1Copyright © Cengage Learning. All rights reserved.

3 Functions and Graphs

3.3 Lines

2

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.

3

Lines

The following concept is fundamental to the study of lines.

4

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)

5

Example 1 – Solution

The lines are sketched in Figure 3. We use the definition ofslope to find the slope of each line.

Figure 3(a) Figure 3(b)

m = m =

6

Example 1 – Solution cont’d

Figure 3(c) Figure 3(d)

m = 0 m undefined

7

Example 1 – Solution

(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.

cont’d

8

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

9

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.

10

Lines

Lines that are horizontal or vertical have simple equations, as indicated in the following chart.

11

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.

12

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

13

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

14

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.

15

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.

16

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

17

Example 4 – Solution

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)

cont’d

point-slope form

multiply by 4

18

Example 4 – Solution

4y – 28 = 5x – 5

–5x + 4y = 23

5x – 4y = –23

The last equation is one of the desired forms for an equation of a line.

Another is 5x – 4y + 23 = 0.

cont’d

multiply by –1

subtract 5x and add 28

multiply factors

19

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

20

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).

21

Lines

We have proved the following result.

22

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

23

Example 5 – Solution

–5y = –2x + 8

y =

y =

The last equation is the slope-intercept form y = mx + b with slope

m = and y-intercept b = .

cont’d

equivalent equation

divide by –5

subtract 2x

24

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.

25

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.

26

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 = .

27

Example 6 – Solution

Plotting the points (4, 0) and and drawing a line through them gives us the graph in Figure 10.

cont’d

Figure 10

28

Lines

The following theorem specifies the relationship between parallel lines (lines in a plane that do not intersect) and slope.

29

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

3y = –6x + 4

y = –2x +

given

subtract 6x

divide by 3

30

Example 7 – Solution

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

cont’d

point-slope form

simplify

subtract 7

31

Example 7 – Solution

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.

cont’d

Figure 12

32

Example 7 – Solution

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 = kgives us 6(5) + 3(–7) = k or, equivalently, k = 9.

The equation 6x + 3y = 9 is equivalent to y = –2x + 3.

cont’d

33

Lines

If the slopes of two nonvertical lines are not the same, thenthe 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.

34

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 foundthat its slope is –2.

Hence, the slope of the required line is the negativereciprocal –[1/(–2)], or .

Using P(5, –7) gives us the following:

y – (–7) = (x – 5)point-slope form

35

Example 8 – Solution

The last equation is in slope-intercept form and shows that the perpendicular line has y-intercept – .

cont’d

put in slope-intercept form

simplify

36

Example 8 – Solution

This line and the given line are sketched in Figure 16.

cont’d

Figure 16

37

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 areshown in Figure 17.

Figure 17

38

Example 9 – Solution

We calculate the following, where M is the midpoint of AB:

Coordinates of M:

Slope of AB:

Slope of l:

cont’d

midpoint formula

slope formula

negative reciprocal of

39

Example 9 – Solution

Using the point M and slope gives us the following equivalent equations for l:

y – = – (x – 1)

6y – 15 = –16(x – 1)

6y – 15 = –16x + 16

16x + 6y = 31

cont’d

point-slope form

multiply by the lcd, 6

multiply

put in general form

40

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 frequentlyin applied problems.

The next example gives one illustration.

41

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°.

42

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.

43

Example 10 – Solution

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

Substituting for a in T = ah + 60 gives us the following formula for T:

cont’d

let T = 42 and h = 5000

solve for a

44

Example 10 – Solution

The graph is sketched in Figure 18, with different scales on the axes.

cont’d

Figure 18

45

Example 10 – Solution

(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:

cont’d

from part (a)

46

Example 10 – Solution cont’d

simplify and approximate

multiply by

add

Let T = 0

47

Lines

A mathematical model is a mathematical description of a problem.

For our purposes, these descriptions will be graphs and equations.