Compositions Answers

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Figure WXYZ is transformed using the rule . Point W of the pre-image is at (1, 6).

Question illustration
A
(–5, 8)
B
(–3, –8)
C
(5, –8)
D
(3, 8)
2
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Triangle A B C is reflected across the line of reflection m to form triangle A prime B prime C prime. Triangle A prime B prime C prime is rotated about point B prime 270 degrees to form triangle A double-prime B double-prime C double-prime.

Question illustration
A
Option
Option A
B
Option
Option B
C
Option
Option C
D
Option
Option D
3

The rule is applied to ΔABC.

Question illustration
A
1
B
2
C
3
D
4
4

Trapezoid E F G H is reflected across line of reflection k to form trapezoid E prime F prime G prime H prime. Trapezoid E prime F prime G prime H prime is shifted down to E double-prime F double-prime G double-prime H double-prime.

Question illustration
A
a reflection across line k followed by a translation down
B
a translation down followed by a reflection across line k
C
a 180° rotation about point G followed by a translation to the right
D
a translation to the right followed by a 180° rotation about point G
5

The rule is applied to ΔFGH to produce ΔF"G"H".

Question illustration
A
(4, –1.5)
B
(4, –0.5)
C
(–1.5, 4)
D
(–0.5, 4)
6

Parallelogram F"G"H"J" is the final image after the rule was applied to parallelogram FGHJ.

Question illustration
A
(–2, 2)
B
(–2, 6)
C
(–3, 4)
D
(–4, 2)
7

On a coordinate plane, 2 triangles are shown. Triangle J K L has points (2, negative 4), (5, negative 4), (2, negative 2). Triangle J double-prime K double-prime L double-prime has points (2, 2), (2, 5), (0, 2).

Question illustration
A
90 degree rotation about point 0 composition translation of 0 units x, negative 2 units y
Option A
B
Translation of 0 units x, negative 2 units y composition 90 degree rotation about point 0
Option B
C
90 degree rotation about point 0 composition translation of negative 2 units x, 0 units y
Option C
D
Translation of negative 2 units x, 0 units y composition 90 degree rotation about point 0
Option D
8

On a coordinate plane, 3 triangles are shown. Triangle B C D has points (1, 4), (1, 2), (5, 3). Triangle B prime C prime D prime has points (negative 1, 4), (negative 1, 2), (negative 5, 3). Triangle B double-prime C double-prime D double-prime has points (5, negative 1), (5, negative 3), (1, negative 2).

Question illustration
A
Translation of 5 units x, negative 6 units y composition reflection across y = negative x
Option A
B
Reflection across y = negative x composition translation of 5 units x, negative 6 units y.
Option B
C
Translation of 6 units x, negative 5 units y composition reflection across the y-axis
Option C
D
Reflection across the y-axis composition translation of 6 units x, negative 5 units y
Option D
9

Square A"B"C"D" is the final image after the rule was applied to square ABCD.

Question illustration
A
(–1, –6)
B
(–1, –2)
C
(–1, 6)
D
(–2, 1)
10

Trapezoid G H J K is rotated about G 90 degrees counterclockwise to form trapezoid G prime H prime J prime K prime. Trapezoid G prime H prime J prime K prime is reflected across the line of reflection m to form trapezoid G double-prime H double-prime J double-prime K double-prime.

Question illustration
A
a translation to the right
B
a reflection across line m
C
a 90° rotation about point G
D
a 180° rotation about point G

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