-
Use the graph below. Write three different translation rules for which the image of
Δ
J
K
L
cap delta j k l has a vertex at the origin.
Find a translation that has the same effect as each composition of translations.
-
(
x
,
y
)
→
(
x
+
2
,
y
+
5
)
open x comma y close rightwards arrow . open x plus 2 comma y plus 5 close followed by
(
x
,
y
)
→
(
x
−
4
,
y
+
9
)
open x comma y close rightwards arrow . open x minus 4 comma y plus 9 close
-
(
x
,
y
)
→
(
x
+
12
,
y
+
0.5
)
open x comma y close rightwards arrow . open x plus 12 comma y plus 0.5 close followed by
(
x
,
y
)
→
(
x
+
1
,
y
−
3
)
open x comma y close rightwards arrow . open x plus 1 comma y minus 3 close
C Challenge
-
Coordinate Geometry
Δ
A
B
C
cap delta eh b c has vertices
A
(
−
2
,
5
)
,
B
(
−
4
,
−
1
)
,
eh open negative 2 comma 5 close comma b open negative 4 comma negative 1 close comma and
C
(
2
,
−
3
)
.
c open 2 comma negative 3 close . Show that the images of the midpoints of the sides of
Δ
A
B
C
cap delta eh b c are the midpoints of the sides of
Δ
A
′
B
′
C
′
cap delta , eh prime , b prime , c prime for the translation
(
x
,
y
)
→
(
x
+
4
,
y
+
2
)
.
open x comma y close rightwards arrow . open x plus 4 comma y plus 2 close .
-
Writing Explain how to use translations to draw a parallelogram.
Standardized Test Prep
SAT/ACT
-
Δ
A
B
C
cap delta eh b c has vertices
A
(
−
5
,
2
)
,
B
(
0
,
−
4
)
,
and
C
(
3
,
3
)
.
eh open negative 5 comma 2 close comma b open 0 comma negative 4 close comma , and c open 3 comma 3 close . What are the vertices of the image of
Δ
A
B
C
cap delta eh b c under the translation
(
x
,
y
)
→
(
x
+
7
,
y
−
5
)
?
open x comma y close rightwards arrow . open x plus 7 comma y minus 5 close question mark
-
A
′
(
2
,
−
3
)
,
B
′
(
7
,
−
9
)
,
C
′
(
10
,
−
2
)
eh prime , open 2 comma negative 3 close comma , b prime , open 7 comma negative 9 close comma , c prime , open 10 comma negative 2 close
-
A
′
(
−
12
,
−
3
)
,
B
′
(
−
7
,
−
9
)
,
C
′
(
−
4
,
−
2
)
eh prime , open negative 12 comma negative 3 close comma , b prime , open negative 7 comma negative 9 close comma , c prime , open negative 4 comma negative 2 close
-
A
′
(
−
12
,
7
)
,
B
′
(
−
7
,
1
)
,
C
′
(
−
4
,
8
)
eh prime , open negative 12 comma 7 close comma , b prime , open negative 7 comma 1 close comma , c prime , open negative 4 comma 8 close
-
A
′
(
2
,
−
3
)
,
B
′
(
10
,
−
2
)
,
C
′
(
7
,
−
9
)
eh prime , open 2 comma negative 3 close comma , b prime , open 10 comma negative 2 close comma , c prime , open 7 comma negative 9 close
-
What is the value of x in the figure below?
- 4.5
- 16
- 18
- 18.5
- In
Δ
P
Q
R
,
P
Q
=
4.5
,
Q
R
=
4.4
,
and
R
P
=
4.6
.
cap delta p q r comma p q equals 4.5 comma q r equals 4.4 comma , and r p equals 4.6 . Which statement is true?
-
m
∠
P
+
m
∠
Q
<
m
∠
R
m angle p plus m angle q less than m angle r
-
∠
Q
is the largest angle
.
angle q . isthelargestangle . .
-
∠
R
is the largest angle
.
angle r . isthelargestangle . .
-
m
∠
R
<
m
∠
P
m angle r less than m angle p
Short Response
-
▱
white parallelogram
ABCD has vertices
A
(
0
,
−
3
)
,
B
(
−
4
,
−
2
)
,
and
D
(
−
1
,
1
)
.
eh open 0 comma negative 3 close comma b open negative 4 comma negative 2 close comma , and d open negative 1 comma 1 close .
- What are the coordinates of C?
- Is
▱
white parallelogram
ABCD a rhombus? Explain.
Mixed Review
See Lesson 8-5.
-
Navigation An airplane lands at a point 100 km east and 420 km south from where it took off. Describe the magnitude and the direction of its flight vector.
See Lesson 4-7.
-
Given:
B
C
¯
≅
E
F
¯
,
B
C
¯
‖
E
F
¯
,
A
D
¯
≅
D
C
¯
≅
C
F
¯
table with 2 rows and 1 column , row1 column 1 , b c bar , approximately equal to , e f bar , comma , b c bar . double vertical bar . e f bar , comma , row2 column 1 , eh d bar , approximately equal to , d c bar , approximately equal to , c f bar , end table
Prove:
A
B
¯
≅
D
E
¯
eh b bar , approximately equal to , d e bar
Get Ready! To prepare for Lesson 9-2, do Exercises 42-44.
See Lesson 3-8.
Write an equation for the line through A perpendicular to the given line.
-
A
(
1
,
−
2
)
;
x
=
−
2
eh open 1 comma negative 2 close semicolon x equals negative 2
-
A
(
−
1
,
−
1
)
;
y
=
1
eh open negative 1 comma negative 1 close semicolon y equals 1
-
A
(
−
1
,
2
)
;
y
=
x
eh open negative 1 comma 2 close semicolon y equals x