College algebra 11th edition lial solutions manual by qiqi111

Chapter 2 GRAPHS AND FUNCTIONS Section 2.1 Rectangular Coordinates

12. P(–4, 3), Q(2, –5) 2 2 (a) d ( P, Q) = [2 – (– 4)] + (–5 – 3)

and Graphs

1. False. (−1, 3) lies in Quadrant II. 2. False. The expression should be

−x 2

)2 + ( y

−y 2

3. True. The origin has coordinates (0, 0) . So, the distance from (0, 0) to (a, b) is

d = (a − 0) 2 + (b − 0) 2 = a 2 + b 2 4. True. The midpoint has coordinates ⎛ a +3a b +(−3b ) ⎞ ⎜ ⎝

⎛ 4a −2b ⎞

⎟ = ⎜ , 2 ⎟⎠ ⎠ ⎝ 2 = (2a, − b).

5. True. When x = 0, y = 2(0) + 4 = 4, so the y-intercept is 4. When y = 0, 0 = 2x + 4 ⇒ x = −2, so the x-intercept is −2. 6. Answers will vary. 7. Any three of the following: (2, −5) , (−1, 7 ) , (3, −9) , (5, −17 ) , (6, −21) 8. Any three of the following: (3, 3) , (−5, −21) , (8,18) , (4, 6) , (0, −6)

= 6 + (–8) = 100 = 10 (b) The midpoint M of the segment joining points P and Q has coordinates ⎛ – 4 +2 3 +(–5) ⎞ ⎛ −2 −2 ⎞ , = , 2 2 2 2 = (−1, −1) . 13. P(8, 2), Q(3, 5) (a) d ( P, Q) = (3 – 8)2 + (5 – 2)2 =

(−5)2

+ 32

= 25 + 9 = 34 (b) The midpoint M of the segment joining points P and Q has coordinates ⎛ 8 +3 2 +5 ⎞ ⎛ 11 7 ⎞ , = , . 2 2 2 2 14. P (−8, 4), Q (3, −5) (a) d ( P, Q) = ⎡⎣3 – (−8)⎤⎦ + (−5 − 4 ) 2

= 112 + (–9)2 = 121 + 81 = 202

9. Any three of the following: (1997, 36), (1999, 35), (2001, 29), (2003, 22), (2005, 23), (2007, 20)

(b) The midpoint M of the segment joining points P and Q has coordinates ⎛ –8 +3 4 +(–5) ⎞ ⎛ 5 1 ⎞ , ⎜ ⎟ = ⎜− , − . ⎟ ⎝ 2 2 ⎠ ⎝ 2 2⎠

10. Any three of the following:

(1998, 90.0) , (2000, 88.5) , (2002, 86.8) , (2004, 89.8) , ( 2006, 90.7) , (2008, 97.4) , (2010, 106.5)

(a) d ( P, Q) = [−13 – (–5)]2 + [1 – (–7)]2

(−8)2

+ 82 = 128 = 8 2

(b) The midpoint M of the segment joining points P and Q has coordinates ⎛ –5 +(–13) –7 +1 ⎞ ⎛ −18 −6 ⎞ , = , 2

166

2 2 (a) d ( P, Q) = [6 – (– 6)] + [10 – (–5)]

= 12 2 + 152 = 144 + 225 = 369 = 3 41

11. P(–5, –7), Q(–13, 1)

15. P(–6, –5), Q(6, 10)

2 2 = (−9, −3) .

(b) The midpoint M of the segment joining points P and Q has coordinates ⎛ – 6+6 –5 +10 ⎞ ⎛0 5 ⎞ ⎛ 5 ⎞ , ⎟ = ⎜ , ⎟ = ⎜ 0, . ⎝ 2 2 ⎠ ⎝2 2⎠ ⎝ 2⎠

Section 2.1 Rectangular Coordinates and Graphs

16. P(6, –2), Q(4, 6) (a) d ( P, Q) = (4 – 6) 2 + [6 – (–2)]2

Since

= 4 + 64 = 68 = 2 17

(b) The midpoint M of the segment joining points P and Q has coordinates ⎛ 6 +4 −2 +6 ⎞ ⎛ 10 4 ⎞ , = , = (5, 2) 2

(

2 2

) (

2, – 5

17. P 3 2 , 4 5 , Q

= (– 4)2 + 122 = 16 + 144 = 160

= (–2) 2 + 82

167

d ( A, C ) = [–10 – (– 6)] + [8 – (– 4)] 2

(

) + ( 160) = ( 2

200

)

, triangle

ABC is a right triangle. 20. Label the points A(–2, –8), B(0, –4), and C(–4, –7). Use the distance formula to find the length of each side of the triangle. d ( A, B) = [0 – (–2)]2 + [– 4 – (–8)]2

)

= 22 + 42 = 4 + 16 = 20 d ( B, C ) = (– 4 – 0)2 + [–7 – (– 4)]2

(a) d ( P, Q) 2

(

( –2 2 ) + ( –5 5 )

2 –3 2

) + (–

5 –4 5

)

= (– 4)2 + (–3) 2 = 16 + 9 = 25 = 5

d ( A, C ) = [– 4 – (–2)] + [–7 – (–8)] = 8 + 125 = 133

= (–2)2 + 12 = 4 + 1 = 5

(b) The midpoint M of the segment joining points P and Q has coordinates ⎛ 3 2 + 2 4 5 +(– 5 ) ⎞ , ⎜⎝ ⎟⎠ 2 2 =

⎛4 2 3 5⎞ ⎛ 3 5⎞ , = 2 2, . ⎜ 2 ⎝

(

2 ⎟ ⎠

) (

⎜ ⎝

18. P – 7 , 8 3 , Q 5 7 , – 3

2 ⎟ ⎠

)

(a) d ( P, Q) = [5 7 – (– 7 )]2 + (– 3 – 8 3 ) 2 = (6 7 )2 + (–9 3 ) 2 = 252 + 243

Since ( 5 )2 + ( 20 )2 = 5 + 20 = 25 = 52 , triangle ABC is a right triangle. 21. Label the points A(–4, 1), B(1, 4), and C(–6, –1). 2

d ( A, B) = [1 – (– 4)] + (4 – 1) = 52 + 32 = 25 + 9 = 34 d ( B, C ) = (– 6 – 1) 2 + (–1 – 4)2 = (–7)2 + (–5) 2 = 49 + 25 = 74 d ( A, C ) = [– 6 – (– 4)]2 + (–1 – 1) 2 2

= (–2) + (–2) = 4 + 4 = 8

= 495 = 3 55 (b) The midpoint M of the segment joining points P and Q has coordinates ⎛ – 7 +5 7 8 3 +(– 3 ) ⎞ , ⎜ ⎟ 2 2 ⎝ ⎠ ⎛4 7 7 3⎞ ⎛ 7 3⎞ = , = 2 7, . ⎜ 2 ⎟ ⎜ ⎟ 2 2

Since ( 8 ) 2 + ( 34 ) 2 ≠ ( 74 ) 2 because 8 + 34 = 42 ≠ 74, triangle ABC is not a right triangle. 22. Label the points A(–2, –5), B(1, 7), and C(3, 15). d ( A, B) = [1 − (−2)]2 + [7 − (−5)]2

⎝

⎠

⎝

⎠

19. Label the points A(–6, –4), B(0, –2), and C(–10, 8). Use the distance formula to find the

= 32 + 12 2 = 9 + 144 = 153 d ( B, C ) = (3 − 1) 2 + (15 − 7) 2 = 2 2 + 82 = 4 + 64 = 68

length of each side of the triangle. d ( A, B) = [0 – (– 6)]2 + [–2 – (– 4)]2

d ( A, C ) = [3 − (−2)]2 + [15 − (−5)]2

= 6 2 + 22 = 36 + 4 = 40

= 52 + 20 2 = 25 + 400 = 425

Since d ( B, C ) = (–10 – 0) 2 + [8 – (–2)]2 = (−10)2 + 10 2 = 100 + 100 = 200

( 68 ) 2 + ( 153 ) 2 ≠

(

425

)

because

68 + 153 = 221 ≠ 425 , triangle ABC is not a right triangle.

168

Chapter 2 Graphs and Functions

23. Label the points A(–4, 3), B(2, 5), and C(–1, –6). 2

⎡⎣ 2 – ( –4)

d ( A, B) =

+ (5 − 3)

= 2 + ( –8)

= 9 + 121 = 130 + ( −6 − 3)

) + ( 90 ) = ( 2

130

)

, triangle

24. Label the points A(–7, 4), B(6, –2), and C(0, –15). 2

d ( A, B) = ⎡⎣ 6 – ( –7 )

d ( B, C ) = =

(0 − 6)

+ ⎡⎣ –15 – ( –2)

+ (14 – 4)

= 265 ≈ 16.279 d ( B, C ) = ⎡⎣ 2 – ( –3)⎤⎦ + ⎡⎣19 – ( –7 )⎤⎦ 2

) =( 2

410

)

, triangle

= 52 + 26 2 = 25 + 676 = 701 ≈ 26.476 d ( A, C ) =

ABC is a right triangle. 25. Label the given points A(0, –7), B(–3, 5), and C(2, –15). Find the distance between each pair of points. 2

= (–3) 2 + (–16)2 = 9 + 256

+ ( −15 − 4)

205

d ( A, C ) = ⎡⎣1 – ( –1)

d ( A, B) = (–3 – 0) 2 + (–7 – 9) 2

= 7 2 + ( −19) = 49 + 361 = 410 2

27. Label the points A(0, 9), B(–3, –7), and C(2, 19).

d ( A, C ) = ⎡⎣ 0 – ( –7 )

) +(

= 26

Because 26 + 2 26 = 3 26 , the points are collinear.

( – 6)2 + ( –13)2

205

= 2 + 10 = 104 = 2 26

+ (−2 − 4 )

= 36 + 169 = 205

(

( –1)2 + ( –52 )2

Since

+ ( –1 – 4)

d ( B, C ) = ⎡⎣1 – ( –2)⎤⎦ + ⎡⎣14 – ( –1)⎤⎦

= 169 + 36 = 205 2

= 32 + 152 = 234 = 3 26

ABC is a right triangle.

= 132 + (−6)

Since d ( A, B) + d ( A, C ) = d (B, C ) or

d ( A, B) = ⎡⎣ –2 – ( –1)

(

= 68 = 2 17

points. 2

= 32 + (−9) = 9 + 81 = 90

Since

26. Label the points A(–1, 4), B(–2, –1), and C(1, 14). Apply the distance formula to each pair of

= (– 3) 2 + (–11) 2

d ( A, C ) = ⎡⎣ – 1 – ( –4 )

3 17 + 2 17 = 5 17 , the points are collinear.

(−1 − 2)2 + (−6 − 5)2

(2 − 0)2 + ⎡⎣ −15 – (−7 )⎤⎦ 2

= 6 2 + 2 2 = 36 + 4 = 40 d ( B, C ) =

d ( A, C ) =

(2 – 0)2 + (19 – 9 )2

= 2 2 + 10 2 = 4 + 100 = 104 ≈ 10.198 Since d ( A, B) + d ( A, C ) ≠ d (B, C )

d ( A, B) = =

(−3 − 0) ( –3)2

+ ⎡⎣5 – ( –7 )⎤⎦

+ 12 2 = 9 + 144

= 153 = 3 17 d ( B, C ) = ⎡⎣ 2 – (−3)

+ ( –15 – 5)

= 52 + ( –20) = 25 + 400 2

= 425 = 5 17

265 + 104 ≠ 701

16.279 + 10.198 ≠ 26.476, 26.477 ≠ 26.476, the three given points are not collinear. (Note, however, that these points are very close to lying on a straight line and may appear to lie on a straight line when graphed.)

Section 2.1 Rectangular Coordinates and Graphs

28. Label the points A(–1, –3), B(–5, 12), and C(1, –11). 2

d ( A, B) = ⎡⎣ –5 – ( –1) =

( – 4)2 + 152

+ ⎡⎣12 – ( –3) = 16 + 225

+ ( –11 – 12) 2

= 565 ≈ 23.7697 d ( A, C ) = ⎡⎣1 – ( –1)

)

5 3

d ( B, C ) = (−1 − 2)2 + (4 − 5) 2

= 62 + ( –23) = 36 + 529

+( −

= 6 2 + 22 = 36 + 4 = 40 = 2 10

d ( B, C ) = ⎡⎣1 – ( –5)

⎡2 – –4 ⎤

d ( A, B)

( )⎦

= ⎣

= 241 ≈ 15.5242

169

30. Label the given points A(–4, 3), B(2, 5), and C(–1, 4). Find the distance between each pair of points.

(−3)2 + (–1) 2

d ( A, C ) = ⎡⎣ −1 – (−4)

+ ⎡⎣ –11 – ( –3)

= 9 + 1 = 10 + ( 4 − 3)

= 3 + 1 = 9 + 1 = 10 = 2 + ( –8) = 4 + 64 2

Since d ( B, C ) + d ( A, C ) = d ( A, B) or

= 68 ≈ 8.2462

10 + 10 = 2 10 , the points are collinear.

Since d(A, B) + d(A, C) ≠ d(B, C) or 241 + 68 ≠ 565

31. Midpoint (5, 8), endpoint (13, 10)

15.5242 + 8.2462 ≠ 23.7697

13 +x

23.7704 ≠ 23.7697, the three given points are not collinear. (Note,

2 13 + x = 10

however, that these points are very close to lying on a straight line and may appear to lie on a straight line when graphed.) 29. Label the points A(–7, 4), B(6,–2), and

d ( A, B) = ⎡6 ⎣ – ( –7 )

+ ( −2 − 4)

= 132 + ( −6) = 169 + 36 2

2 10 + y = 16

x = –3 and

y = 6.

The other endpoint has coordinates (–3, 6). 32. Midpoint (–7, 6), endpoint (–9, 9) = –7

and

2 –9 + x = –14 and x = –5

and

9 +y

2 9 + y = 12 y = 3.

The other endpoint has coordinates (–5, 3).

= 205 ≈ 14.3178 2

10 +y

and

–9 +x

C(–1,1).

and

33. Midpoint (12, 6), endpoint (19, 16)

d ( B, C ) = =

(−1 − 6) (−7)2

+ ⎡⎣1 − ( –2 )

19 +x

= 12

2 19 + x = 24 and

+ 32 = 49 + 9

= 58 ≈ 7.6158

x=5 2

d ( A, C ) = ⎡⎣ −1 – ( –7 )⎤⎦ + (1 − 4)

16 +y

2 16 + y = 12

and

y = – 4.

= 6 + ( –3) = 36 + 9 2

and

= 45 ≈ 6.7082 Since d(B, C) + d(A, C) ≠ d(A, B) or 58 + 45 ≠ 205 7.6158 + 6.7082 ≠ 14.3178 14.3240 ≠ 14.3178, the three given points are not collinear. (Note, however, that these points are very close to lying on a straight line and may appear to lie on a straight line when graphed.)

The other endpoint has coordinates (5, –4). 34. Midpoint (–9, 8), endpoint (–16, 9) –16 + x

= –9

and

2 –16 + x = –18 and x = –2

and

9+ y

2 9 + y = 16 y=7

The other endpoint has coordinates (–2, 7). 35. Midpoint (a, b), endpoint (p, q) p+x q+ y =a and =b 2 2 p + x = 2a and x = 2a − p and

q + y = 2b y = 2b − q

T h e 41. The midpoint M has coordinates o t h e r 2 e n d p o i n t h a s c o o r d i n a t e s ( 2 a − p , 2 b − q ) .

170

Chapter 2 Graphs and Functions

36. Midpoint

⎛ a +b c +d ⎞ , , endpoint (b, d ) 2

b +x

a +b

2 and

b+ x = a+b x=a

d +y

c +d

2 and and

⎛ x1 +x2 y1 +y2 ⎞ , . 2 2 d ( P, M )

d + y =c+d y=c

The other endpoint has coordinates (a, c). 37. The endpoints of the segment are (1990, 21.3) and (20086, 29.4).

⎛ x +x = ⎜ ⎝ =

⎛ y +y

⎞ 2

–x ⎟ +⎜ 1 ⎠ ⎝

⎞ 2

–y⎟ 1 ⎠

2y ⎞ ⎛ x1 +x2 2 x1 ⎞ ⎛ y +y – + 1 2 – 1 2 2 2 2 2

⎛ x −x ⎞

⎛ y −y ⎞

⎛ 1990 +2008 21.3 +29.4 ⎞ ,

M =

= ⎜ ⎝

= (1999, 25.35)

38. The endpoints are (2000, 354) and (2008, 620)

2 2 = ( 2004, 487 ) The average payment to families in 2004 was $487.

⎟ +⎜ ⎠ ⎝

−x 2

⎟ ⎠

(y +

4 x (2 −x1

−y 2

)2 1

4 2 + y − ) (2 y 4

)

⎛ 2000 +2008 354 +620 ⎞ ,

The estimate is 25.35%. This is close to the actual figure of 25.2%.

M =

( x2 − x1 )

+ ( y2 − y1 )

d (M , Q) x +x ⎞ y +y ⎞ ⎛ ⎛ = ⎜ x2 − 1 2 ⎟ + ⎜ y2 − 1 2 ⎟ ⎝ ⎝ 2 ⎠ 2 ⎠ 2

39. The points to use would be (2004, 19,307) and (2008, 22,025). Their midpoint is ⎛ 2004 +2008

⎛ 2 x2 x1 +x2 ⎞ ⎛ 2 y2 y1 +y2 ⎞ − + − 2 2 2 2

19, 307 +22, 025 ⎞

2 = (2006, 20666).

⎛ x −x ⎞ = ⎜ ⎝

40. (a) To estimate the enrollment for 2002, use the points (2000, 11,753) and (2004, 12,980) M =

⎛ 2000 +2004 11, 753 +12, 980 ⎞ ,

2 = ( 2002, 12366.5)

The enrollment for 2002 was about 12,366.5 thousand.

In 2006, the poverty level cutoff was approximately $20,666.

(2x

= ( 2006, 13, 476) The enrollment for 2006 was about 13,476 thousand.

⎟ +⎜ ⎠ ⎝

4 −x1

−y 2

⎟ ⎠

)

) +(y 2 −y 1) 2

4 2

( x2 − x1 )

+ ( y2 − y1 )

( x2 − x1 ) 2 + ( y2 − y1 ) 2

d (P, Q) =

+ Since

1 2

−x 2

)2 + ( y 2− y )2

1 1 2

(b) To estimate the enrollment for 2006, use the points (2004, 12,980) and (2008, 13,972) ⎛ 2004 +2008 12, 980 +13, 972 ⎞ M = , 2 2

⎛ y −y ⎞

)

−x 2

( x2 − x1 ) + ( y2 − y1 ) ( )2 ( )2

x2 − x1

+ y2 − y1 ,

this shows d ( P, M ) + d (M , Q) = d (P, Q)

and

d ( P, M ) = d (M , Q). 42. The distance formula, 2

d = (x2 – x1 ) + ( y2 – y1 ) , can be written as d = [(x2 – x1 ) 2 + ( y2 – y1 ) 2]1 / 2.

Section 2.1 Rectangular Coordinates and Graphs

In exercises 43−54, other ordered pairs are possible. 43. (a)

x 0

y −2

−1

x-intercept: y=0⇒ 0 = 12 x − 2 ⇒ 1 2

y 5 3

y-intercept: x=0⇒ 2 (0 ) + 3 y = 5 ⇒ 3 y = 5 ⇒ y = 53

5 2

x-intercept: y=0⇒ 2 x + 3( 0 ) = 5 ⇒ 2 x = 5 ⇒ x = 25

−1

additional point

x 0

y −3

x-intercept: y=0⇒ 3x 0) =x 6= ⇒ 3x − = 26(⇒ 2

additional point

x⇒ 4= x

additional point (b)

(b)

44. (a)

x 0

y-intercept: x=0⇒ y = 12 (0) − 2 = −2

45. (a)

x 0

y 3

46. (a) y-intercept: x=0⇒ y = −0 + 3 ⇒ y = 3 x-intercept: y=0⇒ 0 = −x + 3 ⇒ −3 = − x ⇒ x = 3 additional point

(b) (b)

y-intercept: x=0⇒ 3 (0) − 2 y = 6 ⇒ −2 y = 6 ⇒ y = −3

171

47. (a)

172

x 0

y x- and y-intercept: 0 = 02

additional point

−2

additional point

Chapter 2 Graphs and Functions

(b)

48. (a)

50. (a)

x 0

−1

y 2

y-intercept: x=0⇒ y = 02 + 2 ⇒ y = 0+2⇒ y = 2

x 0

y −3

−1

additional point

x-intercept: y=0⇒ 0 = x −3⇒ 3= x ⇒ 9= x

y-intercept: x=0⇒ y = 0−2 ⇒ y = −2 ⇒ y = 2

x-intercept: y=0⇒ 0= x−2 ⇒ 0= x−2⇒ 2= x

−2 4

4 2

additional point additional point

y-intercept: x=0⇒ y = 0 −3⇒ y = 0 − 3 ⇒ y = −3

(b)

additional point

2 6 additional point no x-intercept: y = 0 ⇒ 0 = x2 + 2 ⇒ −2 = x 2 ⇒ ± −2 = x (b)

49. (a)

51. (a)

x 3

y 0

x-intercept: y=0⇒ 0 = x−3 ⇒ 0 = x−3⇒ 3= x

additional point

additional

point no y-intercept:

(b) x = 0 ⇒ y = 0 − 3 ⇒ y = −3 (b)

Section 2.1 Rectangular Coordinates and Graphs

52. (a)

x −2

y −2

−4

additional point x-intercept: y=0⇒ 0=− x+4 ⇒ 0= x+4 ⇒ 0 = x + 4 ⇒ −4 = x y-intercept: x=0⇒ y =− 0+4 ⇒ y = − 4 ⇒ y = −4

(b)

173

55. Points on the x-axis have y-coordinates equal to 0. The point on the x-axis will have the same x-coordinate as point (4, 3). Therefore, the line will intersect the x-axis at (4, 0). 56. Points on the y-axis have x-coordinates equal to 0. The point on the y-axis will have the same y-coordinate as point (4, 3). Therefore, the line will intersect the y-axis at (0, 3). 57. Since (a, b) is in the second quadrant, a is negative and b is positive. Therefore, (a, – b) will have a negative x–coordinate and a negative y-coordinate and will lie in quadrant III. (–a, b) will have a positive x-coordinate and a positive y-coordinate and will lie in quadrant I. Also, (–a, – b) will have a positive x-coordinate and a negative y-coordinate and will lie in quadrant IV. Finally, (b, a) will have a positive x-coordinate and a negative y-coordinate and will lie in quadrant IV. 58. Label the points A(−2, 2), B(13,10),

53. (a)

x 0

y 0

−1 2

−1 8

x- and y-intercept: 0 = 03 additional point additional point

C (21, −5), and D(6, −13). To determine which points form sides of the quadrilateral (as opposed to diagonals), plot the points.

(b)

Use the distance formula⎣ to find the⎦length of each side. d ( A, B) = ⎡⎣13 − (−2)⎤⎦ + (10 − 2) 2

54. (a)

1 2

−1 −8

= 152 + 82 = 225 + 64 = 289 = 17 x- and y-intercept: 0 = −03 additional point additional point

d ( B, C ) =

(21 − 13)2 + (−5 − 10)2

= 82 + ( −15) = 64 + 225 2

= 289 = 17

(b)

d (C, D) = =

(6 − 21)2

+ ⎡−13 − (−5)⎤

(−15)2 + (−8)2

= 225 + 64 = 289 = 17

(continued on next page)

174

Chapter 2 Graphs and Functions

(continued) d ( D, A) = =

(−2 − 6)2 + ⎡⎣ 2 − (−13) (−8)2

We check these by showing that d(A, B) = d(B, C) = d(C, D) and that d(A, D) = d(A, B) + d(B, C) + d(C, D). 2

d ( A, B) =

+ 152

(6 − 4 ) 2

d ( B, C ) =

(8 − 6)2 + (11 − 8)2

= 22 + 32 = 4 + 9 = 13

59. To determine which points form sides of the quadrilateral (as opposed to diagonals), plot

+ (8 − 5)

= 2 + 3 = 4 + 9 = 13

= 64 + 225 = 289 = 17 Since all sides have equal length, the four points form a rhombus.

d (C, D) =

the points.

(10 − 8)2 + (14 − 11)2

= 22 + 32 = 4 + 9 = 13 d ( A, D) =

(10 − 4)2 + (14 − 5)2

= 6 2 + 9 2 = 36 + 81 = 117 = 9(13) = 3 13

Use the distance formula to find the length of each side. d ( A, B) =

d(A, B), d(B, C), and d(C, D) all have the same measure and d(A, D) = d(A, B) + d(B, C) + d(C, D) since 3 13 = 13 + 13 + 13.

Section 2.2

(5 − 1)2 + (2 − 1)2

Circles

1. (a) Center (0, 0), radius 6

= 4 + 1 = 16 + 1 = 17 2

d ( B, C ) = =

( x − 0)

( 3 − 5) 2 + ( 4 − 2 ) 2 (−2)

+ ( y − 0) = 6

( x − 0)2 + ( y − 0)2

+2 = 4+4 = 8 2

= 62 ⇒ x 2 + y 2 = 36

(b) d (C, D) = =

(−1 − 3)2 + (3 − 4)2 (−4)2 + (−1)2

= 16 + 1 = 17 d ( D, A) = ⎡1 ⎣ − ( −1)⎤⎦ + (1 − 3) 2

= 2 2 + (−2) = 4 + 4 = 8 2

Since d(A, B) = d(C, D) and d(B, C) = d(D, A), the points are the vertices of a parallelogram. Since d(A, B) ≠ d(B, C), the points are not the vertices of a rhombus.

60. For the points A(4, 5) and D(10, 14), the difference of the x-coordinates is 10 – 4 = 6 and the difference of the y-coordinates is 14 – 5 = 9. Dividing these differences by 3, we obtain 2 and 3, respectively. Adding 2 and 3 to the x and y

coordinates of point A, respectively, we obtain B(4 + 2, 5 + 3) or B(6, 8). Adding 2 and 3 to the x- and y- coordinates of point B, respectively, we obtain C(6 + 2, 8 + 3) or C(8, 11). The desired points are B(6, 8) and C(8, 11).

2. (a) Center (0, 0), radius 9

( x − 0 )2 + ( y − 0 )2 ( x − 0 )2 + ( y − 0 ) 2 (b)

=9 = 92 ⇒ x 2 + y 2 = 81

Section 2.2 Circles

3. (a) Center (2, 0), radius 6

( x − 2 ) + ( y − 0) ( x − 2 )2 + ( y − 0 ) 2 2

(b)

=6 = 62

( x – 2)2 + y 2 = 36 (b)

7. (a) Center (–2, 5), radius 4 ⎡⎣ x − (−2)

+ ( y − 5) = 4 2

[ x – (–2)]2 + ( y – 5)2 = 4 2 ( x + 2) 2 + ( y – 5)2 = 16 (b) 4. (a) Center (3, 0), radius 3

( x − 3)2 + ( y − 0)2 = 3 ( x − 3)2 + y 2 = 9 (b)

8. (a) Center (4, 3), radius 5

( x − 4)2 + ( y − 3)2 = 5 ( x 2 2 − 4) + ( y − 3) = 52 ( x − 2 2 4) + ( y − 3) = 25

5. (a) Center (0, 4), radius 4

( x − 0)2 + ( y − 4)2 2 x 2 + ( y − 4)

=4 = 16

(b)

9. (a) Center (5, –4), radius 7

( x − 5)2 + ⎡⎣ y − (−4)⎤⎦

6. (a) Center (0, –3), radius 7

( x − 0)

+ y − ( −3) + y − (−3)

( x – 5) 2 + [ y – (– 4)]2 = 7 2 ( x – 5)2 + ( y + 4) 2 = 49

( x − 0)

175

x 2 + ( y + 3) 2 = 49

176

Chapter 2 Graphs and Functions

(b)

13. (a) The center of the circle is located at the midpoint of the diameter determined by the points (1, 1) and (5, 1). Using the

10. (a) Center (–3, –2), radius 6 ⎡⎣ x − (−3) ⎡⎣ x − (−3)

+ y − (−2) + y − (−2)

midpoint formula, we have

= 62

( x + 3) 2 + ( y + 2) 2 = 36 (b)

⎛ 1 +5 1 + 1 ⎞ , = (3,1) . The radius is

2 2 one-half the length of the diameter: 1 r= (5 − 1)2 + (1 − 1)2 = 2 2 The equation of the circle is

( x − 3)2 + ( y − 1)2

(b) Expand ( x − 3) + ( y − 1) = 4 to find the equation of the circle in general form: 2

11. (a) Center

(

)

( x − 3)2 + ( y − 1)2

=4 x − 6x + 9 + y 2 − 2 y + 1 = 4 x 2 + y 2 − 6x − 2 y + 6 = 0 2

2, 2 , radius

(x − 2 ) + ( y − 2 )

(x − 2 ) + ( y − 2 ) 2

14. (a) The center of the circle is located at the midpoint of the diameter determined by

the points (−1, 1) and (−1, −5). Using the midpoint formula, we have ⎛ −1 +(−1) 1 +(−5) ⎞ C= , = ( −1, −2) .

(b)

2 2 The radius is one-half the length of the diameter: 1 2 2 r= ⎡ −1 − (−1) + ( −5 − 1) = 3 2 ⎣ The equation of the circle is

( x + 1)2 + ( y + 2)2

(

)

12. (a) Center − 3, − 3 , radius

(

)

(

)

⎡x − − 3 ⎤ + ⎡ y − − 3 ⎤ = 3 2

(b) Expand ( x + 1) + ( y + 2) = 9 to find the 2

equation of the circle in general form: 2

⎣

⎦

(

)

⎣

(

( x + 1)

⎦

)

⎡x − − 3 ⎤ + ⎡ y − − 3 ⎤ = ⎣ ⎦ ⎣ ⎦ 2

( 3)

(x + 3 ) + ( y + 3 )

+ ( y + 2) = 9

x 2 + 2x + 1 + y 2 + 4 y + 4 = 9 2

x + y + 2x + 4 y − 4 = 0

Section 2.2 Circles

15. (a) The center of the circle is located at the midpoint of the diameter determined by the points (−2, 4) and (−2, 0). Using the midpoint formula, we have ⎛ −2 +(−2) 4 +0 ⎞ C= , = (−2, 2 ) . 2 2 The radius is one-half the length of the diameter: 1 r=

⎡ −2 − ( −2) 2 ⎣

( x + 2)

+ ( 4 − 0) = 2

) ( x + 6x + 9) + ( y + 8y + 16) = –9 + 9 + 16 2

( x + 3) 2 + ( y + 4) 2 = 16

2 2 20. x + y + 8x – 6 y + 16 = 0

(x 2

( x + 2) + ( y − 2) = 4 x 2 + 4x + 4 + y 2 − 4 y + 4 = 4 x 2 + y 2 + 4x − 4 y + 4 = 0 2

16. (a) The center of the circle is located at the midpoint of the diameter determined by the points (0, −3) and (6, −3). Using the midpoint formula, we have ⎛ 0 +6 −3 +(−3) ⎞ C= , = (3, −3) . 2 2 The radius is one-half the length of the diameter:

(6 − 0)2 + ⎡⎣ −3 − (−3)

The equation of the circle is

( x − 3)2 + ( y + 3)2

) (

)

+ 8x + y 2 – 6 y = −16

) (

)

+ 8x + 16 + y 2 – 6 y + 9 = –16 + 16 + 9

( x + 4)

+ ( y – 3) = 9 Yes, it is a circle. The circle has its center at (–4, 3) and radius 3. 2

2 2 21. x + y − 4x + 12 y = −4 Complete the square on x and y separately.

) (

) ( x – 4 x + 4) + ( y + 12 y + 36) = – 4 + 4 + 36 – 4 x + y 2 + 12 y = – 4

( x – 2 ) 2 + ( y + 6 )2

22. x 2 + y 2 – 12 x + 10 y = –25 Complete the square on x and y separately.

) (

) ( x – 12x + 36) + ( y + 10 y + 25) =

– 12 x + y 2 + 10 y = –25

– 25 + 36 + 25

( x − 3)2 + ( y + 3)2

=9 x 2 − 6x + 9 + y 2 + 6 y + 9 = 9 x 2 + y 2 − 6x + 6 y + 9 = 0 17. Since the center (–3, 5) is in quadrant II, choice B is the correct graph. 18. Answers will vary. If m > 0, the graph is a circle. If m = 0, the graph is a point. If m < 0, the graph does not exist.

= 36

Yes, it is a circle. The circle has its center at (2, –6) and radius 6.

(b) Expand ( x − 3) + ( y + 3) = 9 to find the equation of the circle in general form: 2

) (

+ 6x + y 2 + 8 y = –9

Complete the square on x and y separately.

+ ( y − 2) = 4

1 2

(b) Expand ( x + 2) + ( y − 2) = 4 to find the equation of the circle in general form:

Yes, it is a circle. The circle has its center at (–3, –4) and radius 4.

The equation of the circle is 2

177

19. x + y + 6x + 8 y + 9 = 0 Complete the square on x and y separately. 2

( x – 6)

+ ( y + 5) = 36 Yes, it is a circle. The circle has its center at (6, –5) and radius 6. 2

2 2 23. 4 x + 4 y + 4 x – 16 y – 19 = 0 Complete the square on x and y separately.

(

) ( ) ) + 4 ( y – 4 y + 4) =

4 x 2 + x + 4 y 2 – 4 y = 19

4 x +x+ 2

1 4

19 + 4

(

4 x+

)

1 2 2

+ 4 ( y – 2) = 36

( 14 ) + 4 ( 4)

( x + 12 )2 + ( y – 2)2 = 9 Yes, it is a circle with center (− 21 , 2 ) and radius 3.

178

Chapter 2 Graphs and Functions

24. 9 x 2 + 9 y 2 + 12 x – 18 y – 23 = 0 Complete the square on x and y separately.

(

9 x2 + 9 x + 3 x+

) ( ) ) + 9 ( y – 2 y + 1) = 4 3

2 3

)

( ) + 9 (1) 4

(

x + 1 + y2 +

2 3

)

, 1 and

25. x 2 + y 2 + 2 x – 6 y + 14 = 0 Complete the square on x and y separately.

) (

)

x 2 + 2 x + y 2 – 6 y = –14

) (

radius

9 5 2

(3)

( 1 , − 1 ) and 3

(

) ( ) )+ 4(y – y + ) =

4 x2 + x + 4 y2 − y = 7

(

= –4

(

4 x+

1 2

)

(

+4 y–

(1) + 4(1 ) )

1 2

4 2

( x + 12 )2 + ( y – 12 )2 = 94

) ( x + 4 x + 4) + ( y – 8 y + 16) =

+ 4 x + y – 8 y = –32 2

(

)

Yes, it is a circle with center − 1 , 1 and radius

3 2

– 32 + 4 + 16

( x + 2 )2 + ( y – 4 ) 2

= –12

31. The midpoint M has coordinates ⎛ –1+5 3 +( –9 ) ⎞ ⎛ 4 −6 ⎞ ⎜ ⎝

, 2

⎟ = ⎜ , ⎟ = (2, – 3). ⎠ ⎝2 2 ⎠

The graph is nonexistent. 32. Use points C(2, –3) and P(–1, 3).

27. x 2 + y 2 − 6 x − 6 y + 18 = 0 Complete the square on x and y separately.

) ( ) ( x − 6 x + 9) + ( y − 6 y + 9) = −18 + 9 + 9 2

+1+1

1 2

7+4

Complete the square on x and y separately.

) (

)

26. x 2 + y 2 + 4 x – 8 y + 32 = 0

30. 4 x 2 + 4 y 2 + 4 x − 4 y − 7 = 0 Complete the square on x and y separately.

The graph is nonexistent.

3 2

4 x2 + x +

)

( x + 1)2 + ( y – 3)2

y+1

(x − 3 ) + ( y + 3)

+ 2 x + 1 + y 2 – 6 y + 9 = –14 + 1 + 9

) (

Yes, it is a circle with center

radius 2.

(

x2 −

+ ( y – 1) = 4

2 3

Yes, it is a circle with center −

2 3

+ 9 ( y – 1) = 36

( x + 23 )

(

(9x − 6 x) + (9 y + 6 y ) = 23 9 ( x − x ) + 9 ( y + y ) = 23

x + 9 y 2 – 2 y = 23

23 + 9

9 x+

29. 9 x 2 + 9 y 2 − 6 x + 6 y − 23 = 0 Complete the square on x and y separately.

− 6 x + y 2 − 6 y = −18 2

( x − 3)2 + ( y − 3)2 The graph is the point (3, 3).

d (C, P) = =

( –1 – 2) ( –3)2

+ ⎡⎣3 – ( –3)

+ 62 = 9 + 36

= 45 = 3 5 The radius is 3 5. 33. Use points C(2, –3) and Q(5, –9). 2

28. x 2 + y 2 + 4 x + 4 y + 8 = 0 Complete the square on x and y separately.

(x (x

) (

)

+ 4 x + y 2 + 4 y = −8

) (

)

+ 4 x + 4 + y 2 + 4 y + 4 = −8 + 4 + 4

( x + 2 ) 2 + ( y + 2 )2

d (C, Q) =

(5 – 2 )

+ ⎡⎣ –9 – ( –3)

= 3 + ( –6) = 45 = 3 5 The radius is 3 5.

=0 The graph is the point (−2, −2).

= 9 + 36

Section 2.2 Circles

( 92 ) . To find the radius, we

34. Use the points P(–1, 3) and Q(5, –9).

The center is C 5,

Since d ( P, Q) = ⎡⎣5 – ( –1)

179

+ ( –9 – 3)

(

can use points C 5,

) and P(−1, 2). 2

= 62 + ( –12) = 36 + 144 = 180 2

= 6 5 ,the radius is

d (C, P) = ⎡⎣5 – ( –1)⎤⎦ +

d (P, Q). Thus

= 6

2 r=

(5) +

(

−2

169

2 9

)2 13 2

( 9 ) and

( )

1 6 5 = 3 5. 2

We could also use points C 5, Q(11, 7).

35. The center-radius form for this circle is ( x – 2) + ( y + 3) = (3 5 ) ⇒ 2

d (C, Q) = =

( x – 2) 2 + ( y + 3) 2 = 45.

( 29 – 7)

(5 − 11)

(−6)2

( 5)

+ −

169

36. Label the endpoints of the diameter P(3, –5) and Q(–7, 3). The midpoint M of the segment joining P and Q has coordinates ⎛ 3 +(–7) –5 +3 ⎞ ⎛ −4 –2 ⎞ , = , = (–2, – 1). 2

d ( P, Q ) = =

= 52 + ( –4) = 25 + 16 = 41 2

We could also use points C(–2, –1).and Q(–7, 3). d (C, Q) = ⎡−7 ⎣ – ( –2 ) =

(−5)2

+ ⎡⎣3 – ( –1)⎤⎦

+ 42 = 25 + 16 = 41

(−12 )

+ (−5)

= 169 = 13 1 1 13 d ( P, Q ) = (13) = 2 2 2 The center-radius form of the equation of the circle is 2 2

( x − 5) 2 + ( y − 9 )

2 2

(−1 − 11)2 + (2 − 7)2 2

+ ⎣⎡ –5 – ( –1)

Using the points P and Q to find the length of the diameter, we have

The center is C(–2, –1). To find the radius, we can use points C(–2, –1) and P(3, –5) d (C, P) = ⎡⎣3 – ( –2 )

= 13

( x − 5)

(

+ y−

)

( 13 )

169

38. Label the endpoints of the diameter P(5, 4) and Q(−3, −2). The midpoint M of the

We could also use points P(3, –5) and Q(–7, 3) to find the length of the diameter. The length of the radius is one-half the length of the

segment joining P and Q has coordinates ⎛ 5 +(−3) 4 +(−2) ⎞ , = (1, 1) .

diameter.

2 2 The center is C(1, 1). To find the radius, we can use points C(1, 1) and P(5, 4).

d (P, Q) =

( −7 – 3)2 + ⎡⎣3 – ( –5) (−10)2

+ 82 = 100 + 64

d (C, P) =

(5 − 1)2 + (4 − 1)2

= 164 = 2 41

(

)

1 1 d (P, Q) = 2 41 = 41 2 2

= 4 2 + 32 = 25 = 5 We could also use points C(1, 1) and Q(−3, −2). 2

The center-radius form of the equation of the circle is [ x – (–2)]2 + [ y – (–1)]2 = ( 41) 2 ( x + 2) 2 + ( y + 1)2 = 41

d (C, Q) = ⎡⎣1 − (−3)⎤⎦ + ⎡⎣1 − ( –2 )⎤⎦ = 4 2 + 32 = 25 = 5 Using the points P and Q to find the length of the diameter, we have

37. Label the endpoints of the diameter P(–1, 2) and Q(11, 7). The midpoint M of the segment joining P and Q has coordinates ⎛ −1 +11 2 +7 ⎞ ⎛ 9 ⎞ , = 5, . 2

d ( P, Q ) = ⎡⎣ 5 − (−3)⎤⎦ + ⎡⎣ 4 − (−2)⎤⎦ = 82 + 6 2 = 100 = 10 1 1 d ( P, Q) = (10) = 5 2 2

(continued on next page)

180

Chapter 2 Graphs and Functions

(continued) The center-radius form of the equation of the circle is

( x − 1)2 + ( y − 1)2 ( x − 1)2 + ( y − 1)2

= 52 = 25

39. Label the endpoints of the diameter P(1, 4) and Q(5, 1). The midpoint M of the segment joining P and Q has coordinates ⎛ 1 +5 4 + 1 ⎞ , = 3, 52 . 2 2

( ) The center is C (3, 25 ) .

Check algebraically:

The length of the diameter PQ is

( x + 9) 2 + ( y + 4)2 = 169

(1 − 5)2 + (4 − 1)2

(−4)2

+ 32 = 25 = 5.

( x − 7) 2 + ( y − 4) 2 = 25 (3 − 7) 2 + (1 − 4) 2 = 25 4 2 + 32 = 25 ⇒ 25 = 25 (3 + 9)2 + (1 + 4)2 = 169 2

The length of the radius is

( 5) =

The center-radius form of the equation of the

12 + 5 = 169 ⇒ 169 = 169 ( x + 3)2 + ( y − 9)2 = 100 (3 + 3) 2 + (1 − 9)2 = 100

circle is 2

( x − 3)

(

+ y−

5 2 2

5 2

) == ( 2 )

( x − 3) 2 + ( y − 25 )

6 + (−8) = 100 ⇒ 100 = 100 (3, 1) satisfies all three equations, so the epicenter is at (3, 1).

25 4

40. Label the endpoints of the diameter P(−3, 10) and Q(5, −5). The midpoint M of the segment joining P and Q has coordinates ⎛ −3 +5 10 +(−5) ⎞ , = 1, 52 . 2 2

( )

( 52 ) .

42. The three equations are ( x − 3)2 + ( y − 1)2 = 5 , ( x − 5) 2 + ( y + 4) 2 = 36 , and ( x + 1)2 + ( y − 4) 2 = 40 . From the graph of the three circles, it appears that the epicenter is located at (5, 2).

The center is C 1,

The length of the diameter PQ is

(−3 − 5)2 + ⎡⎣10 − (−5)

(−8)2 + 152

= 289 = 17. The length of the radius is

1 2

(17) = 172 .

The center-radius form of the equation of the circle is

(

5 2 2

(

5 2 2

( x − 1)

+ y−

( x − 1)

+ y−

) = (172 ) )

289 4

41. The equations of the three circles are ( x − 7)2 + ( y − 4) 2 = 25 ,

Check algebraically: ( x − 3)2 + ( y − 1) 2 = 5 (5 − 3) 2 + (2 − 1) 2 = 5 ( x + 9)2 + ( y + 4) 2 = 169 , and

( x + 3)2 + ( y − 9) 2 = 100 . From the graph of the three circles, it appears that the epicenter is located at (3, 1).

22 + 12 = 5 ⇒ 5 = 5 ( x − 5) + ( y + 4)2 = 36 (5 − 5)2 + (2 + 4)2 = 36 62 = 36 ⇒ 36 = 36 2

( x + 1) 2 + ( y − 4)2 = 40 (5 + 1)2 + (2 − 4)2 = 40 6 2 + (−2)2 = 40 ⇒ 40 = 40 (5, 2) satisfies all three equations, so the epicenter is at (5, 2).

College Algebra 11th Edition Lial SOLUTIONS MANUAL Full download: http://testbanklive.com/download/college-algebra-11th-edition-lial-solutionsmanual/ College Algebra 11th Edition Lial TEST BANK Full download: http://testbanklive.com/download/college-algebra-11th-edition-lial-test-bank/ People also search: college algebra 11th edition pdf free download essentials of college algebra 11th edition ebook essentials of college algebra pdf college algebra textbook college algebra book pearson