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Precalculus Enhanced with Graphing Utilities 4th Edition Solution Manual

Solve your textbook questions with ease using Precalculus Enhanced with Graphing Utilities 4th Edition Solution Manual, a comprehensive and easy-to-follow guide.

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    1 Chapter 1 Graphs 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. x - coordinate, or abscissa; y-coordinate, or ordinate. 6. quadrants 7. midpoint 8. False; the distance between two points is never negative. 9. 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. 10. True; 1 2 1 2 , 2 2 x x y y M + + ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ 11. (a) Quadrant II (b) Positive x-axis (c) Quadrant III (d) Quadrant I (e) Negative y-axis (f) Quadrant IV 12. (a) Quadrant I (b) Quadrant III (c) Quadrant II (d) Quadrant I (e) Positive y-axis (f) Negative x-axis 13. The points will be on a vertical line that is two units to the right of the y-axis. 14. The points will be on a horizontal line that is three units above the x-axis.

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    Chapter 1: Graphs ISM: Precalculus EGU 2 15. ( ) 1, 4 − ; Quadrant II 16. (3, 4); Quadrant I 17. (3, 1); Quadrant I 18. ( ) 6, 4 − − ; Quadrant III 19. min 11 max 5 scl 1 min 3 max 6 scl 1 X X X Y Y Y = − = = = − = = 20. min 3 max 7 scl 1 min 4 max 9 scl 1 X X X Y Y Y = − = = = − = = 21. min 30 max 50 scl 10 min 90 max 50 scl 10 X X X Y Y Y = − = = = − = = 22. min 90 max 30 scl 10 min 50 max 70 scl 10 X X X Y Y Y = − = = = − = = 23. min 10 max 110 scl 10 min 10 max 160 scl 10 X X X Y Y Y = − = = = − = = 24. min 20 max 110 scl 10 min 10 max 60 scl 10 X X X Y Y Y = − = = = − = = 25. min 6 max 6 scl 2 min 4 max 4 scl 2 X X X Y Y Y = − = = = − = = 26. min 3 max 3 scl 1 min 2 max 2 scl 1 X X X Y Y Y = − = = = − = = 27. min 6 max 6 scl 2 min 1 max 3 scl 1 X X X Y Y Y = − = = = − = = 28. min 9 max 9 scl 3 min 12 max 4 scl 4 X X X Y Y Y = − = = = − = = 29. min 3 max 9 scl 1 min 2 max 10 scl 2 X X X Y Y Y = = = = = =

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    ISM: Precalculus EGU Chapter 1: Graphs 3 30. min 22 max 10 scl 2 min 4 max 8 scl 1 X X X Y Y Y = − = − = = = = 31. 2 2 1 2 ( , ) (2 0) (1 0) 4 1 5 d P P = − + − = + = 32. 2 2 1 2 ( , ) ( 2 0) (1 0) 4 1 5 d P P = − − + − = + = 33. 2 2 1 2 ( , ) ( 2 1) (2 1) 9 1 10 d P P = − − + − = + = 34. ( ) 2 2 1 2 ( , ) 2 ( 1) (2 1) 9 1 10 d P P = − − + − = + = 35. ( ) ( ) ( ) 2 2 2 2 1 2 ( , ) (5 3) 4 4 2 8 4 64 68 2 17 d P P = − + − − = + = + = = 36. ( ) ( ) ( ) ( ) 2 2 2 2 1 2 ( , ) 2 1 4 0 3 4 9 16 25 5 d P P = − − + − = + = + = = 37. ( ) 2 2 1 2 2 2 ( , ) 6 ( 3) (0 2) 9 ( 2) 81 4 85 d P P = − − + − = + − = + = 38. ( ) ( ) 2 2 2 2 1 2 ( , ) 4 2 2 ( 3) 2 5 4 25 29 d P P = − + − − = + = + = 39. ( ) 2 2 2 2 1 2 ( , ) (6 4) 4 ( 3) 2 7 4 49 53 d P P = − + − − = + = + = 40. ( ) ( ) 2 2 1 2 2 2 ( , ) 6 ( 4) 2 ( 3) 10 5 100 25 125 5 5 d P P = − − + − − = + = + = = 41. ( ) 2 2 1 2 2 2 ( , ) 2.3 ( 0.2 (1.1 0.3) (2.5) (0.8) 6.25 0.64 6.89 2.625 d P P = − − + − = + = + = ≈ 42. ( ) 2 2 1 2 2 2 ( , ) 0.3 1.2 (1.1 2.3) ( 1.5) ( 1.2) 2.25 1.44 3.69 1.92 d P P = − − + − = − + − = + = ≈ 43. 2 2 2 2 1 2 ( , ) (0 ) (0 ) d P P a b a b = − + − = + 44. 2 2 2 2 1 2 2 ( , ) (0 ) (0 ) 2 2 d P P a a a a a a = − + − = + = = 45. ( ) ( ) 1 2 1,3 ; 5,15 P P = = ( ) ( ) ( ) ( ) ( ) 2 2 1 2 2 2 , 5 1 15 3 4 12 16 144 160 4 10 d P P = − + − = + = + = = 46. ( ) ( ) 1 2 8, 4 ; 2,3 P P = − − = ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 1 2 2 2 , 2 8 3 4 10 7 100 49 149 d P P = − − + − − = + = + = 47. ( ) ( ) 1 2 4, 6 ; 4, 8 P P = − = − ( ) ( ) ( ) ( ) ( ) ( ) 2 2 1 2 2 2 , 4 4 8 6 8 14 64 196 260 2 65 d P P = − − + − − = + − = + = = 48. ( ) ( ) 1 2 0, 6 ; 3, 8 P P = = − ( ) ( ) ( ) ( ) ( ) 2 2 1 2 2 2 , 3 0 8 6 3 14 9 196 205 d P P = − + − − = + − = + =

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    Chapter 1: Graphs ISM: Precalculus EGU 4 49. ( 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, [ ] [ ] 1 ( , ) ( , ) 2 1 13 13 2 1 13 2 13 square units 2 A d A B d B C = ⋅ ⋅ = ⋅ ⋅ = ⋅ = 50. ( 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 2 100 square units A d A B d B C = ⋅ ⋅ = ⋅ ⋅ = ⋅ ⋅ =

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    ISM: Precalculus EGU Chapter 1: Graphs 5 51. ( 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 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 = ⋅ ⋅ = ⋅ ⋅ = ⋅ ⋅ = ⋅ ⋅ = 52. ( 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 ( , ) 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 ( , ) ( , ) ( , ) 29 116 145 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 29 116 2 1 29 2 29 2 1 2 29 2 29 square units A d A C d B C = ⋅ ⋅ = ⋅ ⋅ = ⋅ ⋅ = ⋅ ⋅ =

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    Chapter 1: Graphs ISM: Precalculus EGU 6 53. (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 problem, [ ] [ ] 1 ( , ) ( , ) 2 1 4 5 2 10 square units A d A B d A C = ⋅ ⋅ = ⋅ ⋅ = 54. (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 = ⋅ ⋅ = ⋅ ⋅ =

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    ISM: Precalculus EGU Chapter 1: Graphs 7 55. 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 + + ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ − + + ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ = 56. 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 + + ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ − + + ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ = 57. 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 + + ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ − + + ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ 58. 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 + + ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ + − + ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ − ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ = − ⎜ ⎟ ⎝ ⎠ 59. 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 + + ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ + − + ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ − ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ = − 60. 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 + + ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ − + − + ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ − − ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ = − − ⎜ ⎟ ⎝ ⎠ 61. The coordinates of the midpoint are: 1 2 1 2 ( , ) , 2 2 0.2 2.3 0.3 1.1 , 2 2 2.1 1.4 , 2 2 (1.05, 0.7) x x y y x y + + ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ − + + ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ = 62. The coordinates of the midpoint are: ( ) 1 2 1 2 ( , ) , 2 2 1.2 ( 0.3) 2.3 1.1 , 2 2 0.9 3.4 , 2 2 0.45, 1.7 x x y y x y + + ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ + − + ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ = 63. 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 + + ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ + + ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠

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    Chapter 1: Graphs ISM: Precalculus EGU 8 64. 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 + + ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ + + ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ 65. Consider points of the form ( ) 2, y that are a distance of 5 units from the point ( ) 2, 1 − − . ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 1 2 1 2 2 2 2 2 2 2 2 1 4 1 16 1 2 2 17 d x x y y y y y y y y = − + − = − − + − − = − + − − = + + + = + + ( ) ( )( ) 2 2 2 2 2 2 5 2 17 5 2 17 25 2 17 0 2 8 0 4 2 y y y y y y y y y y = + + = + + = + + = + − = + − 4 0 4 y y + = = − or 2 0 2 y y − = = Thus, the points ( ) 2, 4 − and ( ) 2, 2 are a distance of 5 units from the point ( ) 2, 1 − − . 66. Consider points of the form ( ) , 3 x − that are a distance of 13 units from the point ( ) 1, 2 . ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 1 2 1 2 2 2 2 2 2 1 2 3 2 1 5 2 1 25 2 26 d x x y y x x x x x x x = − + − = − + − − = − + + = − + + = − + ( ) ( )( ) 2 2 2 2 2 2 13 2 26 13 2 26 169 2 26 0 2 143 0 13 11 x x x x x x x x x x = − + = − + = − + = − − = − + 13 0 13 x x − = = or 11 0 11 x x + = = − Thus, the points ( ) 13, 3 − and ( ) 11, 3 − − are a distance of 13 units from the point ( ) 1, 2 . 67. 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 5 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 5 8 25 5 8 25 25 8 25 0 8 0 8 x x x x x x x x x x = − + = − + = − + = − = − 0 x = or 8 0 8 x x − = = Thus, the points ( ) 0, 0 and ( ) 8, 0 are on the x- axis and a distance of 5 units from the point ( ) 4, 3 − .

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    ISM: Precalculus EGU Chapter 1: Graphs 9 68. 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 5 units from the point ( ) 4, 4 . ( ) ( ) ( ) ( ) 2 2 2 1 2 1 2 2 2 2 2 2 4 0 4 4 16 8 16 16 8 8 32 d x x y y y y y y y y y = − + − = − + − = + − + = + − + = − + ( ) ( )( ) 2 2 2 2 2 2 5 8 32 5 8 32 25 8 32 0 8 7 0 7 1 y y y y y y y y y y = − + = − + = − + = − + = − − 7 0 7 y y − = = or 1 0 1 y y − = = Thus, the points ( ) 0, 7 and ( ) 0,1 are on the y- axis and a distance of 5 units from the point ( ) 4, 4 . 69. 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 = − + − = + = + = 70. 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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    Chapter 1: Graphs ISM: Precalculus EGU 10 71. Let ( ) 1 0, 0 P = , ( ) 2 0, P s = , ( ) 3 , 0 P s = , and ( ) 4 , P s s = . (0,s) (s,0) (0,0) (s,s) X Y 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. 72. 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 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 D M a a a a a E M a a a a F M + + ⎛ ⎞ ⎛ ⎞ = = = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ ⎞ ⎜ ⎟ ⎛ ⎞ + + ⎜ ⎟ = = = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ + + ⎜ ⎟ ⎛ ⎞ ⎜ ⎟ = = = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎜ ⎟ ⎝ ⎠ ( ) 2 2 2 2 2 2 3 3 , 0 4 2 4 3 4 4 3 16 16 2 a a a d D E a a a a a ⎛ ⎞ ⎛ ⎞ = − + − ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ ⎞ ⎛ ⎞ = + ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ = + = ( ) 2 2 2 2 2 2 3 , 0 4 2 4 3 4 4 3 16 16 2 a a a d D F a a a a a ⎛ ⎞ ⎛ ⎞ = − + − ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ ⎞ ⎛ ⎞ = − + ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ = + = ( ) 2 2 2 2 2 3 3 3 , 4 4 4 4 0 2 4 2 a a a a d E F a a a ⎛ ⎞ ⎛ ⎞ = − + − ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ ⎞ = + ⎜ ⎟ ⎝ ⎠ = = Since the sides are the same length, the triangle is equilateral.

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    ISM: Precalculus EGU Chapter 1: Graphs 11 73. 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. 74. ( ) 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. 75. ( ) ( ) 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. 76. ( ) ( ) 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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    Chapter 1: Graphs ISM: Precalculus EGU 12 77. 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 78. 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 79. 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 232.4 feet d = − + − = + − = ≈ c. Using the distance formula: 2 2 2 2 (300 0) (300 90) 300 210 134100 366.2 feet d = − + − = + = ≈ 80. 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 126.5 feet d = − + − = + − = ≈ c. Using the distance formula: 2 2 2 2 (220 0) (220 60) 220 160 74000 272.0 feet d = − + − = + = ≈ 81. The Neon 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 d t t t t t t = + = + = = d 40t 30t

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    ISM: Precalculus EGU Chapter 1: Graphs 13 82. 15 miles 5280 ft 1 hr 22 ft/sec 1 hr 1 mile 3600 sec ⋅ ⋅ = ( ) 2 2 2 100 22 10000 484 d t t = + = + 100 22 t d Section 1.2 1. ( ) ( ) 2 3 1 7 2 3 6 3 3 6 x x x x + − = − + = − + = − = − The solution set is { } 6 − . 2. ( )( ) 2 2 9 5 0 2 1 5 0 x x x x − − = + − = 2 1 0 2 1 1 2 x x x + = = − = − or 5 0 5 x x − = = The solution set is { } 1 ,5 2 − . 3. intercepts 4. zeros; roots 5. y-axis 6. 4 7. ( ) 3, 4 − 8. True 9. False; we solve for y in terms of x . 10. False; a graph can be symmetric with respect to both coordinate axes (in such cases it will also be symmetric with respect to the origin). For example: 2 2 1 x y + = 11. 12. 13. 14.

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    Chapter 1: Graphs ISM: Precalculus EGU 14 15. 5 − 5 y 5 − 5 a = (5, 2) b = ( 5, 2) − − c = ( 5, 2) − (5, −2) 16. 17. 18. 19. 20. 21. 4 y x x = − 4 0 0 0 0 0 = − = 4 1 1 1 1 0 = − ≠ 4 0 ( 1) 1 0 1 1 = − − − ≠ − − (0, 0) is on the graph of the equation. 22. 3 2 y x x = − 3 0 0 2 0 0 0 = − = 3 1 1 2 1 1 1 = − ≠ − 3 1 1 2 1 1 1 − = − − = − (0, 0) and (1, –1) are on the graph of the equation. 23. 2 2 9 y x = + 2 2 3 0 9 9 9 = + = 2 2 0 3 9 0 18 = + ≠ 2 2 0 ( 3) 9 0 18 = − + ≠ (0, 3) is on the graph of the equation. 24. 3 1 y x = + 3 2 1 1 8 2 = + ≠ 3 1 0 1 1 1 = + = 3 0 1 1 0 0 = − + = (0, 1) and (–1, 0) are on the graph of the equation.

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    ISM: Precalculus EGU Chapter 1: Graphs 15 25. 2 2 4 x y + = 2 2 0 2 4 4 4 + = = 2 2 ( 2) 2 4 8 4 − + = ≠ 2 2 2 2 4 4 4 + = = ( ) (0, 2) and 2, 2 are on the graph of the equation. 26. 2 2 4 4 x y + = 2 2 0 4 1 4 4 4 + ⋅ = = 2 2 2 4 0 4 4 4 + ⋅ = = ( ) 2 2 1 2 2 4 4 5 4 + = ≠ (0, 1) and (2, 0) are on the graph of the equation. 27. a. Intercepts: ( ) 1, 0 − and ( ) 1, 0 b. Symmetric with respect to the x-axis, y-axis, and the origin. 28. a. Intercepts: ( ) 0,1 b. Not symmetric to x-axis, y-axis, or origin 29. a. Intercepts: ( ) 2 0 , π − , ( ) 0,1 , and ( ) 2 , 0 π b. Symmetric with respect to the y-axis. 30. a. Intercepts: ( ) 2, 0 − , ( ) 0, 3 − , and ( ) 2, 0 b. Symmetric with respect to the y-axis. 31. a. Intercepts: ( ) 0, 0 b. Symmetric with respect to the x-axis. 32. a. Intercepts: ( ) 2, 0 − , ( ) 0, 2 , ( ) 0, 2 − , and ( ) 2, 0 b. Symmetric with respect to the x-axis, y-axis, and the origin. 33. a. Intercepts: ( ) 2, 0 − , ( ) 0, 0 , and ( ) 2, 0 b. Symmetric with respect to the origin. 34. a. Intercepts: ( ) 4, 0 − , ( ) 0, 0 , and ( ) 4, 0 b. Symmetric with respect to the origin. 35. − 10 y x 5 − 5 (0, 9) − 10 (2, 5) − ( 2, 5) − − 36. y x 5 − 5 (5, 3) (5, 3) − ( 4, 0) − − 5 5 (0, 2) − (0, 2) 37. − 5 y x 5 π −π , 2 2 π ⎛ ⎞ − − ⎜ ⎟ ⎝ ⎠ , 2 2 π ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ (π, 0) (−π, 0) 38.

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