Since the slope is ^17 , the graph of the line rises
from left to right at a rate of one vertical unit
up per seven horizontal units right. The y-
intercept is (0,–2), and the correct graph is
given by choice d.
- c.The graph ofy= cis a horizontal line that is
either above or below the x-axis. If it lies above
the x-axis, the graph crosses into only Quad-
rant I and Quadrant II, while if it lies below
the x-axis, it crosses into only Quadrant III
and Quadrant IV.
- b.The graph ofy= c,where c 0, is a hori-
zontal line that lies either above or below the
x-axis, and must cross the y-axis.
- a.For instance, consider the line whose equa-
tion is y= –x– 1. Its graph is shown here:
Observe that the graph does indeed cross into
three of the four quadrants.
- e.A line perpendicular to the given line must
have a slope m= ^32 . So, the line given in choice
eis the correct choice.
- e.Two lines are parallel if and only if they
have the same slope. This is true for the line
provided in choice esince the slopes of both
this lineand the given one are 6.
275.b. The line provided in choice bis equivalent
to y = –2x+6. Since this has the same slope as
the given line, namely –2, we conclude that the
correct answer is b.
Set 18 (Page 42)
- d.Since we want a line perpendicular to a line
with slope m 1 = ^34 , we must use m 2 = – = –^43
as the slope. Since the point (–6,4) must be on
the line, the point-slope formula for the line is
y– 4 = –^43 (x+ 6). This is equivalent to the
equation y= –^43 x– 4.
- b.A line parallel to y= 3x+ 8 must have
slope 3. Using the point-slope formula for a
line with the point (4,4), we see that the
equation of the line we seek is y– 4 = 3(x– 4),
which simplifies to y = 3x– 8.
- b.The slope of the line passing through the
two given points is m= –^65 – –^24 = –^49 . This is
actually the slope of the line we seek because
the line parallel to the one containing the two
given points. Using this slope with the point
(0,12), we see that the point-slope form of the
equation of the lines is y– 12 = –^49 (x– 0),
which simplifies to y = –^49 x+ 12.
279 c.A line perpendicular to the given line must
have slope ^1183 . Using this slope with the point
(0,0), we see that the point-slope form
of the equation of the line we seek is y– 0 =
^1183 (x– 0), which simplifies to y= ^1183 x.
- a.Only vertical lines have undefined slopes.
The only vertical line among the choices
provided is given by choice a.
- b.Only horizontal lines have zero slopes. The
only horizontal line among the choices pro-
vided is given by choice b.
^1
^34
x
y
–10 –8 –6 –4 –2 2 4 6 8 10
–2
–4
–6
–8
–10
10
8
6
4
2
ANSWERS & EXPLANATIONS–
