Solutions Manual Precalculus 11th Edition By Ron Larson

Solutions Manual For
Precalculus 11th 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 11th 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 ....................................................................................................859

Section 10.2 Introduction to Conics: Parabolas ......................................................873

Section 10.3 Ellipses ................................................................................................883

Section 10.4 Hyperbolas ..........................................................................................896

Section 10.5 Rotation of Conics ..............................................................................911

Section 10.6 Parametric Equations..........................................................................929

Section 10.7 Polar Coordinates ...............................................................................946

Section 10.8 Graphs of Polar Equations .................................................................957

Section 10.9 Polar Equations of Conics ..................................................................968

Review Exercises ........................................................................................................978

Problem Solving .........................................................................................................995

Practice Test..............................................................................................................1003




© 2022 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

,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
3. The tangent of the angle between two nonperpendicular
lines with slopes m1 and m2 is given by the formula π
= radians = 60°
3
m2 − m1
tan θ = .
1 + m1m2 2
19. m =
3
4. The distance d between the point ( x1 , y1 ) and the line 2
= tan θ
Ax1 + By1 + C 3
Ax + By + C = 0 is given by d = .  2
A2 + B 2 θ = arctan  
 3
π 3 ≈ 0.5880 radian ≈ 33.7°
5. m = tan =
6 3
1
20. m =
π 4
6. m = tan = 1 1
4 = tan θ
4
3π 1
7. m = tan = −1 θ = tan −1  
4  4
≈ 0.2450 radian ≈ 14.0°
2π
8. m = tan = − 3
3 21. m = −1
π −1 = tan θ
9. m = tan = 3
3 θ = 180° + arctan ( −1)
3π
5π 3 = radians = 135°
10. m = tan = − 4
6 3
22. m = − 3
11. m = tan 0.39 ≈ 0.4111
− 3 = tan θ
12. m = tan 0.63 ≈ 0.7291 θ = tan −1 − 3 + π( )
13. m = tan 1.27 ≈ 3.2236 2π
= radians = 120°
3
14. m = tan 1.35 ≈ 4.4552
3
23. m = −
2
15. m = tan 1.81 ≈ − 4.1005
3
− = tan θ
16. m = tan 2.88 ≈ −0.2677 2
 3
θ = tan −1  −  + π
17. m = 1  2
1 = tan θ ≈ 2.1588 radians ≈ 123.7°
π
θ = radian = 45°
4




© 2022 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 859

, 860 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 ≈ 94.8°
− = tan θ
3
 1  5π  2 3   11 1 
θ = arctan −  = radians = 150° 34.  , − ,  − , − 
 3 6  5 4   10 4 

( )
− 14 − − 43 1
1
28. 3,( )(
3 , 6, − 2 3 ) m = 11
− 10 − 2
= − 2
− 23
= −
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



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