Section 3.4 The Slope-Intercept Form of the Equation of a Line
63. If the line rises from left to right, it has a positive
slope. It passes through the origin, (0, 0) and a
second point with equal x– and y-coordinates. The
64. If the line falls from left to right, the slope must be
negative. It passes through the origin,
0, 0 , and
has a second point with opposite x– and y–
65. a. The y–intercept is 64 and the slope is
Change in 46 64 18 0.45
y
66. a. The y–intercept is 16 and the slope is
Change in 30 16 14 0.35.
y
mx
67. – 69. Answers will vary.
70. makes sense
71. does not make sense; Explanations will vary.
Sample explanation: The slope can be determined if
74. false; Changes to make the statement true will vary.
A sample change is: It is possible for m to equal b.
A sample change is: By solving the equation for y,
you can determine that the y–intercept is 7.
2
5
79. 713
xx
324
8
x
x
The solution is {8}.
2
30 30 0
81. A = 14, P = 25% = 0.25
14 0.25
APB
B
Chapter 3 Linear Equations in Two Variables
82. 34( 1)
34 4
47
yx
yx
yx
−= +
−= +
=+
83. 3
3(4)
2
yx
+=− −
Mid-Chapter Check Point – Chapter 3
1. a. The x-intercept is 4.
b. The y-intercept is 2.
c. The points (4, 0) and (0, 2) lie on the line.
3. a. The x-intercept is 0.
b. The y-intercept is 0.
c. The points (0, 0) and (5, 3) lie on the line.
5. 2y=−
7.
12
3
yx=−
Mid-Chapter Check Point
10. 32yx=+
11.
30
3
xy
yx
+=
=−
13. 4yx=−
14.
53
3
5
yx
yx
=−
=−
15.
520
4
y
y
=
=
16.
52 10
2510
55
2
xy
yx
yx
−=
−=−+
=−
The slope is
5
2
and the y–intercept is −5.
17. Line through (–5,–3) and (0,–4):
lines are perpendicular.
18. Line through (–4,1) and (2,7):
Change in 7 1 6 1
Change in 2 ( 4) 6
y
mx
−
====
−−
parallel.
Since the product of their slopes is
()
12 2 1,−=−≠− the lines are not perpendicular.
19. Line through (2,–4) and (7,0):
Chapter 3 Linear Equations in Two Variables
20. a. The y-intercept is 29 and the slope is Change in 49 29 20 5
Change in 4 0 4
y
mx
−
====
−.
3.5 Check Points
1. Begin with the point-slope equation of a line.
()
11
yy mxx
−= −
2. a. Begin by finding the slope: 6(1) 5 5
1(2) 1
m−−− −
===−
−−−
Using the slope and either point, find the point-slope equation of a line.
3. Since the line is parallel to 31,yx=+
we know it will have slope
3.m=
We are given that it passes through
(2,5).−
We use the slope and point to write the equation in point-slope form.
()
()
()
11
53 2
yy mxx
yx
−= −
−= −−
Section 3.5 The Point-Slope Form of the Equation of a Line
4. a. Solve the given equation for y to obtain slope-
intercept form.
312
xy
+=
b. We use the slope of 3 and the point
(2,6)−−
to
write the equation in point-slope form.
()
11
yy mxx
−= −
5. Find slope:
32.8 30.0 2.8 0.28
20 10 10
m−
===
−
Use the point-slope form to write the equation. Then
solve for y to obtain slope-intercept form.
()
11
yy mxx
−= −
3.5 Concept and Vocabulary Check
1.
12
()yy mxx−= −
2. standard
6. vertical
7.
4−
;
4−
1. Begin with the point-slope equation of a line.
()
()
11
53 2
yy mxx
yx
−= −
−= −
Now solve this equation for y to write the equation
Now solve this equation for y to write the equation
in slope-intercept form.
()
16 3
16 18
617
yx
yx
yx
−= −
−= −
=−
4. Begin with the point-slope equation of a line.
()
()
()
()
11
97 4
97 4
yy mxx
yx
yx
−= −
−= −−
−= +
Chapter 3 Linear Equations in Two Variables
5. Begin with the point-slope equation of a line.
()
() ()
()
11
28 3
yy mxx
yx
−= −
−− =− −−
6. Begin with the point-slope equation of a line.
()
11
yy mxx
−= −
7. Begin with the point-slope equation of a line.
()
()
()
11
012 8
yy mxx
yx
−= −
−=− −−
8. Begin with the point-slope equation of a line.
()
() ( )
11
3110
yy mxx
yx
−= −
−− =− −
11 3
yx
=− −
9. Begin with the point-slope equation of a line.
()
11
1
yy mxx
−= −
1
22
yx
+=−−
()
4
114
4
yx
+=− +
Now solve this equation for y to write the equation
in slope-intercept form.
11. Begin with the point-slope equation of a line.
()
()
11
1
00
yy mxx
yx
−= −
−= −
Section 3.5 The Point-Slope Form of the Equation of a Line
12. Begin with the point-slope equation of a line.
()
()
11
1
00
yy mxx
yx
−= −
−= −
3
13. Begin with the point-slope equation of a line.
()
()
11
2
(2) 6
yy mxx
yx
−= −
−− =− −
14. Begin with the point-slope equation of a line.
()
()
()
11
3
(4) 10
5
3
410
yy mxx
yx
yx
−= −
−− =− −
+=− −
15. slope =
10 2 8 2
51 4
−==
−
Using the slope and either point, find the point-
slope equation of a line.
22 2
2
yx
yx
−= −
=
16. slope =
15 5 10 2
83 5
−==
−
17. slope =
30 3 1
03 3
−==
+
Using the slope and either point, find the point-
slope equation of a line.
()
() ()
11
01 3 or 31 0
yy mxx
yx yx
−= −
−= + −= −
Chapter 3 Linear Equations in Two Variables
19. slope =
41 5 1
23 5
+==
+
Using the slope and either point, find the point-
slope equation of a line.
20. slope =
()
()
14
31
12 3
−−− ==
−−
Using the slope and either point, find the point-
slope equation of a line.
()
yy mxx
−= −
21. slope =
()
()
41
5
347
−− =
−−
Using the slope and either point, find the point-
slope equation of a line.
22. slope =
()
51 6 3
268 4
−− −
==−
−−
Using the slope and either point, find the point-
39
142
37
42
yx
yx
−=− −
=− −
11 0 0
−+ ==
in slope-intercept form.
()
10 3
10
1
yx
y
y
+= +
+=
=−
Section 3.5 The Point-Slope Form of the Equation of a Line
25. Use the points (2,4) and (−2,0) to find the slope.
slope =
04 4 1
22 4
+−
==
−− −
26. Use the points (1,–3) and (−1,0) to find the slope.
slope =
()
0333
11 2 2
−− ==−
−− −
Find the point-slope equation of a line.
27. Use the points
1,0
2
−
and (0,4) to find the slope.
slope = 40 4 8
11
022
−==
+
Find the point-slope equation of a line.
()
11
yy mxx
−= −
28. Use the points
()
4, 0 and
()
0, 2−to find the slope.
slope = 20 1
−− =
12
2
yx
=−
29. For 5,yx= 5.
m
=
slope, 7.
m
=−
b. A line perpendicular to it would have slope
1.
7
m
=
32. a. Parallel: 9
m
=−
Chapter 3 Linear Equations in Two Variables
34. a. Parallel: 1
m
=
35. For 21,
5
yx
=− − 2.
5
m
=−
a. A line parallel to this line would have the same
2
36. a. Parallel: 3
7
m
=−
b. Perpendicular: 7
3
m
=
37. To find the slope, we rewrite the equation in slope-
intercept form.
47
xy
+=
38. 811
811
xy
yx
+=
=− +
a. Parallel: 8
m
=−
39. To find the slope, we rewrite the equation in slope-
intercept form.
12
2
yx
=− +
So, 1.
2
m
=−
2.
m
=
40. 32 6
236
33
2
xy
yx
yx
+=
=− +
=− +
a. Parallel: 3
2
m
=−
b. Perpendicular: 2
3
m
=
a. A line parallel to this line would have the same
slope, 2.
3
m
=
Section 3.5 The Point-Slope Form of the Equation of a Line
42. 34 7
437
xy
yx
−=−
−=−−
43. We know that 6
x
= is a vertical line with
undefined slope.
a. A line parallel to it would also be vertical with
undefined slope.
45. Since L is parallel to 2,yx= we know it will have
slope
2.m
= We are given that it passes through
(4, 2). We use the slope and point to write the
equation in point-slope form.
46. L will have slope
2m
=− . Using the point and the
slope, we have
()
42 3.
yx−=− −
Solve for y to
obtain slope-intercept form.
426
210
yx
yx
−=− +
=− +
47. Since L is perpendicular to 2,yx= we know it will
Solve for y to obtain slope-intercept form.
1
42
yx
−=− −
(–1, 2). Use the slope and point to write the
equation in point-slope form.
()
()
1
21
2
yx
−= −−
15
22
yx
=+
49. Since the line is parallel to 43,yx=− + we know it
will have slope
4.
m=− We are given that it passes
()
10 4 8
10 4 32
442
yx
yx
yx
+=− +
+=−−
=− −
50. L will have slope
5.
m=− The line passes through
(–2, –7). Use the slope and point to write the
equation in point-slope form.
() ()
()
75 2
yx
−− =− −−
Chapter 3 Linear Equations in Two Variables
51. Since the line is perpendicular to 16,
5
yx=+
we
know it will have slope 5.m=− We are given that
it passes through (2, –3). We use the slope and
52. L will have slope 3.m=− The line passes through
(–4, 2 ). We use the slope and point to write the
equation in point-slope form.
53. To find the slope, we rewrite the equation in slope-
intercept form.
23 7
xy
−=
will have slope 2.
3
m= We are given that it passes
through (–2, 2). We use the slope and point to write
the equation in point-slope form.
()
yy mxx
−= −
54. Find the slope.
32 5
235
35
xy
yx
−=
−=−+
Solve for y to obtain slope-intercept form.
33
322
yx
−= +
55. To find the slope, we rewrite the equation in slope-
intercept form.
23
23
xy
yx
−=
−=−+
it passes through (4, –7). We use the slope and
point to write the equation in point-slope form.
()
() ( )
()
11
724
72 4
yy mxx
yx
yx
−= −
−− =− −
+=− −
Section 3.5 The Point-Slope Form of the Equation of a Line
56. Find the slope.
712
712
xy
yx
+=
=− +
57. Through (2, 4) and same y-intercept as x − 4y = 8.
Solve the equation to obtain the y-intercept.
48
48
xy
yx
−=
−=−+
58. Through
()
2, 6
with the same y-intercept as the
graph of 318xy−=
.
Find the y-intercept:
03 18
318
y
y
−=
−=
59. x-intercept at −4 and parallel to the line containing
(3, 1) and (2, 6)
First, find the slope of the line going through the
60. x-intercept at
6−
and parallel to the line containing
()
4, 3−
and
()
2, 2
.
Find the slope of the line.
()
2355
m−−
===−
515
2
yx
=− −
61. Since the line is perpendicular to
6x=
which is a
vertical line, we know the graph is a horizontal line
with 0 slope. The graph passes through
()
1, 5−
, so