Solutions Manual Precalculus 10th Edition By Ron Larson

Solutions Manual For
Precalculus 10th Edition By Ron
Larson


(All Chapters 1-10, 100%
Original Verified, A+ Grade)


All Chapters Arranged Reverse:
10-1
This is The Original Solutions
Manual For 10th Edition, All
other Files in The Market are
Fake/Old/Wrong Edition.

, C H A P T E R 1 0
Topics in Analytic Geometry

Section 10.1 Lines ....................................................................................................821

Section 10.2 Introduction to Conics: Parabolas ......................................................832

Section 10.3 Ellipses ................................................................................................842

Section 10.4 Hyperbolas ..........................................................................................852

Section 10.5 Rotation of Conics ..............................................................................865

Section 10.6 Parametric Equations..........................................................................883

Section 10.7 Polar Coordinates ...............................................................................900

Section 10.8 Graphs of Polar Equations .................................................................910

Section 10.9 Polar Equations of Conics ..................................................................921

Review Exercises ........................................................................................................930

Problem Solving .........................................................................................................946

Practice Test................................................................................................................954

,C H A P T E R 1 0
Topics in Analytic Geometry
Section 10.1 Lines
1. inclination 18. m = 3
2. tan θ 3 = tan θ
θ = tan −1 3
m2 − m1
3. π
1 + m1m2 = radians = 60°
3
Ax1 + By1 + C 2
4. . 19. m =
A2 + B 2 3
2
π 3 = tan θ
5. m = tan = 3
6 3  2
θ = arctan  
π  
3
6. m = tan =1 ≈ 0.5880 radian ≈ 33.7°
4

3π 1
7. m = tan = −1 20. m =
4 4
1
= tan θ
2π 4
8. m = tan = − 3
3 1
θ = tan −1  
 4
π
9. m = tan = 3 ≈ 0.2450 radian ≈ 14.0°
3
21. m = −1
5π 3
10. m = tan = − −1 = tan θ
6 3
θ = 180° + arctan ( −1)
11. m = tan 0.39 ≈ 0.4111 3π
= radians = 135°
4
12. m = tan 0.63 ≈ 0.7291
22. m = − 3
13. m = tan 1.27 ≈ 3.2236
− 3 = tan θ
14. m = tan 1.35 ≈ 4.4552 θ = tan −1 − 3 + π( )
15. m = tan 1.81 ≈ − 4.1005 2π
= radians = 120°
3
16. m = tan 2.88 ≈ −0.2677
3
23. m = −
17. m = 1 2
1 = tan θ 3
− = tan θ
2
π
θ = radian = 45°  3
4 θ = tan −1  −  + π
 2
≈ 2.1588 radians ≈ 123.7°

,822 Chapter 10 Topics in Analytic Geometry


5 30. (12, 8), ( −4, − 3)
24. m = −
9
8 − ( −3) 11
5 m = =
− = tan θ 12 − ( −4) 16
9
 5 11
θ = tan −1  −  + π = tan θ
  9 16
11
≈ 2.6345 radians ≈ 150.9° θ = tan −1 ≈ 0.6023 radian ≈ 34.5°
16
25. ( )
3, 2 , (0, 1)
31. ( −2, 20), (10, 0)
1− 2 −1 1 0 − 20 20 5
m = = = m = = − = −
0 − 3 − 3 3 10 − ( −2) 12 3
1
= tan θ 5
3 − = tan θ
3
1
θ = arctan  5
3 θ = π + arctan  −  ≈ 2.1112 radians ≈ 121.0°
 3
π
= radian = 30°
6 32. (0, 100), (50, 0)

(
26. 1, 2 3 , 0, )( 3 ) m =
100 − 0
= −2
0 − 50
3 − 2 3 − 3 −2 = tan θ
m = = = 3
0 −1 −1 θ = tan −1 ( −2) + π ≈ 2.0344 radians ≈ 116.6°
3 = tan θ
π 1 3 1 1
θ = arctan 3 = radians = 60° 33.  , ,  , 
3  4 2 3 2
1 − 3
27. − ( )
3, −1 , (0, − 2) m = 2
1 −
2
1
= −
1
1
= −12
3 4 12
−2 − ( −1) −1 −12 = tan θ
m = =
0− −( 3 ) 3
θ = arctan ( −12) + π ≈ 1.6539 radians
1
− = tan θ ≈ 94.8°
3
 1  5π  2 3   11 1 
θ = arctan  −  = radians = 150° 34.  , − ,  − , − 
 3 6  5 4   10 4 

( )
− 14 − − 34 1
28. 3,( )(
3 , 6, − 2 3 ) m = 11
− 10 − 2
= − 2
− 32
= −
1
3
5
−2 3 − 3 −3 3 1
m = = = − 3 − = tan θ
6 −3 3 3
− 3 = tan θ  1
θ = arctan  −  + π ≈ 2.8198 radians
2π  3
θ = arctan − ( 3 = )3
radians = 120°
≈ 161.6°

29. (6, 1), (10, 8) 35. 2 x + 2 y − 5 = 0
8 −1 7 5
m = = y = −x +  m = −1
10 − 6 4 2
7 −1 = tan θ
= tan θ
4 3π
θ = arctan ( −1) = radians = 135°
7 4
θ = arctan ≈ 1.0517 radians ≈ 60.3°
4

, Section 10.1 Lines 823


36. x − 3y + 1 = 0 43. 4 x + 5 y − 9 = 0
9
1 1 1 y = − 54 x +  m = − 54
y = x +  m = 5
3 3 3 − 54 = tan θ
1
= tan θ
3 ( )
θ = tan −1 − 54 + π
1 π ≈ 2.4669 radians ≈ 141.3°
θ = arctan = radian = 30°
3 6
44. 5 x + 3 y = 0
37. 3x − 3 y + 1 = 0 y = − 53 x  m = − 53
1
y = x +  m =1 − 53 = tan θ
3
1 = tan θ ( )
θ = π + arctan − 53 ≈ 2.1112 radians ≈ 121.0°
π
θ = arctan 1 = radian = 45°
4 45. 3x + y = 3  y = −3 x + 3  m1 = −3
x − y = 2  y = x − 2  m2 = 1
38. 3x − y + 2 = 0
1 − ( −3)
y = 3x + 2  m = 3 tan θ = = 2
1 + ( −3)(1)
3 = tan θ
θ = arctan 2 ≈ 1.1071 radians ≈ 63.4°
π
θ = arctan 3 = radians = 60°
3 1 2 1
46. x + 3 y = 2  y = − x +  m1 = −
3 3 3
39. x + 3y + 2 = 0
1 3 1
1 2 1 x − 2 y = −3  y = x +  m2 =
y = − x −  m = − 2 2 2
3 3 3
1 tan θ =
(1 2) − ( −1 3) =1
− = tan θ 1 + ( −1 3)(1 2)
3
 1  5π π
θ = arctan  − θ = tan −1 1 = 45° = radian
 = radians = 150° 4
 3 6

47. x − y = 0  y = x  m1 = 1
40. −2 3x − 2 y = 0
3 1 3
y = − 3x  m = − 3 3x − 2 y = −1  y = x +  m2 =
2 2 2
− 3 = tan θ 3
−1
2π 1
θ = arctan − ( 3 =) 3
radians = 120° tan θ = 2
 3
=
5
1 +  (1)
 2
41. 6 x − 2 y + 8 = 0
1
y = 3x + 4  m = 3 θ = arctan ≈ 0.1974 radian ≈ 11.3°
5
3 = tan θ
θ = arctan 3 ≈ 1.2490 radians ≈ 71.6° 48. 2x − y = 2  y = 2x − 2  m1 = 2
4 4
4 x + 3 y = 24  y = − x + 8  m2 = −
42. 2 x − 6 y − 12 = 0 3 3
1x 1
y = 3
− 2  m = 3 (−4 3) − 2
tan θ = = 2
1
= tan θ 1 + ( 2)( −4 3)
3

θ = arctan 1
≈ 0.3218 radian ≈ 18.4° θ = tan −1 2 ≈ 63.4° ≈ 1.1071 radians
3

,824 Chapter 10 Topics in Analytic Geometry


1 7 1 1 1
49. x − 2y = 7  y = x −  m1 = 51. x + 2 y = 8  y = − x + 4  m1 = −
2 2 2 2 2
5 1 1
6x + 2 y = 5  y = −3 x +  m2 = −3 x − 2y = 2  y = x −1  m2 =
2 2 2

1 1  1
−3 − − − 
2 2  2 4
tan θ = = 7 tan θ = =
1  1  1  3
1 +  ( −3) 1 +  −  
2  2  2 
θ = arctan 7 ≈ 1.4289 radians ≈ 81.9° 4
θ = arctan   ≈ 0.9273 radian ≈ 53.1°
3
5 5
50. 5 x + 2 y = 16  y = − x + 8  m1 = −
2 2 3 3 3
52. 3 x − 5 y = 3  y = x −  m1 =
3 1 3 5 5 2
3x − 5 y = −1  y = x +  m2 =
5 5 5 3 12 3
3x + 5 y = 12  y = − x +  m2 = −
5 5 5
tan θ =
(−5 2) − (3 5) =
31
1 + ( −5 2)(3 5) 5
tan θ =
(3 5) − (−3 5) =
15
31 1 + (3 5)( −3 5) 8
θ = tan −1 ≈ 80.8° ≈ 1.4109 radians
5 15
θ = tan −1 ≈ 61.9° ≈ 1.0808 radians
8

5 5
53. 0.05 x − 0.03 y = 0.21  y = x − 7  m1 =
3 3
7 7
0.07 x + 0.02 y = 0.16  y = − x + 8  m2 = −
2 2
 7 5
−  −   31
tan θ =     =
2 3
 5  7  29
1 +   − 
 3  2 
 31 
θ = arctan   ≈ 0.8187 radian ≈ 46.9°
 29 

2 19 2
54. 0.02 x − 0.05 y = −0.19  y = x +  m1 =
5 5 5
3 3
0.03 x + 0.04 y = 0.52  y = − x + 13  m2 = −
4 4

tan θ =
(−3 4) − (2 5) ≈
23
1 + ( 2 5)( −3 4) 14
 23 
θ = tan −1   ≈ 58.7° ≈ 1.0240 radians
 14 