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EXAMPLE 1.3
Find the set of points that satisfy the condition
|x + 3| ¤ 2 (*)
and exhibit it on a number line.

SOLUTION
In case x + 3 ≥ 0 then |x + 3| = x + 3 and we may write condition (—) as
x+3¤2
or
x ¤ ’1.
Combining x + 3 ≥ 0 and x ¤ ’1 gives ’3 ¤ x ¤ ’1.
On the other hand, if x + 3 < 0 then |x + 3| = ’(x + 3). We may then write
condition (—) as
’(x + 3) ¤ 2
or
’5 ¤ x.
Combining x + 3 < 0 and ’5 ¤ x gives ’5 ¤ x < ’3.
We have found that our inequality |x + 3| ¤ 2 is true precisely when either
’3 ¤ x ¤ ’1 or ’5 ¤ x < ’3. Putting these together yields ’5 ¤ x ¤ ’1.
We display this set in Fig. 1.5.

_9 _6 _3 0 3 6 9

Fig. 1.5

You Try It: Solve the inequality |x’4| > 1. Exhibit your answer on a number line.
You Try It: On a real number line, sketch the set {x: x 2 ’ 1 < 3}.
CHAPTER 1 Basics 5

1.3 Coordinates in Two Dimensions
We locate points in the plane by using two coordinate lines (instead of the single
line that we used in one dimension). Refer to Fig. 1.6. We determine the coordinates
of the given point P by ¬rst determining the x-displacement, or (signed) distance
from the y-axis and then determining the y-displacement, or (signed) distance from
the x-axis. We refer to this coordinate system as (x, y)-coordinates or Cartesian
coordinates. The idea is best understood by way of some examples.

y



P



x



Fig. 1.6



EXAMPLE 1.4
Plot the points P = (3, ’2), Q = (’4, 6), R = (2, 5), S = (’5, ’3).

SOLUTION
The ¬rst coordinate 3 of the point P tells us that the point is located 3 units
to the right of the y-axis (because 3 is positive). The second coordinate ’2 of
the point P tells us that the point is located 2 units below the x-axis (because
’2 is negative). See Fig. 1.7.
The ¬rst coordinate ’4 of the point Q tells us that the point is located 4 units
to the left of the y-axis (because ’4 is negative). The second coordinate 6 of
the point Q tells us that the point is located 6 units above the x-axis (because
6 is positive). See Fig. 1.7.
The ¬rst coordinate 2 of the point R tells us that the point is located 2 units
to the right of the y-axis (because 2 is positive). The second coordinate 5 of the
point R tells us that the point is located 5 units above the x-axis (because 5 is
positive). See Fig. 1.7.
The ¬rst coordinate ’5 of the point S tells us that the point is located 5 units
to the left of the y-axis (because ’5 is negative). The second coordinate ’3 of
the point S tells us that the point is located 3 units below the x-axis (because
’3 is negative). See Fig. 1.7.
6 CHAPTER 1 Basics

y


Q
R
4



1

x
1 4

S P




Fig. 1.7

EXAMPLE 1.5
Give the coordinates of the points X, Y, Z, W exhibited in Fig. 1.8.

y




Z



Y

x

X

W




Fig. 1.8


SOLUTION
The point X is 1 unit to the right of the y-axis and 3 units below the x-axis.
Therefore its coordinates are (1, ’3).
The point Y is 2 units to the left of the y-axis and 1 unit above the x-axis.
Therefore its coordinates are (’2, 1).
CHAPTER 1 Basics 7

The point Z is 5 units to the right of the y-axis and 4 units above the x-axis.
Therefore its coordinates are (5, 4).
The point W is 6 units to the left of the y-axis and 5 units below the x-axis.
Therefore its coordinates are (’6, ’5).
You Try It: Sketch the points (3, ’5), (2, 4), (π, π/3) on a set of axes. Sketch
the set {(x, y): x = 3} on another set of axes.
EXAMPLE 1.6
= {(x, y): y = 3}. Sketch the set of points k =
Sketch the set of points
{(x, y): x = ’4}.
SOLUTION
The set consists of all points with y-coordinate equal to 3. This is the set
of all points that lie 3 units above the x-axis. We exhibit in Fig. 1.9. It is a
horizontal line.




l




Fig. 1.9

The set k consists of all points with x-coordinate equal to ’4. This is the set
of all points that lie 4 units to the left of the y-axis. We exhibit k in Fig. 1.10.
It is a vertical line.

EXAMPLE 1.7
Sketch the set of points S = {(x, y): x > 2} on a pair of coordinate axes.
SOLUTION
Notice that the set S contains all points with x-coordinate greater than 2.
These will be all points to the right of the vertical line x = 2. That set is
exhibited in Fig. 1.11.

You Try It: Sketch the set {(x, y): x + y < 4}.
You Try It: Identify the set (using set builder notation) that is shown in Fig. 1.12.
8 CHAPTER 1 Basics




k




Fig. 1.10

y




Y
FL
AM

x
TE




Fig. 1.11



1.4 The Slope of a Line in the Plane
A line in the plane may rise gradually from left to right, or it may rise quite steeply
from left to right (Fig. 1.13). Likewise, it could fall gradually from left to right,
or it could fall quite steeply from left to right (Fig. 1.14). The number “slope”
differentiates among these different rates of rise or fall.
Look at Fig. 1.15. We use the two points P = (p1 , p2 ) and Q = (q1 , q2 ) to
calculate the slope. It is
q 2 ’ p2
m= .
q1 ’ p1
CHAPTER 1 Basics 9

y




1

x
1




Fig. 1.12

y




x




Fig. 1.13

It turns out that, no matter which two points we may choose on a given line, this
calculation will always give the same answer for slope.
EXAMPLE 1.8
Calculate the slope of the line in Fig. 1.16.

SOLUTION
We use the points P = (’1, 0) and Q = (1, 3) to calculate the slope of
this line:
3’0 3
m= =.
1 ’ (’1) 2
10 CHAPTER 1 Basics

y




x




Fig. 1.14

y




Q



P

x




Fig. 1.15


We could just as easily have used the points P = (’1, 0) and R = (3, 6) to
calculate the slope:
6’0 6 3
m= ==.
3 ’ (’1) 4 2
If a line has slope m, then, for each unit of motion from left to right, the line
rises m units. In the last example, the line rises 3/2 units for each unit of motion to
the right. Or one could say that the line rises 3 units for each 2 units of motion to
the right.
CHAPTER 1 Basics 11

y


R = (3,6)



Q = (1,3)

P = (_ 1,0)
x




Fig. 1.16

y



R = (_ 2,10) 10
8
6
S = (_ 1,5)
4
2

x
246

T = (1,_ 5)




Fig. 1.17


EXAMPLE 1.9
Calculate the slope of the line in Fig. 1.17.

SOLUTION
We use the points R = (’2, 10) and T = (1, ’5) to calculate the slope of
this line:
10 ’ (’5)
m= = ’5.
(’2) ’ 1
12 CHAPTER 1 Basics

We could just as easily have used the points S = (’1, 5) and T = (1, ’5):
5 ’ (’5)
m= = ’5.
’1 ’ 1
In this example, the line falls 5 units for each 1 unit of left-to-right motion. The
negativity of the slope indicates that the line is falling.
The concept of slope is unde¬ned for a vertical line. Such a line will have any
two points with the same x-coordinate, and calculation of slope would result in
division by 0.

You Try It: What is the slope of the line y = 2x + 8?

You Try It: What is the slope of the line y = 5? What is the slope of the line
x = 3?
Two lines are perpendicular precisely when their slopes are negative reciprocals.
This makes sense: If one line has slope 5 and the other has slope ’1/5 then we
see that the ¬rst line rises 5 units for each unit of left-to-right motion while the
second line falls 1 unit for each 5 units of left-to-right motion. So the lines must be
perpendicular. See Fig. 1.18(a).

y




x




Fig. 1.18(a)


You Try It: Sketch the line that is perpendicular to x +2y = 7 and passes through
(1, 4).
Note also that two lines are parallel precisely when they have the same slope.
See Fig. 1.18(b).
CHAPTER 1 Basics 13

y




x


Fig. 1.18(b)



1.5 The Equation of a Line
The equation of a line in the plane will describe”in compact form”all the points
that lie on that line. We determine the equation of a given line by writing its
slope in two different ways and then equating them. Some examples best illustrate
the idea.
EXAMPLE 1.10
Determine the equation of the line with slope 3 that passes through the point
(2, 1).

SOLUTION
Let (x, y) be a variable point on the line. Then we can use that variable point
together with (2, 1) to calculate the slope:
y’1
m= .
x’2
On the other hand, we are given that the slope is m = 3. We may equate the

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