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3 Functions and Graphs
3.3 Lines
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Lines
One of the basic concepts in geometry is that of a line. In
this section we will restrict our discussion to lines that lie in
a coordinate plane.
This will allow us to use algebraic methods to study their
properties. Two of our principal objectives may be stated as
follows:
(1) Given a line l in a coordinate plane, find an equation
whose graph corresponds to l.
(2) Given an equation of a line l in a coordinate plane,
sketch the graph of the equation.
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Lines
The following concept is fundamental to the study of lines.
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Example 1 – Finding slopes
Sketch the line through each pair of points, and find its
slope m:
(a) A(–1, 4) and B(3, 2)
(b) A(2, 5) and B(–2, –1)
(c) A(4, 3) and B(–2, 3)
(d) A(4, –1) and B(4, 4)
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Example 1 – Solution
The lines are sketched in Figure 3. We use the definition of
slope to find the slope of each line.
m=
Figure 3(a)
m=
Figure 3(b)
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Example 1 – Solution
m=0
Figure 3(c)
cont’d
m undefined
Figure 3(d)
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Example 1 – Solution
cont’d
(a)
(b)
(c)
(d) The slope is undefined because the line is parallel to
the y-axis. Note that if the formula for m is used, the
denominator is zero.
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Lines
The diagram in Figure 5 indicates the slopes of several
lines through the origin. The line that lies on the x-axis has
slope m = 0.
Figure 5
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Lines
If this line is rotated about O in the counterclockwise
direction (as indicated by the blue arrow), the slope is
positive and increases, reaching the value 1 when the line
bisects the first quadrant and continuing to increase as the
line gets closer to the y-axis.
If we rotate the line of slope m = 0 in the clockwise direction
(as indicated by the red arrow), the slope is negative,
reaching the value –1 when the line bisects the second
quadrant and becoming large and negative as the line gets
closer to the y-axis.
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Lines
Lines that are horizontal or vertical have simple equations,
as indicated in the following chart.
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Lines
A common error is to regard the graph of y = b as
consisting of only the one point (0, b).
If we express the equation in the form 0  x + y = b, we see
that the value of x is immaterial; thus, the graph of y = b
consists of the points (x, b) for every x and hence is a
horizontal line.
Similarly, the graph of x = a is the vertical line consisting of
all points (a, y), where y is a real number.
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Example 3 – Finding equations of horizontal and vertical lines
Find an equation of the line through A(–3, 4) that is
parallel to
(a) the x-axis
(b) the y-axis
Solution:
The two lines are sketched in
Figure 6. As indicated in the
preceding chart, the equations
are y = 4 for part (a) and
x = –3 for part (b).
Figure 6
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Lines
Let us next find an equation of a line l through a point
P1(x1, y1) with slope m. If P(x, y) is any point with x  x1
(see Figure 7), then P is on l if and only if the slope of the
line through P1 and P is m—that is, if
Figure 7
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Lines
This equation may be written in the form
y – y1 = m(x – x1).
Note that (x1, y1) is a solution of the last equation, and
hence the points on l are precisely the points that
correspond to the solutions.
This equation for l is referred to as the point-slope form.
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Lines
The point-slope form is only one possibility for an equation
of a line. There are many equivalent equations.
We sometimes simplify the equation obtained using the
point-slope form to either
ax + by = c
or
ax + by + d = 0,
where a, b, and c are integers with no common factor,
a > 0, and d = –c.
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Example 4 – Finding an equation of a line through two points
Find an equation of the line through A(1, 7) and B(–3, 2).
Solution:
The line is sketched in Figure 8.
Figure 8
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Example 4 – Solution
cont’d
The formula for the slope m gives us
We may use the coordinates of either A or B for (x1, y1) in
the point-slope form.
Using A(1, 7) gives us the following:
y–7=
(x – 1)
4(y – 7) = 5(x – 1)
point-slope form
multiply by 4
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Example 4 – Solution
4y – 28 = 5x – 5
–5x + 4y = 23
5x – 4y = –23
cont’d
multiply factors
subtract 5x and add 28
multiply by –1
The last equation is one of the desired forms for an
equation of a line.
Another is 5x – 4y + 23 = 0.
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Lines
The point-slope form for the equation of a line may be
rewritten as y = mx – mx1 + y1, which is of the form
y = mx + b
with b = –mx1 + y1. The real number b is the y-intercept of
the graph, as indicated in Figure 9.
Figure 9
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Lines
Since the equation y = mx + b displays the slope m and
y-intercept b of l, it is called the slope-intercept form for
the equation of a line.
Conversely, if we start with y = mx + b , we may write
y – b = m(x – 0).
Comparing this equation with the point-slope form, we see
that the graph is a line with slope m and passing through
the point (0, b).
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Lines
We have proved the following result.
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Example 5 – Expressing an equation in slope-intercept form
Express the equation 2x – 5y = 8 in slope-intercept form.
Solution:
Our goal is to solve the given equation for y to obtain the
form
y = mx + b.
We may proceed as follows:
2x – 5y = 8
given
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Example 5 – Solution
cont’d
–5y = –2x + 8
subtract 2x
y=
divide by –5
y=
equivalent equation
The last equation is the slope-intercept form y = mx + b
with slope
m=
and y-intercept b =
.
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Lines
It follows from the point-slope form that every line is a
graph of an equation
ax + by = c,
where a, b, and c are real numbers and a and b are not
both zero. We call such an equation a linear equation in
x and y.
Let us show, conversely, that the graph of ax + by = c, with
a and b not both zero, is always a line.
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Lines
If b  0, we may solve for y, obtaining
y=
which, by the slope-intercept form, is an equation of a line
with slope –a/b and y-intercept c/b. If b = 0 but a  0, we
may solve for x, obtaining x = c/a, which is the equation of
a vertical line with x-intercept c/a.
This discussion establishes the following result.
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Example 6 – Sketching the graph of a linear equation
Sketch the graph of 2x – 5y = 8.
Solution:
We know from the preceding discussion that the graph is a
line, so it is sufficient to find two points on the graph.
Let us find the x- and y-intercepts by substituting y = 0 and
x = 0, respectively, in the given equation, 2x – 5y = 8.
x-intercept: If y = 0, then 2x = 8, or x = 4.
y-intercept: If x = 0, then –5y = 8, or y =
.
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Example 6 – Solution
cont’d
Plotting the points (4, 0) and
and drawing a line
through them gives us the graph in Figure 10.
Figure 10
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Lines
The following theorem specifies the relationship between
parallel lines (lines in a plane that do not intersect) and
slope.
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Example 7 – Finding an equation of a line parallel to a given line
Find an equation of the line through P(5, –7) that is parallel
to the line 6x + 3y = 4.
Solution:
We first express the given equation in slope-intercept form:
6x + 3y = 4
given
3y = –6x + 4
subtract 6x
y = –2x +
divide by 3
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Example 7 – Solution
cont’d
The last equation is in slope-intercept form, y = mx + b,
with slope m = –2 and y-intercept .
Since parallel lines have the same slope, the required line
also has slope –2.
Using the point P(5, –7) gives us the following:
y – (–7) = –2(x – 5)
y + 7 = –2x + 10
y = –2x + 3
point-slope form
simplify
subtract 7
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Example 7 – Solution
cont’d
The last equation is in slope-intercept form and shows that
the parallel line we have found has y-intercept 3. This line
and the given line are sketched in Figure 12.
Figure 12
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Example 7 – Solution
cont’d
As an alternative solution, we might use the fact that lines
of the form 6x + 3y = k have the same slope as the given
line and hence are parallel to it.
Substituting x = 5 and y = –7 into the equation 6x + 3y = k
gives us 6(5) + 3(–7) = k or, equivalently, k = 9.
The equation 6x + 3y = 9 is equivalent to y = –2x + 3.
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Lines
If the slopes of two nonvertical lines are not the same, then
the lines are not parallel and intersect at exactly one point.
The next theorem gives us information about
perpendicular lines (lines that intersect at a right angle).
A convenient way to remember the conditions on slopes of
perpendicular lines is to note that m1 and m2 must be
negative reciprocals of each other—that is, m1 = –1/m2 and
m2 = –1/m1.
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Example 8 – Finding an equation of a line perpendicular to a given line
Find the slope-intercept form for the line through P(5, –7)
that is perpendicular to the line 6x + 3y = 4.
Solution:
We considered the line 6x + 3y = 4 in Example 7 and found
that its slope is –2.
Hence, the slope of the required line is the negative
reciprocal –[1/(–2)], or .
Using P(5, –7) gives us the following:
y – (–7) =
(x – 5)
point-slope form
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Example 8 – Solution
cont’d
simplify
put in slope-intercept form
The last equation is in slope-intercept form and shows that
the perpendicular line has y-intercept – .
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Example 8 – Solution
cont’d
This line and the given line are sketched in Figure 16.
Figure 16
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Example 9 – Finding an equation of a perpendicular bisector
Given A(–3, 1) and B(5, 4), find the general form of the
perpendicular bisector l of the line segment AB.
Solution:
The line segment AB and its perpendicular bisector l are
shown in Figure 17.
Figure 17
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Example 9 – Solution
cont’d
We calculate the following, where M is the midpoint of AB:
Coordinates of M:
midpoint formula
Slope of AB:
slope formula
Slope of l:
negative reciprocal of
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Example 9 – Solution
Using the point M
and slope
equivalent equations for l:
y–
=–
(x – 1)
cont’d
gives us the following
point-slope form
6y – 15 = –16(x – 1)
multiply by the lcd, 6
6y – 15 = –16x + 16
multiply
16x + 6y = 31
put in general form
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Lines
Two variables x and y are linearly related if y = ax + b,
where a and b are real numbers and a  0.
Linear relationships between variables occur frequently
in applied problems.
The next example gives one illustration.
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Example 10 – Relating air temperature to altitude
The relationship between the air temperature T (in °F) and
the altitude h (in feet above sea level) is approximately
linear for 0  h  20,000.
If the temperature at sea level is 60°, an increase of 5000
feet in altitude lowers the air temperature about 18°.
(a) Express T in terms of h, and sketch the graph on an
hT-coordinate system.
(b) Approximate the air temperature at an altitude of 15,000
feet.
(c) Approximate the altitude at which the temperature is 0°.
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Example 10 – Solution
(a) If T is linearly related to h, then
T = ah + b
for some constants a and b (a represents the slope and
b the T-intercept).
Since T = 60° when h = 0 ft (sea level), the T-intercept
is 60, and the temperature T for 0  h  20,000 is given
by
T = ah + 60.
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Example 10 – Solution
cont’d
From the given data, we note that when the altitude
h = 5000 ft, the temperature T = 60° – 18° = 42°.
Hence, we may find a as follows:
42 = a(5000) + 60
let T = 42 and h = 5000
solve for a
Substituting for a in T = ah + 60 gives us the following
formula for T:
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Example 10 – Solution
cont’d
The graph is sketched in Figure 18, with different scales on
the axes.
Figure 18
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Example 10 – Solution
cont’d
(b) Using the last formula for T obtained in part (a), we find
that the temperature (in °F) when h = 15,000 is
T=–
(15,000) + 60
= –54 + 60
= 6.
(c) To find the altitude h that corresponds to T = 0°,
we proceed as follows:
from part (a)
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Example 10 – Solution
cont’d
Let T = 0
add
multiply by
simplify and approximate
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Lines
A mathematical model is a mathematical description of a
problem.
For our purposes, these descriptions will be graphs and
equations.
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