 2 ( , ) , 2 2 4 4 3 5 , 2 2 8 0 , 2 2 (4, 0) x x y y x y + +   =     − + +   =       =     = 38. The coordinates of the midpoint are: ( ) 1 2 1 2 ( , ) , 2 2 2 2 0 4 , 2 2 0 4 , 2 2 0, 2 x x y y x y + +   =     − + +   =       =     = 39. The coordinates of the midpoint are: 1 2 1 2 ( , ) , 2 2 3 6 2 0 , 2 2 3 2 , 2 2 3 ,1 2 x x y y x y + +   =     − + +   =       =       =     40. The coordinates of the midpoint are: 1 2 1 2 ( , ) , 2 2 2 4 3 2 , 2 2 6 1 , 2 2 1 3, 2 x x y y x y + +   =     + − +   =     −   =       = −    

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    Chapter 1: Graphs and Functions 6 41. The coordinates of the midpoint are: 1 2 1 2 ( , ) , 2 2 4 6 3 1 , 2 2 2 10 , 2 2 (5, 1) x x y y x y + +   =     + − +   =     −   =     = − 42. The coordinates of the midpoint are: 1 2 1 2 ( , ) , 2 2 4 2 3 2 , 2 2 2 1 , 2 2 1 1, 2 x x y y x y + +   =     − + − +   =     − −   =       = − −     43. The coordinates of the midpoint are: 1 2 1 2 ( , ) , 2 2 0 0 , 2 2 , 2 2 x x y y x y a b a b + +   =     + +   =       =     44. The coordinates of the midpoint are: 1 2 1 2 ( , ) , 2 2 0 0 , 2 2 , 2 2 x x y y x y a a a a + +   =     + +   =       =     45. The x coordinate would be 2 3 5 + = and the y coordinate would be 5 2 3 − = . Thus the new point would be ( ) 5,3 . 46. The new x coordinate would be 1 2 3 − − = − and the new y coordinate would be 6 4 10 + = . Thus the new point would be ( ) 3,10 − 47. a. If we use a right triangle to solve the problem, we know the hypotenuse is 13 units in length. One of the legs of the triangle will be 2+3=5. Thus the other leg will be: 2 2 2 2 2 5 13 25 169 144 12 b b b b + = + = = = Thus the coordinates will have an y value of 1 12 13 − − = − and 1 12 11 − + = . So the points are ( ) 3,11 and ( ) 3, 13 − . b. Consider points of the form ( ) 3, y that are a distance of 13 units from the point ( ) 2, 1 − − . ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 1 2 1 2 2 2 2 2 2 3 ( 2) 1 5 1 25 1 2 2 26 d x x y y y y y y y y = − + − = − − + − − = + − − = + + + = + + ( ) ( )( ) 2 2 2 2 2 2 13 2 26 13 2 26 169 2 26 0 2 143 0 11 13 y y y y y y y y y y = + + = + + = + + = + − = − + 11 0 11 y y − = = or 13 0 13 y y + = = − Thus, the points ( ) 3,11 and ( ) 3, 13 − are a distance of 13 units from the point ( ) 2, 1 − − . 48. a. If we use a right triangle to solve the problem, we know the hypotenuse is 17 units in length. One of the legs of the triangle will be 2+6=8. Thus the other leg will be: 2 2 2 2 2 8 17 64 289 225 15 b b b b + = + = = = Thus the coordinates will have an x value of 1 15 14 − = − and 1 15 16 + = . So the points are ( ) 14, 6 − − and ( ) 16, 6 − .

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    Section 1.1: The Distance and Midpoint Formulas 7 b. Consider points of the form ( ) , 6 x − that are a distance of 17 units from the point ( ) 1, 2 . ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 1 2 1 2 2 2 2 2 2 1 2 6 2 1 8 2 1 64 2 65 d x x y y x x x x x x x = − + − = − + − − = − + + = − + + = − + ( ) ( )( ) 2 2 2 2 2 2 17 2 65 17 2 65 289 2 65 0 2 224 0 14 16 x x x x x x x x x x = − + = − + = − + = − − = + − 14 0 14 x x + = = − or 16 0 16 x x − = = Thus, the points ( ) 14, 6 − − and ( ) 16, 6 − are a distance of 13 units from the point ( ) 1, 2 . 49. Points on the x -axis have a y -coordinate of 0. Thus, we consider points of the form ( ) , 0 x that are a distance of 6 units from the point ( ) 4, 3 − . ( ) ( ) ( ) ( ) ( ) 2 2 2 1 2 1 2 2 2 2 2 2 4 3 0 16 8 3 16 8 9 8 25 d x x y y x x x x x x x = − + − = − + − − = − + + − = − + + = − + ( ) 2 2 2 2 2 2 2 6 8 25 6 8 25 36 8 25 0 8 11 ( 8) ( 8) 4(1)( 11) 2(1) 8 64 44 8 108 2 2 8 6 3 4 3 3 2 x x x x x x x x x = − + = − + = − + = − − − − ± − − − = ± + ± = = ± = = ± 4 3 3 x = + or 4 3 3 x = − Thus, the points ( ) 4 3 3, 0 + and ( ) 4 3 3, 0 − are on the x -axis and a distance of 6 units from the point ( ) 4, 3 − . 50. Points on the y -axis have an x -coordinate of 0. Thus, we consider points of the form ( ) 0, y that are a distance of 6 units from the point ( ) 4, 3 − . ( ) ( ) ( ) ( ) 2 2 2 1 2 1 2 2 2 2 2 2 4 0 3 4 9 6 16 9 6 6 25 d x x y y y y y y y y y = − + − = − + − − = + + + = + + + = + + ( ) 2 2 2 2 2 2 2 6 6 25 6 6 25 36 6 25 0 6 11 ( 6) (6) 4(1)( 11) 2(1) 6 36 44 6 80 2 2 6 4 5 3 2 5 2 y y y y y y y y y = + + = + + = + + = + − − ± − − = − ± + − ± = = − ± = = − ± 3 2 5 y = − + or 3 2 5 y = − − Thus, the points ( ) 0, 3 2 5 − + and ( ) 0, 3 2 5 − − are on the y- axis and a distance of 6 units from the point ( ) 4, 3 − . 51. a. To shift 3 units left and 4 units down, we subtract 3 from the x -coordinate and subtract 4 from the y -coordinate. ( ) ( ) 2 3,5 4 1,1 − − = − b. To shift left 2 units and up 8 units, we subtract 2 from the x -coordinate and add 8 to the y -coordinate. ( ) ( ) 2 2,5 8 0,13 − + =

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    Chapter 1: Graphs and Functions 8 52. Let the coordinates of point B be ( ) , x y . Using the midpoint formula, we can write ( ) 1 8 2,3 , 2 2 x y − + +   =     . This leads to two equations we can solve. 1 2 2 1 4 5 x x x − + = − + = = 8 3 2 8 6 2 y y y + = + = = − Point B has coordinates ( ) 5, 2 − . 53. ( ) 1 2 1 2 , , 2 2 x x y y M x y + +   = =     . ( ) 1 1 1 , ( 3, 6) P x y = = − and ( , ) ( 1, 4) x y = − , so 1 2 2 2 2 2 3 1 2 2 3 1 x x x x x x + = − + − = − = − + = and 1 2 2 2 2 2 6 4 2 8 6 2 y y y y y y + = + = = + = Thus, 2 (1, 2) P = . 54. ( ) 1 2 1 2 , , 2 2 x x y y M x y + +   = =     . ( ) 2 2 2 , (7, 2) P x y = = − and ( , ) (5, 4) x y = − , so 1 2 1 1 1 2 7 5 2 10 7 3 x x x x x x + = + = = + = and 1 2 1 1 1 2 ( 2) 4 2 8 ( 2) 6 y y y y y y + = + − − = − = + − − = Thus, 1 (3, 6) P = − . 55. The midpoint of AB is: ( ) 0 6 0 0 , 2 2 3, 0 D + +   =     = The midpoint of AC is: ( ) 0 4 0 4 , 2 2 2, 2 E + +   =     = The midpoint of BC is: ( ) 6 4 0 4 , 2 2 5, 2 F + +   =     = ( ) 2 2 2 2 ( , ) 0 4 (3 4) ( 4) ( 1) 16 1 17 d C D = − + − = − + − = + = ( ) 2 2 2 2 ( , ) 2 6 (2 0) ( 4) 2 16 4 20 2 5 d B E = − + − = − + = + = = 2 2 2 2 ( , ) (2 0) (5 0) 2 5 4 25 29 d A F = − + − = + = + = 56. Let 1 2 (0, 0), (0, 4), ( , ) P P P x y = = = ( ) ( ) ( ) 2 2 1 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 , (0 0) (4 0) 16 4 , ( 0) ( 0) 4 16 , ( 0) ( 4) ( 4) 4 ( 4) 16 d P P d P P x y x y x y d P P x y x y x y = − + − = = = − + − = + = → + = = − + − = + − = → + − = Therefore, ( ) 2 2 2 2 4 8 16 8 16 2 y y y y y y y = − = − + = = which gives 2 2 2 2 16 12 2 3 x x x + = = = ± Two triangles are possible. The third vertex is ( ) ( ) 2 3, 2 or 2 3, 2 − .

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    Section 1.1: The Distance and Midpoint Formulas 9 57. Let ( ) 1 0, 0 P = , ( ) 2 0, P s = , ( ) 3 , 0 P s = , and ( ) 4 , P s s = . y x (0, ) s (0, 0) ( , 0) s ( , ) s s The points 1 P and 4 P are endpoints of one diagonal and the points 2 P and 3 P are the endpoints of the other diagonal. 1,4 0 0 , , 2 2 2 2 s s s s M + +     = =         2,3 0 0 , , 2 2 2 2 s s s s M + +     = =         The midpoints of the diagonals are the same. Therefore, the diagonals of a square intersect at their midpoints. 58. Let ( ) 1 0, 0 P = , ( ) 2 , 0 P a = , and 3 3 , 2 2 a a P   =       . To show that these vertices form an equilateral triangle, we need to show that the distance between any pair of points is the same constant value. ( ) ( ) ( ) ( ) ( ) 2 2 1 2 2 1 2 1 2 2 2 , 0 0 0 d P P x x y y a a a = − + − = − + − = = ( ) ( ) ( ) 2 2 2 3 2 1 2 1 2 2 2 2 2 2 , 3 0 2 2 3 4 4 4 4 d P P x x y y a a a a a a a a = − + −     = − + −           = + = = = ( ) ( ) ( ) 2 2 1 3 2 1 2 1 2 2 2 2 2 2 , 3 0 0 2 2 3 4 4 4 4 d P P x x y y a a a a a a a = − + −     = − + −           = + = = = Since all three distances have the same constant value, the triangle is an equilateral triangle. Now find the midpoints: 1 2 2 3 1 3 4 5 6 0 0 0 , , 0 2 2 2 3 3 3 0 , 2 2 , 4 4 2 2 3 0 0 3 2 2 , , 2 2 4 4 P P P P P P a a P M a a a a a P M a a a a P M + +     = = =               + +   = = =               + +       = = =             ( ) 2 2 4 5 2 2 2 2 3 3 , 0 4 2 4 3 4 4 3 16 16 2 a a a d P P a a a a a     = − + −               = +           = + = ( ) 2 2 4 6 2 2 2 2 3 , 0 4 2 4 3 4 4 3 16 16 2 a a a d P P a a a a a     = − + −               = − +           = + = ( ) 2 2 5 6 2 2 2 3 3 3 , 4 4 4 4 0 2 4 2 a a a a d P P a a a     = − + −             = +     = = Since the sides are the same length, the triangle is equilateral.

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    Chapter 1: Graphs and Functions 10 59. 2 2 1 2 2 2 ( , ) ( 4 2) (1 1) ( 6) 0 36 6 d P P = − − + − = − + = = ( ) 2 2 2 3 2 2 ( , ) 4 ( 4) ( 3 1) 0 ( 4) 16 4 d P P = − − − + − − = + − = = 2 2 1 3 2 2 ( , ) ( 4 2) ( 3 1) ( 6) ( 4) 36 16 52 2 13 d P P = − − + − − = − + − = + = = Since [ ] [ ] [ ] 2 2 2 1 2 2 3 1 3 ( , ) ( , ) ( , ) d P P d P P d P P + = , the triangle is a right triangle. 60. ( ) 2 2 1 2 2 2 ( , ) 6 ( 1) (2 4) 7 ( 2) 49 4 53 d P P = − − + − = + − = + = ( ) 2 2 2 3 2 2 ( , ) 4 6 ( 5 2) ( 2) ( 7) 4 49 53 d P P = − + − − = − + − = + = ( ) 2 2 1 3 2 2 ( , ) 4 ( 1) ( 5 4) 5 ( 9) 25 81 106 d P P = − − + − − = + − = + = Since [ ] [ ] [ ] 2 2 2 1 2 2 3 1 3 ( , ) ( , ) ( , ) d P P d P P d P P + = , the triangle is a right triangle. Since ( ) ( ) 1 2 2 3 , , d P P d P P = , the triangle is isosceles. Therefore, the triangle is an isosceles right triangle. 61. ( ) ( ) 2 2 1 2 2 2 ( , ) 0 ( 2) 7 ( 1) 2 8 4 64 68 2 17 d P P = − − + − − = + = + = = ( ) 2 2 2 3 2 2 ( , ) 3 0 (2 7) 3 ( 5) 9 25 34 d P P = − + − = + − = + = ( ) ( ) 2 2 1 3 2 2 ( , ) 3 ( 2) 2 ( 1) 5 3 25 9 34 d P P = − − + − − = + = + = Since 2 3 1 3 ( , ) ( , ) d P P d P P = , the triangle is isosceles. Since [ ] [ ] [ ] 2 2 2 1 3 2 3 1 2 ( , ) ( , ) ( , ) d P P d P P d P P + = , the triangle is also a right triangle. Therefore, the triangle is an isosceles right triangle. 62. ( ) ( ) 2 2 1 2 2 2 ( , ) 4 7 0 2 ( 11) ( 2) 121 4 125 5 5 d P P = − − + − = − + − = + = = ( ) 2 2 2 3 2 2 ( , ) 4 ( 4) (6 0) 8 6 64 36 100 10 d P P = − − + − = + = + = = ( ) ( ) 2 2 1 3 2 2 ( , ) 4 7 6 2 ( 3) 4 9 16 25 5 d P P = − + − = − + = + = = Since [ ] [ ] [ ] 2 2 2 1 3 2 3 1 2 ( , ) ( , ) ( , ) d P P d P P d P P + = , the triangle is a right triangle.

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    Section 1.1: The Distance and Midpoint Formulas 11 63. Using the Pythagorean Theorem: 2 2 2 2 2 90 90 8100 8100 16200 16200 90 2 127.28 feet d d d d + = + = = = = ≈ 90 90 90 90 d 64. Using the Pythagorean Theorem: 2 2 2 2 2 60 60 3600 3600 7200 7200 60 2 84.85 feet d d d d + = + = → = = = ≈ 60 60 60 60 d 65. a. First: (90, 0), Second: (90, 90), Third: (0, 90) (0,0) (0,90) (90,0) (90,90) X Y b. Using the distance formula: 2 2 2 2 (310 90) (15 90) 220 ( 75) 54025 5 2161 232.43 feet d = − + − = + − = = ≈ c. Using the distance formula: 2 2 2 2 (300 0) (300 90) 300 210 134100 30 149 366.20 feet d = − + − = + = = ≈ 66. a. First: (60, 0), Second: (60, 60) Third: (0, 60) (0,0) (0,60) (60,0) (60,60) x y b. Using the distance formula: 2 2 2 2 (180 60) (20 60) 120 ( 40) 16000 40 10 126.49 feet d = − + − = + − = = ≈ c. Using the distance formula: 2 2 2 2 (220 0) (220 60) 220 160 74000 20 185 272.03 feet d = − + − = + = = ≈ 67. The Focus heading east moves a distance 30 t after t hours. The truck heading south moves a distance 40 t after t hours. Their distance apart after t hours is: 2 2 2 2 2 (30 ) (40 ) 900 1600 2500 50 miles d t t t t t t = + = + = = d 40t 30t 68. 15 miles 5280 ft 1 hr 22 ft/sec 1 hr 1 mile 3600 sec ⋅ ⋅ = ( ) 2 2 2 100 22 10000 484 feet d t t = + = +

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