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Kunci Jawaban dan Pembahasan Soal || Buku Calculus 9th Purcell Chapter 0 - 0.3 Number 1 – 28

Pembahasan Soal Buku Calculus 9th Purcell Chapter 0 Section 0.3

◀ CHAPTER 0 PRELIMINARIES ▶

◀ SECTION 0.3 The Rectangular Coordinate System ▶


Concepts Review
  1. The distance between the points (–2, 3) and (x, y) is ______.
  2. The equation of the circle of radius 5 and center (–4, 2) is ______.
  3. The midpoint of the line segment joining (–2, 3) and (5, 7) is ______.
  4. The line through (a, b) and (c, d) has slope m = ______ provided a ≠ c.
Answer:
  1. √[(x + 2)2 + (y – 3)2]
  2. (x + 4)2 + (y – 2)2 = 25
  3. (–2 + 5/2 , 3 + 7/2) = (1.5 , 5)
  4. d – b/c – a
Calculus 9th Purcell Chapter 0 - 0.3

Problem Set 0.1, Number 1 – 28.
    In Problems 1-4, plot the given points in the coordinate plane and then find the distance between them.
  1. (3, 1), (1, 1)
    2. Answer:
    Calculus 9th Purcell Chapter 0 - 0.3 Number 1
    d = √[(3 – 1)2 + (1 – 1)2] = √4 = 2

  2. (–3, 5), (2, –2)
    3. Answer:
    Calculus 9th Purcell Chapter 0 - 0.3 Number 2
    d = √[(–3 – 2)2 + (5 + 2)2] = √74 ≈ 8.60

  3. (4, 5), (5, –8)
    4. Answer:
    Calculus 9th Purcell Chapter 0 - 0.3 Number 3
    d = √[(4 – 5)2 + (5 + 8)2] = √170 ≈ 13.04

  4. (–1, 5), (6, 3)
    5. Answer:
    Calculus 9th Purcell Chapter 0 - 0.3 Number 4
    d = √[(–1 – 6)2 + (5 – 3)2] = √[49 + 4] = √53 ≈ 7.28

  5. Show that the triangle whose vertices are (5, 3), (–2, 4), and (10,8) is isosceles.
    6. Answer:
    d1 = √[(5 + 2)2 + (3 – 4)2] = √[49 + 1] = √50
    d2 = √[(5 – 10)2 + (3 – 8)2] = √[25 + 25] = √50
    d3 = √[(–2 – 10)2 + (4 – 8)2] = √[144 + 16] = √160
    d1 = d2 so the triangle is isosceles.

  6. Show that the triangle whose vertices are (2, –4), (4, 0), and (8, –2) is a right triangle.
    7. Answer:
    a = √[(2 – 4)2 + (–4 – 0)2] = √4 + 16 = √20
    b = √[(4 – 8)2 + (0 + 2)2] = √16 + 4 = √20
    c = √[(2 – 8)2 + (–4 + 2)2] = √36 + 4 = √40
    a2 + b2 = c2, so the triangle is a right triangle.

  7. The points (3, –1) and (3, 3) are two vertices of a square. Give three other pairs of possible vertices.
    8. Answer:
    (–1, –1), (–1, 3); (7, –1), (7, 3); (1, 1), (5, 1)

  8. Find the point on the x-axis that is equidistant from (3, 1) and (6, 4).
    9. Answer:
    √[(x – 3)2 + (0 – 1)2]
    =
    √[(x – 6)2 + (0 – 4)2] ;
    x2 – 6x + 10
    =
    x2 – 12x + 52
    6x
    =
    42
    x
    =
    7 ⇒ (7, 0)

  9. Find the distance between (–2, 3) and the midpoint of the segment joining (–2, –2) and (4, 3).
    10. Answer:
    (–2 + 4/2 , –2 + 3/2) = (1 , 1/2) ;
    d = √[(1 + 2)2 + (1/2 – 3)2] = √[9 + 25/4] ≈ 3.91

  10. Find the length of the line segment joining the midpoints of the segments AB and CD, where A = (1, 3), B = (2, 6), C = (4, 7), and D = (3, 4).
    11. Answer:
    midpoint of AB = (1 + 2/2 , 3 + 6/2) = (3/2 , 3/2)
    midpoint of CD = (4 + 3/2 , 7 + 4/2) = (7/2 , 11/2)
    d = √[(3/2 – 7/2)2 + (9/2 – 11/2)2]
    d = √[4 + 1] = √5 ≈ 2.24

    In Problems 11-16, find the equation of the circle satisfying the given conditions.
  11. Center (1, 1), radius 1
    12. Answer:
    (x – 2)2 + (y – 1)2 = 1

  12. Center (–2, 3), radius 4
    13. Answer:
    (x + 2)2 + (y – 3)2 = 42
    (x + 2)2 + (y – 3)2 = 16

  13. Center (2, –1), goes through (5, 3)
    14. Answer:
    (x – 2)2 + (y + 1)2 = r2
    (5 – 2)2 + (3 + 1)2 = r2
    r2 = 9 + 16 = 25
    (x – 2)2 + (y + 1)2 = 25

  14. Center (4, 3), goes through (6, 2)
    15. Answer:
    (x – 4)2 + (y – 3)2 = r2
    (6 – 4)2 + (2 – 3)2 = r2
    r2 = 4 + 1 = 5
    (x – 4)2 + (y – 3)2 = 5

  15. Diameter AB, where A = (1, 3) and B = (3, 7)
    16. Answer:
    center = (1 + 3/2 , 3 + 7/2) = (2, 5)
    radius = ½ √[(1 – 3)2 + (3 – 7)2] = ½ √[4 + 16]
    radius = ½ √20 = √5
    (x – 2)2 + (y – 5)2 = 5

  16. Center (3, 4) and tangent to x-axis
    17. Answer:
    Since the circle is tangent to the x-axis, r = 4.
    (x – 3)2 + (y – 4)2 = 16

    In Problems 17-22, find the center and radius of the circle with the given equation.
  17. x2 + 2x + 10 + x2 – 6y – 10 = 0
    18. Answer:
    x2 + 2x + 10 + y2 – 6y – 10
    =
    0
    x2 + 2x + y2 – 6y
    =
    0
    (x2 + 2x + 1) + (y2 – 6y + 9)
    =
    1 + 9
    (x + 1)2 + (y – 3)2
    =
    10
    Center = (–1, 3); radius = √10

  18. x2 + y2 – 6y = 16
    19. Answer:
    x2 + y2 – 6y
    =
    16
    x2 + (y2 – 6y + 9)
    =
    16 + 9
    x2 + (y – 3)2
    =
    25
    Center = (0, 3); radius = 5

  19. x2 + y2 – 12x + 35 = 0
    20. Answer:
    x2 + y2 – 12x + 35
    =
    0
    x2 – 12x + y2
    =
    –35
    (x2 – 12x + 36) + y2
    =
    –35 + 36
    (x – 6)2 + y2
    =
    1
    Center = (6, 0); radius = 1

  20. x2 + y2 – 10x + 10y = 0
    21. Answer:
    x2 + y2 – 10x + 10y
    =
    0
    (x2 – 10x + 25) + (y2 + 10y + 25)
    =
    25 + 25
    (x – 5)2 + (y + 5)2
    =
    50
    Center = (5, –5); radius = √50 = 5√2

  21. 4x2 + 16x + 15 + 4y2 + 6y = 0
    22. Answer:
    4x2 + 16x + 15 + 4y2 + 6y
    =
    0
    4(x2 + 4x + 4) + 4(y2 + 3/2 · y + 9/16)
    =
    –15 + 16 + 9/4
    4(x + 2)2 + 4(y + 3/4)2
    =
    13/4
    (x + 2)2 + (y + 3/4)2
    =
    13/16
    Center = (–2, –3/4); radius = √13/4

  22. x2 + 16x + 105/16 + 4y2 + 3y = 0
    23. Answer:
    4x2 + 16x + 105/16 + 4y2 + 3y
    =
    0
    4(x2 + 4x + 4) + 4(y2 + 3/2 · y + 9/64)
    =
    – 105/16 + 16 + 9/16
    4(x + 2)2 + 4(y + 3/8)2
    =
    10
    (x + 2)2 + (y + 3/8)2
    =
    5/2
    Center = (–2, –3/8); radius = √(5/2) = √10/2

    In Problems 23-28, find the slope of the line containing the given two points.
  23. (1, 1) and (2, 2)
    24. Answer:
     
    2 – 1
    =
    1
    2 – 1

  24. (3, 5) and (4, 7)
    25. Answer:
     
    7 – 5
    =
    2
    4 – 3

  25. (2, 3) and (–5, –6)
    26. Answer:
     
    –6 – 3
    =
    9
    –5 – 2
    7

  26. (2, –4) and (0, –6)
    27. Answer:
     
    –6 + 4
    =
    1
    0 – 2

  27. (3, 0) and (0, 5)
    28. Answer:
     
    5 – 0
    =
    –
    5
    0 – 3
    3

  28. (–6, 0) and (0, 6)
    29. Answer:
     
    6 – 0
    =
    1
    0 + 6

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