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Document 1805668
```Chapter 3,
continued
3.4 Exercises (pp. 175–178)
4. H;
Because 1 and 2 are congruent alternate interior
angles, you know that p i q by the Alternate Interior
Angles Converse.
5. A;
Using the Alternate Interior Angles Converse:
(6x 1 5)8 5 238
6x 5 18
x53
Lines * and m are parallel when x 5 3.
6. Using the Linear Pair Postulate, m2 5 1808 2 838 5
978. Using the Corresponding Angles Postulate,
m1 5 m2 5 978.
Lesson 3.4
change to horizontal change between any two points on
the line.
2. When you apply the slope formula to a horizontal line,
the numerator of the resulting fraction is zero, so the
slope is zero. When you apply the slope formula to a
vertical line, the denominator of the resulting fraction is
zero, so the slope is undeﬁned.
3. D; m is positive.
4. A; m is negative.
5. B; m is zero.
y2 2 y1
625
1
7. m 5 } 5 } 5 }
x2 2 x1
523
2
6. C; m is undeﬁned.
1
1. Slope of line b:
y2 2 y1
420
m5}
5}
52
x2 2 x1
624
2. Slope of line c:
y2 2 y1
424
m5}
5}
50
x2 2 x1
620
y2 2 y1
123
2
3. Slope of line m: m1 5 } 5 } 5 2}
5
x2 2 x1
4 2 (21)
y2 2 y1
23 2 (21)
2
Slope of line t: m2 5 }
5}
5 2}5
x2 2 x1
3 2 (22)
Because m and t have the same slope, they are parallel.
1. The slope of a non-vertical line is the ratio of vertical
The slope is }2.
3.4 Guided Practice (pp. 171–174)
y2 2 y1
522
3
1
4. Slope of line n: m1 5 } 5 } 5 } 5 }
x2 2 x1
620
6
2
y2 2 y1
024
24
Slope of line m: m2 5 }
5}
5}
5 22
x2 2 x1
422
2
Because the product of m1 and m2 is 21, n > m.
5. The parachute in jump c was in the air approximately
1.25 seconds longer than the parachutes in jumps a and
b. So, the parachute in jump c was in the air for the
longest time.
6. The x-intercepts represent the time (in minutes) that
it takes each parachute to reach the ground. You can
eliminate choice B, because jumps b and c do not have
the same x-intercept, so the parachutes were not open for
the same amount of time.
y2 2 y1
520
5
7. Slope of line q: m1 5 } 5 } 5 2}
x2 2 x1
24 2 0
4
y2 2 y1
720
7
Slope of line t: m2 5 }
5}
5 2}
x2 2 x1
210 2 0
10
Line q is steeper because the absolute value of the slope
of its line is greater.
rise
Skill Practice
300
350
8. Slope 5 }
run 5 } 5 0.857
The roller coaster is more steep than the Magnum
because the slope of its line is greater (0.875 > 0.5125).
The roller coaster is less steep than the Millenium Force
because the slope of its line is less (0.875 < 1).
y2 2 y1
26 2 2
28
8. m 5 } 5 } 5 } 5 22
x2 2 x1
4
2 2 (22)
The slope is 22.
y2 2 y1
21 2 (21)
0
9. m 5 } 5 } 5 } 5 0
x2 2 x1
8
3 2 (25)
The slope is 0.
y2 2 y1
621
5
5
10. m 5 } 5 } 5 } 5 2}
x2 2 x1
022
22
2
5
The slope is 2}2.
run
11. The slope of the line was computed using } instead
rise
rise
3
.
of }
run . The rise is 3 and the run is 4, so the slope is }
4
12. The values of x2 and x1 were interchanged when
computing the slope.
y2 2 y1
725
2
m5}
5}
5}
5 21
x 2x
224
22
2
1
y2 2 y1
420
2
13. Slope of line 1: m1 5 } 5 } 5 }
x2 2 x1
721
3
y2 2 y1
620
6
3
Slope of line 2: m2 5 }
5}
5}
5 2}2
x2 2 x1
327
24
2
3
Because m1m2 5 }3 + 2}2 5 21, the lines
are perpendicular.
y2 2 y1
3
22 2 1
14. Slope of line 1: m1 5 } 5 } 5 }
x2 2 x1
4
27 2 (23)
y2 2 y1
4 2 (21)
5
Slope of line 2: m2 5 }
5}
5 }6
x2 2 x1
822
Because m1 Þ m2 and m1m2 Þ 21, the lines are neither
parallel nor perpendicular.
y2 2 y1
723
4
15. Slope of line 1: m1 5 } 5 } 5 } 5 1
x2 2 x1
4
25 2 (29)
y2 2 y1
226
24
5}
5}
5 21
Slope of line 2: m2 5 }
x2 2 x1
4
27 2 (211)
Because m1m2 5 1 + (21) 5 21, the lines
are perpendicular.
Geometry
Worked-Out Solution Key
63
Chapter 3,
continued
1
16. P(3, 22), slope 2}
6
y
n y
1
1
h
x
(3, 22)
1
(9, 23)
21
(3, 1)
21
6
x
P(3, 23)
(23, 22)
5
17. P(24, 0), slope }
2
y
1
y2 2 y1
26
22 2 4
24. Slope of line h: m1 5 } 5 } 5 } 5 23
x2 2 x1
523
2
(22, 5)
5
Slope of line n: 23 + m2 5 21
1
25
(24, 0)
x
1
m2 5 }3
y
2
18. P(0, 5), slope }
3
y
(4, 25)
22
2
(3, 4)
3
3
(3, 7)
2
1
1
(0, 5)
(9, 2)
P(6, 1)
3
x
(5, 22)
n
h
21
x
y2 2 y1
523
2
19. Slope of line 1: m1 5 } 5 } 5 }
5
x2 2 x1
3 2 (22)
y2 2 y1
521
4
Slope of line 2: m2 5 }
5}
5 }3
x2 2 x1
623
y2 2 y1
24 2 (23)
1
25. Slope of line h: m1 5 } 5 } 5 2}
7
x2 2 x1
2 2 (25)
1
Slope of line n: 2}7 + m2 5 21
m2 5 7
n
Line 2 is steeper because the slope of its line is
1
4 2
greater 1 }3 > }5 2.
22 2 (21)
1
20. Slope of line 1: m1 5 } 5 } 5 2}
x2 2 x1
3
1 2 (22)
y2 2 y1
24 2 (23)
1
Slope of line 2: m2 5 }
5}
5 2}4
x2 2 x1
21 2 (25)
Line 1 is steeper because the absolute value of the slope
{ { {2
1{
1
1
of its line is greater 2}3 > 2}4 .
y2 2 y1
622
21. Slope of line 1: m1 5 } 5 } 5 4
x2 2 x1
23 2 (24)
y2 2 y1
826
2
21
7
y2 2 y1
y
(23, 1)
1
x
(2, 24)
(25, 23)
h
P(24, 26)
26. If the points all lie on the same line, then the slope of the
line containing any two of the points will be the same.
Use (23, 3) and (1, 22):
y2 2 y1
22 2 3
5
5}
5 2}4
m1 5 }
x 2x
1 2 (23)
2
1
5}
5 }2 5 1
Slope of line 2: m2 5 }
x 2x
321
Use (23, 3) and (4, 0):
Line 1 is steeper because the slope of its line is
greater (4 > 1).
m2 5 }
5}
5 2}7
x 2x
4 2 (23)
2
1
22. You can determine which of two lines is steeper by
comparing the slope of each line. The line that has the
slope with the greater absolute value is steeper.
y2 2 y1
1 2 (22)
3
1
23. Slope of line h: m1 5 } 5 } 5 } 5 }
x2 2 x1
6
2
3 2 (23)
1
Slope of line n: }2 + m2 5 21
m2 5 22
y2 2 y1
2
023
3
1
Because the slopes of lines are different, you know that
the points do not all lie on the same line.
y
(23, 3)
1
(4, 0)
21
x
(1, 22)
27. Parallel lines have the same slope.
Use (22, 4) and (25, 1):
y2 2 y1
124
m2 5 m1 5 }
5}
51
x 2x
252 (22)
2
64
Geometry
Worked-Out Solution Key
1
y
(0, 2)
1
(1, 3)
1
1
1
x
1
Chapter 3,
continued
28. The product of the slopes of two perpendicular lines
34. B; The slope of the line for car B is less steep than the
is 21.
Use (21, 21) and (2, 0):
y2 2 y1
0 2 (21)
y
1
m1 5 }
5}
5 }3
x 2x
2 2 (21)
2
1
1
21
1
} + m2 5 21
3
slope of the line for car A, so the gas remaining in car
B does not deplete as quickly as the distance driven
increases. Car B has the better gas mileage.
(1, 3)
23
35.
(2, 0)
1
a
(0, 4
)
c
m2 5 23
2
1
(0, 1)
2
1
2}2 2 1
y2 2 y1
3
}
}
m2 5 m1 5
5
5 2}
x2 2 x1
423
y
Line b is the most steep because the absolute value of its
slope is the greatest. Line c is the least steep because the
absolute value of its slope is the least.
x
rise
y2 2 y1
The recommended slope for a ladder is 4.
}v
4
m5}
x 2x
b. When h 5 6:
v
}54
h
1
y22
22 5 }
0 2 (23)
y22
v
6
}54
22 5 }
3
v 5 24
22(3) 5 y 2 2
The ladder touches the building 24 feet above
the ground.
26 5 y 2 2
24 5 y
c. When v 5 34:
v
}54
h
y2 2 y1
m5}
x2 2 x1
0 2 (24)
1
}5}
x 2 (27)
3
1
3
34
h
}54
8.5 5 h
4
x17
}5}
The base of the ladder is 8.5 feet away from
the building.
x 1 7 5 12
x55
32.
v
slope 5 }
5}
54
run 5 }
1
h
(0, 22)
2
31.
v
2
37. a.
y2 2 y1
Horizontal (ft)
0
50
100 150 200
x2 2 x1
Vertical (ft)
0
29
58
m5}
1 2 (23)
24 5 }
x24
4
24 5 }
x24
87
116
Horizontal (ft) 250 300 350 400 450
Vertical (ft)
145 174 203 232 261
24(x 2 4) 5 4
Horizontal (ft) 500 550 600 650 700
24x 1 16 5 4
Vertical (ft)
24x 5 212
290 319 348 377 406
At the top, the incline is 406 feet high.
x53
rise
29
b. slope 5 }
run 5 }
50
Problem Solving
rise
6
2
33. slope 5 }
5 }3
run 5 }
9
2
The slope of the slide is }3 .
Vertical dist. (ft)
30.
2
x
1
36. a. h 5 } v
4
1
1
(3, 0)
21
1
Use (3, 1) and 4, 2}2 :
23
(0, 6
)
x
29. Parallel lines have the same slope.
(22, 1)
y
b
600
(700, 406)
400
200 (50, 29)
0
0
200
400
600
800
Horizontal dist. (ft)
Geometry
Worked-Out Solution Key
65
continued
rise
144
c. slope 5 }
run 5 }
271
144
The slope of the Burgenstock Incline is }
.
271
The Duquesne Incline is steeper because the slope of
1
29
50
144
271
42. The slope of the line using (23, 1) and (0, k),
k21
, must be the same as the slope of the line
m1 5 }
3
4
, in order for the
using (23, 1) and (k, 5), m2 5 }
k13
points to be collinear.
2
m1 5 m2
its climb path is greater } > } .
4
k21
}5}
k13
3
38. Given: p i q, r i q
Prove: p ir
It is given that p i q, so by the Slopes of Parallel Lines
Postulate, mp 5 mq. It is also given that q i r, so by the
Slopes of Parallel Lines Postulate, mq 5 mr. By the
Transitive Property of Equality, mp 5 mr. Therefore, by
the Parallel Lines Postulate, p i r.
change in proﬁt
39. average rate of change 5 }}
change in time
15,400 2 8500
6900
1150
The average rate of change is \$1150 per year.
(k 1 5)(k 2 3) 5 0
k1550
k 5 25
or
k2350
or
k53
Mixed Review for TAKS
43. A;
1
V 5 }3 Bh
change in height
change in time
40. average rate of change 5 }}
306
45
k2 1 2k 2 15 5 0
The two values of k are 25 and 3.
5 }}
5}
5}
2006 2 2000
6
1
706 2 400
45
(k 2 1)(k 1 3) 5 12
5 }31 }2 2(12)(5)(10)
1 1
6.8
1
5}5}5}
The averate rate of change is 6.8 feet per minute.
41. a. Because the slope of the graph is steepest during that
period, the NBA attendance increased the most from
1985 to 1990.
change in attendance
rate of change 5 }}
change in time
16 million 2 7 million
5 }}
1990 2 1985
9 million
5}
5 1.8
5
The rate of change from 1985–1990 is about
1.8 million people per year.
b. Because the slope of the graph is steepest during that
period, the NHL attendance increased the most from
1995 to 2000.
change in attendance
rate of change 5 }}
change in time
18 million 2 6 million
5 }}
5
12 million
5}
5 2.4
5
The rate of change from 1995–2000 is about
2.4 million people per year.
c. The graph for the NFL attendance shows that there
was a small but steady increase in attendance from
1985 to 2000.
5 100
The volume of the triangular pyramid is
100 cubic centimeters.
Quiz 3.3–3.4 (p.178)
1. Using the Consecutive Interior Angles Converse,
2x 1 54 5 180
2 x 5 126
x 5 63.
The lines are parallel when x 5 63.
2. Using the Corresponding Angles Converse,
3x 2 5 5 145
3x 5 150
x 5 50.
The lines are parallel when x 5 50.
3. Using the Alternate Exterior Angles Converse,
88 5 4x 2 12
100 5 4x
25 5 x.
The lines are parallel when x 5 25.
y2 2 y1
3 2 (21)
4
4. m 5 } 5 } 5 } 5 2
x2 2 x1
321
2
The slope of the line is 2.
y2 2 y1
522
3
5. m 5 } 5 } 5 } 5 1
x2 2 x1
421
3
The slope of the line is 1.
y2 2 y1
26 2 (22)
24
6. m 5 } 5 } 5 } 5 1
x2 2 x1
24
27 2 (23)
The slope of the line is 1.
66
Geometry
Worked-Out Solution Key
Chapter 3,
Chapter 3,
continued
9. Cost of one month’s rental online: y 5 16.5
Technology Activity 3.4 (p. 179)
Cost of one month’s rental locally: y 5 4x, where
x represents the number of DVDs rented
2. When one of the lines is vertical, its slope is undeﬁned,
so the product of the slopes is also undeﬁned.
Cost (dollars)
Lesson 3.5
3.5 Guided Practice (pp. 181–183)
y2 2 y1
1 2 (21)
2
1. m 5 } 5 } 5 }
x2 2 x1
320
3
y
36
32
y 5 4x
28
24
20 (4.125, 16.5)
16
y 5 16.5
12
8
4
0
y-intercept: 21
y 5 mx 1 b
The point of intersection is (4.125, 16.5). Using the
graph, you can see that it is cheaper to rent online if you
rent 5 or more DVDs per month.
225
23
2. m 5 } 5 } 5 } 5 21
x2 2 x1
3
1 2 (22)
10. With a 2-for-1 coupon, the equation for two or more local
y 5 mx 1 b
rentals is y 5 4(x 2 1) 5 4x 2 4. This graph intersects
the graph of y 5 16.5 at the point (5.125, 16.5). So, you
would have to rent 6 or more DVDs to make the online
2 5 21(1) 1 b
35b
y 5 2x 1 3
3. The line parallel to y 5 3x 2 5 has a slope of 3.
m 5 3, (x, y) 5 (1, 5)
3.5 Exercises (pp. 184–187)
y
Skill Practice
(1, 5)
y 5 mx 1 b
1. In the expression slope-intercept form, the word intercept
3
5 5 3(1) 1 b
refers to the point where the line crosses the y-axis.
(0, 2)
25b
2. To ﬁnd the x-intercept, let y 5 0 and solve for x. To ﬁnd
y 5 3x 1 2
the y-intercept, let x 5 0 and solve for y.
x
1
y 5 3x 1 2
0 2 (24)
4
3. m 5 } 5 }
320
3
(1, 22)
y 5 3x 2 5
0 1 2 3 4 5 6 7 8 9 x
Number of rentals
2
y 5 }3 x 2 1
y2 2 y1
(0, 25)
4. You know x 5 4 and y 5 2 are perpendicular because
x 5 4 is a vertical line and y 5 2 is a horizontal line.
Horizontal and vertical lines are always perpendicular.
22 2 (23)
1
4. m 5 } 5 }
5
0 2 (25)
y-intercept: 24
y-intercept: 22
y 5 mx 1 b
y 5 mx 1 b
4
y 5 }3 x 2 4
y 5 }5 x 2 2
1
26
3
22 2 4
5. m 5 } 5 } 5 2}
4
2
1 2 (23)
5. The slope is the monthly fee, \$50, and the y-intercept is
the cost of joining the gym, \$125.
y 5 mx 1 b
6. 2x 2 3y 5 6
3
4 5 2}2 (23) 1 b
x-intercept:
y-intercept:
2x 2 3(0) 5 6
2(0) 2 3y 5 6
2x 5 6
23y 5 6
x53
9
4 5 }2 1 b
1
2}2 5 b
y 5 22
y
1
3
(3, 0)
21
1
y 5 2}2x 2 }2
23 2 3
6
6. m 5 } 5 2}
5
2 2 (23)
x
(0, 22)
y 5 mx 1 b
6
7. y 5 4
3 5 2}5 (23) 1 b
8. x 5 23
y
(23, 4)
3
2}5 5 b
y
(2, 4)
(23, 5)
6
3
y 5 2}5 x 2 }5
1
21
1
21
x
x
(23, 21)
Geometry
Worked-Out Solution Key
67
```
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