F G

Chapter 9,
continued
FG
4. J;
4.
14
F GF
Women’s Team: [60 18 15] 16
16
GF
E
F G H
E9 F9 G9
23 23 1 2
2 4
4
5
2
4 4 2
3 3 21
0 1
21 0
G
H9
2
22
y
5 [60(14) 1 18(16) 1 15(16)]
F
G
F
E
5 [1368]
FG
E
10
H
1
Men’s Team: [60 18 15] 13
x
1
G
13
H
5 [60(10) 1 18(13) 1 15(13)]
5.
5 [1029]
F GF
1
0
The total cost of equipment for the women’s team is
$1368 and the total cost of equipment for the men’s team
is $1029.
GF
G
E
F G H
E9 F9 G9 H9
23 23 1 2
23 23 1 2
5
2
4 4 2
2
4 4 2
0
1
y
5. A;
6. F(24, 3) l F9(24 1 s, 3 1 t) 5 F9(21, 4)
24 1 s 5 21
E9
31t54
s53
G9
F9
A reflection across the x-axis is illustrated in the photo.
H9
1
21
t51
x
(x, y) l (x 1 3, y 1 1)
G(3, 21) l G9(3 1 3, 21 1 1) 5 G9(6, 0)
6. B; 2s 5 12
The x-coordinate of G9 is 6.
s56
3r 5 2s 1 3 5 2(6) 1 3 5 15
Lesson 9.4
3r 5 15
9.4 Guided Practice (pp. 599–601)
1.
r55
9.4 Exercises (pp. 602–605)
P
D
E9
Skill Practice
F
1. The center of rotation is a fixed point in which a figure is
F9
turned about during a rotation transformation.
2. During a rotation of 908, the x-value becomes the
D9
2.
(a, b) l (2b, a)
opposite of the old y-value and the y-value becomes the
old x-value. The 908 rotation matrix gives the same result.
y L9
3. Reflection; The horses are reflected across the edge of the
J(3, 0) l J9(0, 3)
K(4, 3) l K9(23, 4)
K9
stream which acts like the line of symmetry.
K
J9
L(6, 0) l L9(0, 6)
4. Rotation; The steering wheel turns and everything rotates
1
3.
F GF
0 21
1
0
GF
E
F G H
E9
23 23 1 2
22
5
2
4 4 2
23
around a center point.
x
21
F9
24
23
5. Translation; The train moves horizontally.
L
J
G9
24
1
G
H9
22
2
6. C; The angle is close to a straight angle which is 1808.
7. A; The angle is less than 908.
8. B; The angle is a little greater than 908.
9.
B9
y
C9
F
E
G
H
G
A9
H
1
A
1
x
P
F
278
B
E
Geometry
Worked-Out Solution Key
C
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
E
Chapter 9,
10.
continued
11.
G
P
T9
P
Q
T
R9
F9
Q9
(a, b) l (2b, a)
12.
0 1
21 0
B(2, 4) l B9(24, 2)
C(3, 1) l C9(21, 3)
15.
(a, b) l (2a, 2b)
(a, b) l (b, 2a)
14.
J(1, 4) l J9(21, 24)
Q(26, 23) l Q9(23, 6)
K(5, 5) l K9(25, 25)
R(25, 0) l R9(0, 5)
L(7, 2) l L9(27, 22)
S(23, 0) l S9(0, 3)
M(2, 2) l M9(22, 22)
T(21, 23) l T9(23, 1)
F GF
0 21
1
0
A
1
4
GF
B
5
6
C
4
5
3
A9
24
1
B9
26
5
G
C9
23
4
A9
F
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
0
21
J
1
1
GF
K
2
21
L
0
5
23
y
4
y54
21. C; (a, b) l (2a, 2b)
Q(1, 2) l Q9(21, 22)
J9
21
21
K9
22
1
G
L9
0
3
A rotation of 7208 is equivalent to 7208 2 3608 5 3608
rotation.
908, 4508: (a, b) l (2b, a)
1808, 5408: (a, b) l (2a, 2b)
2708, 6308: (a, b) l (b, 2a)
3608, 7208: (a, b) l (a, b)
L9
908
1808
2708
3608
A9(0, 2)
A9(22, 0)
A9(0, 22)
A9(2, 0)
B(3, 4) B9(24, 3) B9(23, 24) B9(4, 23)
B9(3, 4)
C(5, 2) C9(22, 5) C9(25, 22) C9(2, 25)
C9(5, 2)
J
K9
22
K
A(2, 0)
x
J9
L
17.
3y 5 12
A rotation of 6308 is equivalent to 6308 2 3608 5 2708
rotation.
x
GF
21
0
3y 5 x 1 7 5 5 1 7 5 12
rotation.
C
22
16.
20. A; x 5 5
A rotation of 5408 is equivalent to 5408 2 3608 5 1808
rotation.
B
2
21 2
1 3
22. A rotation of 4508 is equivalent to 4508 2 3608 5 908
A
C9
21 2
1 3
Q9 is in Quadrant III
y
B9
0 1
21 0
19. The rotation matrix is written second. It should come first.
A(23, 2) l A9(22, 23)
13.
F G
F GF G
F GF G
0 1
.
21 0
S
J9
G9
J
18. The rotation matrix for 2708 is incorrect. It should be
S9 R
F
F GF
F
0 1
21 0
P9
24
5
4
P
24
24
Q
2
22
Q9
22
22
R9
25
22
R
2
25
G
S9
27
4
P9
2
28
Q9
R9
P
4508
5408
6308
7208
A9(0, 2)
A9(22, 0)
A9(0, 22)
A9(2, 0)
B(3, 4) B9(24, 3) B9(23, 24) B9(4, 23)
B9(3, 4)
C(5, 2) C9(22, 5) C9(25, 22) C9(2, 25)
C9(5, 2)
A(2, 0)
To find a rotation of 18908, first subtract 3608 until you
get an angle between 08 and 3608.
y
S9
G
S
24
27
Q
x
18908 2 5(3608) 5 908
Use 908 to find the rotation, which is A9(0, 2).
R
S
Geometry
Worked-Out Solution Key
279
Chapter 9,
continued
23. D;
25. (23, 2, 0)
A. (4, 0) l (4 1 4, 0 2 2) l (8, 22)
z
(4, 2) l (4 1 4, 2 2 2) l (8, 0)
(3, 2, 0)
(7, 0) l (7 1 4, 0 2 2) l (11, 22)
(7, 2) l (7 1 4, 2 2 2) l (11, 0)
y
B. (a, b) l (2a, 2b)
(3, 2, 0)
x
(4, 0) l (24, 0)
(4, 2) l (24, 22)
26. y 5 2x 2 3
(7, 0) l (27, 0)
y-intercept: (0, 23)
(7, 2) l (27, 22)
C.
x-intercept: 0 5 2x 2 3
y
3 5 2x
1 }32, 0 2
3
2
1
}5x
x
1
y
2
3
2
( , 0)
New points: (1, 0), (1, 2), (4, 0), (4, 2)
x
1
y
D.
(0, 3)
1
y 2x 3
x
1
a. Rotate 908, so it is >.
1
m1 5 2}2
New points: (0, 0), (0, 2), (23, 0), (23, 2)
(a, b) l (2b, a)
y
y 5 2}2 x 1 b, 1 }2 , 0 2
1
B
C
A(23, 2) l A9(22, 23)
B
B(2, 4) l B9(24, 2)
A
C(3, 1) l C9(21, 3)
0 5 2}2 1 }2 2 1 b
1 3
1
C
1
}
422
2
Slope of AB 5 } 5 }5
3
x
0 5 2}4 1 b
3
4
}5b
A
2 2 (23)
}
5
23 2 2
Slope of A9B9 5 } 5 2}2
1
b.
1
y 5 2}2 x 1 b
} 124
5 23
Slope of BC 5 }
322
23 5 2}2 (0) 1 b
}
322
1
Slope of B9C9 5 } 5 }3
23 5 b
1
21 2 (24)
} }
1
(23) }3 5 21, so BC > B9C9
1 2
}
122
1
Slope of AC 5 } 5 2}6
3 2 (23)
}
3 2 (23)
21 2 (22)
6
Slope of A9C9 5 } 5 }1 5 6
}
AC > A9C9
1 2}16 2(6) 5 21, so }
280
3
y 5 2}2 x 1 }4
22 2 (24)
}
AB > A9B9
1 }25 21 2}52 2 5 21, so }
3
Geometry
Worked-Out Solution Key
1
y 5 2}2 x 2 3
(0, 23)
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
24.
Chapter 9,
continued
27. y 5 2x 1 8
32. If you enter a revolving door from the inside of a
building and rotate 1808; you will be directly opposite
from where you started and would have exited the
building. If you would have rotated 3608, you would
have made a complete circle and would end up back in
the building where you started.
y-intercept: (0, 8)
x-intercept: (8, 0)
y
(0, 8)
33.
y x 8
1
Statements
x
(8, 0)
1
a. Rotate 1808 then same line
y 5 2x 1 8
28.
b. y 5 2x 1 8
1
y 5 }2 x 1 5
y-intercept: (0, 5)
1
x-intercept: 0 5 }2 x 1 5
1
25 5 }2 x
210 5 x
(210, 0)
1. PQ 5 PQ9, PR 5 PR9,
mŽ QPQ9 5 mŽ RPR9
1. Definition of a rotation
about a point
2. mŽ QPQ9 5 mŽ QPR9
1 mŽ R9PQ9
mŽ RPR9 5 mŽ RPQ
1 mŽ QPR9
2. Angle Addition Postulate
3. mŽ QPR9 1 mŽ R9PQ9
5 mŽ RPQ 1 mŽ QPR9
3. Transitive Property of
Equality
4. mŽ QPR 5 mŽ Q9PR9
4. Subtraction Property of
Equality
5. nRPQ > nR9PQ9
5. SAS Congruence
Postulate
y
} }
6. QR > Q9R9
(0, 5)
7. QR 5 Q9R9
1
y 2x 5
6. Corresponding parts of
congruent triangles are
congruent.
7. Definition of congruent
segments
2
(10, 0)
Reasons
x
2
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
34. Q9
a. Rotate 2708 so line is >.
m1 5 22
Q
R
P
y 5 22x 1 b
0 5 22(210) 1 b
0 5 20 1 b
Statements
Reasons
y 5 22x 2 20
1. A rotation about P maps
1. Given
Q to Q9 and R to R9. P, Q,
and R are collinear.
y 5 22x 1 b
2. PQ 5 PQ9, PR 5 PR9
2. Definition of a rotation
about a point
3. PQ 5 PR 1 RQ,
PQ9 5 PR9 1 R9Q9
3. Segment Addition
Postulate
4. PR 1 RQ 5 PR9 1 R9Q9
4. Transitive Property of
Equality
220 5 b
b.
R
5 5 22(0) 1 b
55b
y 5 22x 1 5
Problem Solving
29. 2708; The line segment joining A to the center of rotation
is perpendicular to the line segment joining A9 to the
center of rotation. Then, A travels 908 1 908 1 908 which
equals 2708.
5. PR9 1 RQ 5 PR9 1 R9Q9 5. Substitution Property of
Equality
6. QR 5 Q9R9
6. Subtraction Property of
Equality
30. 1808; A9 is directly opposite A from the center of rotation
forming a straight angle.
31. 1208; The line segment joining A to the center of
3608
apart from the line segment
rotation is 1208 or }
3
1
2
joining A9 to the center of rotation.
Geometry
Worked-Out Solution Key
281
Chapter 9,
continued
b. Yes; If you choose any two points on a line and
35. Q9
rotate the two points 908, 1808, 2708, and 3608, then
determine the slope of each image and compare it to
its preimage. The same relationship will hold true.
Q
37. a. y 5 x 2 1 1
R
P R
x 5 y2 1 1
original graph:
Statements
1. A rotation about P maps
1. Given
Q to Q9 and R to R9. P and
R are the same point.
2. PQ 5 PQ9
2. Definition of a rotation
about a point
3. P 5 R 5 R9
3. Given
4. QR 5 Q9R9
4. Substitution Property of
Equality
new graph:
x
5
2
1 2 5
y
1
(5, 2)
(2, 1)
(1, 0)
21
x
c. No, the image does not pass the vertical line test.
38. F(1, 2), G(3, 4)
3
A }2, 0
2
y 2 intercept: y 5 2x 2 3 5 2(0) 2 3 5 23
B(0, 23)
(a, b) l (2b, a)
A1 }2 , 0 2 l A91 0, }2 2
3
3
B(0, 23) l B9(3, 0)
(a, b) l (2a, 2b)
(a, b) l (2b, a)
F(1, 2) l F9(22, 1)
G(3, 4) l G9(24, 3)
}
F9F 0 lies on the straight line through (0, 4). From F9 to
(0, 4), you move 2 units to the right and 3 units up. So,
}
the coordinates of F 0 are (2, 7). G9G0 lies on the straight
line through (0, 4). From G9 to (0, 4), you move 4 units
to the right and 1 unit up. So, the coordinates of G0 are
(4, 5).
y
A1 }2 , 0 2 l A91 2}2 , 0 2
3
3
G0
B(0, 23) l B9(0, 3)
(0, 4
)
(a, b) l (b, 2a)
G9
A1 }2 , 0 2 l A91 0, 2}2 2
3
F0
3
B(0, 23) l B9(23, 0)
(a, b) l (a, b)
1
F9
21
39. F(1, 2), G(3, 4)
(a, b) l (b, 2a)
A1 }2 , 0 2 l A91 }2 , 0 2
F(1, 2) l F9(2, 21)
B(0, 23) l B9(0, 23)
G(3, 4) l G9(4, 23)
3
3
y
180
90
360
1
x
2
270
y 2x 3
The 908 and 2708 rotation is prependicular to the
preimage. The 1808 rotation is parallel to the preimage.
The 3608 rotation is the same line as the preimage.
Geometry
Worked-Out Solution Key
x
From (22, 0) to F9 you move 4 units to the right and
1 unit down. For a 908 rotation of F9 about (22, 0),
you move 1 unit right and 4 units up. So, F 0(22 1 1,
0 1 4) 5 F 0(21, 4). From (22, 0) to G9, you move
6 units to the right and 3 units down for a 908 rotation of
G9 about (22, 0), you move 3 units right and 6 units up.
So, G0(22 1 3, 0 1 6) 5 G0(1, 6).
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
1
3
}5x
2
282
1 2 5
b. The angle of rotation is 2708.
3 5 2x
3608:
2
(5,2)
0 5 2x 2 3
2708:
5
(2,1)
x-intercept: y 5 2x 2 3
1808:
y
y 22 21 0 1 2
36. a. y 5 2x 2 3
908:
x 22 21 0 1 2
Reasons
Chapter 9,
continued
5. The x-coordinate is now 24 and the y-coordinate is
y G0
now 3.
6. J9
F0
y
K9
J
1
(2, 0)
L9
F9
22
L
K
x
P
G9
40. The distance from the origin to A never changes. It will
always equal 2. For a 908 rotation, you would add 908
to the angle. You would do the same for 1808 and 2708
rotations.
(2, 1208), (2, 2108), (2, 3008)
1
21
x
J9(25, 9), K9(23, 9), L9(23, 7)
Lesson 9.5
Mixed Review for TAKS
Investigating Geometry Activity 9.5 (p. 607)
41. C;
Explore 1
The maximum value of the parabola occurs at (0, 3).
42. H;
Each number shown can be written as mm.
Step 3: nD0E0F 0 > nDEF; DD0E0F 0 is a translation
of nDEF.
Step 4: Yes; The relationship hold true.
1 5 11
Explore 2
4 5 22
Step 2: n A0B0C 0 > n ABC; n A0B0C 0 is a rotation
of n ABC.
27 5 3
3
256 5 44
Because 5 5 3125, the next number that would be in the
sequence is 3125.
5
Step 3: The measure of the acute angle is half the measure of
Ž APA0. Yes, the relationship hold true.
1. A translation can map a figure onto the same image as a
reflection in two parallel planes.
Problem Solving Workshop 9.4 (p. 606)
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
1.
reflection in two intersecting lines.
y
C9
2. A rotation can map a figure onto the same image as a
D9
9.5 Guided Practice (pp. 608–611)
B9
1. A(3, 2), B(6, 3), C(7, 1)
A9
1
21
x
A
B
A(3, 2) l A9(3, 2 2 4) 5 A9(3, 22)
B(6, 3) l B9(6, 3 2 4) 5 B9(6, 21)
C(7, 1) l C9(7, 1 2 4) 5 C9(7, 23)
D
C
2.
Translation: (x, y) l (x, y 2 4)
Reflection: y-axis, (x, y) l (2x, y)
A9(3, 22) l A0(23, 22)
y R
B9(6, 21) l B0(26, 21)
S
C9(7, 23) l C 0(27, 23)
T
T9
2. A0(29, 22) l (29 1 x, y) l A(3, 2)
1
R9
21
x
S9
3. 908 clockwise and 2708 counterclockwise are rotations
29 1 x 5 3
x 5 12
(x, y) l (x 1 12, y) followed by a reflection in the x-axis.
in opposite directions placing you at 908 below the
reference line.
4. Draw the preimage and trace the preimage on the tracing
paper. To reflect in the y-axis, flip the paper vertically and
line up the axes and trace. To reflect in the x-axis, flip the
paper horizontally and line up the axes and trace.
Geometry
Worked-Out Solution Key
283