01_02_Linear_Functions_Straight_lines (1).ppt

Copyright © Cengage Learning. All rights reserved.
1
STRAIGHT LINES
AND LINEAR
FUNCTIONS

Copyright © Cengage Learning. All rights reserved.
1.2 Straight Lines

3
Straight Lines
In computing income tax, business firms are allowed by law
to depreciate certain assets such as buildings, machines,
furniture, and automobiles over a period of time.
Linear depreciation, or the straight-line method, is often
used for this purpose.

4
Straight Lines
The graph of the straight line shown in Figure 10 describes
the book value V of a network server that has an initial
value of $10,000 and that is being depreciated linearly over
5 years with a scrap value of $3000.
Figure 10
Linear depreciation of an asset

5
Straight Lines
Note that only the solid portion of the straight line is of
interest here.
The book value of the server at the end of year t, where t
lies between 0 and 5, can be read directly from the graph.
But there is one shortcoming in this approach: The result
depends on how accurately you draw and read the graph.
A better and more accurate method is based on finding an
algebraic representation of the depreciation line.
To see how a straight line in the xy-plane may be described
algebraically, we need first to recall certain properties of
straight lines.

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Slope of a Line
Let L denote the unique straight line that passes through
the two distinct points (x1, y1) and (x2, y2). If x1  x2, then we
define the slope of L as follows.
Figure 11

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Slope of a Line
If x1 = x2, then L is a vertical line (Figure 12).
Figure 12
The slope is undefined if x1 = x2.

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Slope of a Line
Its slope is undefined since the denominator in Equation (3)
will be zero and division by zero is proscribed.
Observe that the slope of a straight line is a constant
whenever it is defined.
The number y = y2 – y1 (y is read “delta y”) is a measure
of the vertical change in y, and x = x2 – x1 is a measure of
the horizontal change in x as shown in Figure 11.
From this figure we can see that the slope m of a straight
line L is a measure of the rate of change of y with respect
to x.

10
Slope of a Line
Furthermore, the slope of a nonvertical straight line is
constant, and this tells us that this rate of change is
constant.
Figure 13a shows a straight line L1 with slope 2.
Figure 13(a)
The line rises (m > 0).

11
Slope of a Line
Observe that L1 has the property that a 1-unit increase in x
results in a 2-unit increase in y.
To see this, let x = 1 in Equation (3) so that m = y.
Since m = 2, we conclude that y = 2.

12
Slope of a Line
Similarly, Figure 13b shows a line L2 with slope –1.
Figure 13(b)
The line falls (m < 0).

13
Slope of a Line
Observe that a straight line with positive slope slants
upward from left to right (y increases as x increases),
whereas a line with negative slope slants downward from
left to right (y decreases as x increases).
Finally, Figure 14 shows a family of straight lines passing
through the origin with indicated slopes.
Figure 14
A family of straight lines

14
Example 1
Sketch the straight line that passes through the point (–2, 5)
and has slope – .
Solution:
First, plot the point (–2, 5) (Figure 15).
Figure 15

15
Example 1 – Solution
Next, recall that a slope of – indicates that an increase of
1 unit in the x-direction produces a decrease of units in
the y-direction, or equivalently, a 3-unit increase in the
x-direction produces a 3( ), or 4-unit, decrease in the
y-direction.
Using this information, we plot the point (1, 1) and draw the
line through the two points.
cont’d

16
Example 2
Find the slope m of the line that passes through the
points (–1, 1) and (5, 3).
Solution:
Choose (x1, y1) to be the point (–1, 1) and (x2, y2) to be the
point (5, 3).
Then, with x1 = –1, y1 = 1, x2 = 5, and y2 = 3, we find, using
Equation (3),

17
See Figure 16.
You may verify that the result obtained would be the same
had we chosen the point (–1, 1) to be (x2, y2) and the
point (5, 3) to be (x1, y1).
cont’d
Figure 16
L passes through (5, 3) and (–1, 1).

18
Slope of a Line
Note:
The slope of a horizontal line is zero.
We can use the slope of a straight line to determine
whether a line is parallel to another line.

19
Example 4
Let L1 be a line that passes through the points (–2, 9) and
(1, 3), and let L2 be the line that passes through the points
(–4, 10) and (3, –4).
Determine whether L1 and L2 are parallel.
Solution:
The slope m1 of L1 is given by
m1 =
= –2

20
The slope m2 of L2 is given by
m2 =
Since m1 = m2, the lines L1 and L2 are in fact parallel.
cont’d
= –2

21
See Figure 18.
cont’d
Figure 18
L1 and L2 have the same slope and hence are parallel.

23
Equations of Lines
We now show that every straight line lying in the xy-plane
may be represented by an equation involving the variables
x and y.
One immediate benefit of this is that
problems involving straight lines may
be solved algebraically.
Let L be a straight line parallel to the
y-axis (perpendicular to the x-axis)
(Figure 19).
Figure 19
The vertical line x = a

24
Equations of Lines
Then L crosses the x-axis at some point (a, 0) with the
x-coordinate given by x = a, where a is some real number.
Any other point on L has the form (a, y), where y is an
appropriate number.
Therefore, the vertical line L is described by the sole
condition
x = a
and this is accordingly an equation of L.

25
Equations of Lines
For example, the equation x = –2 represents a vertical line
2 units to the left of the y-axis, and the equation x = 3
represents a vertical line 3 units to the right of the y-axis
(Figure 20).
Figure 20
The vertical lines x = –2 and x = 3

26
Equations of Lines
Next, suppose L is a nonvertical line, so it has a
well-defined slope m. Suppose (x1, y1) is a fixed point lying
on L and (x, y) is a variable point on L distinct from (x1, y1)
(Figure 21).
Figure 21
L passes through (x1, y1) and has slope m.

27
Equations of Lines
Using Equation (3) with the point (x2, y2) = (x, y), we find
that the slope of L is given by
m =
Upon multiplying both sides of the equation by x – x1, we
obtain Equation (4).

28
Equations of Lines
Equation (4) is called the point-slope form of an equation of
a line because it uses a given point (x1, y1) on a line and the
slope m of the line.

29
Example 5
Find an equation of the line that passes through the point
(1, 3) and has slope 2.
Solution:
Using the point-slope form of the equation of a line with the
point (1, 3) and m = 2, we obtain
y – 3 = 2(x – 1)
which, when simplified, becomes
2x – y + 1 = 0
y – y1 = m(x – x1)

30
See Figure 22.
cont’d
Figure 22
L passes through (1, 3) and has slope 2.

31
Equations of Lines
We can use the slope of a straight line to determine
whether a line is perpendicular to another line.
If the line L1 is vertical (so that its slope is undefined), then
L1 is perpendicular to another line, L2, if and only if L2 is
horizontal (so that its slope is zero).

32
Example 7
Find an equation of the line that passes through the point
(3, 1) and is perpendicular to the line 2x – y + 1 = 0
Solution:
Since the slope of the line 2x – y + 1 = 0 is 2, it follows that
the slope of the required line is given by m = – , the
negative reciprocal of 2.
Using the pointslope form of the equation of a line, we
obtain
y – 1 = – (x – 3) y – y1 = m(x – x1)

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2y – 2 = –x + 3
x + 2y – 5 = 0
See Figure 24.
cont’d
Figure 24
L2 is perpendicular to L1 and passes through (3,
1).

34
Equations of Lines
A straight line L that is neither horizontal nor vertical cuts
the x-axis and the y-axis at, say, points (a, 0) and (0, b),
respectively (Figure 25).
Figure 25
The line L has x-intercept a and y-intercept b.

35
Equations of Lines
The numbers a and b are called the x-intercept and
y-intercept, respectively, of L.
Now, let L be a line with slope m and y-intercept b. Using
Equation (4), the pointslope form of the equation of a line,
with the point given by (0, b) and slope m, we have
y – b = m(x – 0)
y = mx + b

36
Example 9
Determine the slope and y-intercept of the line whose
equation is 3x – 4y = 8.
Solution:
Rewrite the given equation in the slope-intercept form.
Thus,
3x – 4y = 8
–4y = –3x + 8
y = x – 2

37
Comparing this result with Equation (5),
we find m = and b = –2,
and we conclude that the slope and y-intercept of the given
line are and –2, respectively.
cont’d

38
Applied Example 10 – Sales of a Sporting Goods Store
The sales manager of a local sporting goods store plotted
sales versus time for the last 5 years and found the points
to lie approximately along a straight line (Figure 26).
Figure 26
Sales of a sporting goods store

39
Applied Example 10 – Sales of a Sporting Goods Store
By using the points corresponding to the first and fifth
years, find an equation of the trend line. What sales figure
can be predicted for the sixth year?
Solution:
Using Equation (3) with the points (1, 20) and (5, 60), we
find that the slope of the required line is given by
m = = 10
cont’d

40
Next, using the point-slope form of the equation of a line
with the point (1, 20) and m = 10, we obtain
y – 20 = 10(x – 1)
y = 10x + 10
as the required equation.
cont’d
y – y1 = m(x – x1)

41
The sales figure for the sixth year is obtained by letting
x = 6 in the last equation, giving
y = 10(6) + 10 = 70
or $700,000.
cont’d

42
General Form of an Equation
of a Line

43
General Form of an Equation of a Line
We have considered several forms of the equation of a
straight line in the plane. These different forms of the
equation are equivalent to each other.
In fact, each is a special case of the following equation.

44
We now state an important result concerning the algebraic
representation of straight lines in the plane.
This result justifies the use of the adjective linear in
describing Equation (6).

45
Example 12
Sketch the straight line represented by the equation
3x – 4y – 12 = 0
Solution:
Since every straight line is uniquely determined by two
distinct points, we need to find only two points through
which the line passes in order to sketch it.
For convenience, let’s compute the points at which the line
crosses the x- and y-axes.
Setting y = 0, we find x = 4, the x-intercept, so the line
crosses the x-axis at the point (4, 0).

46
Setting x = 0 gives y = –3, the y-intercept, so the line
crosses the y-axis at the point (0, –3).
A sketch of the line appears in Figure 27.
cont’d
Figure 27
To sketch 3x – 4y – 12 = 0, first find the
x-intercept, 4, and the y-intercept, –3.

47

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