slide 3- 1 copyright © 2007 pearson education, inc. publishing as pearson addison-wesley

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Graph of Linear Equations

3.1 Graphs and Applications of Linear Equations3.2 More with Graphing and Intercepts3.3 Slope and Applications3.4 Equations of Lines3.5 Graphing Using the Slope and the y-intercept3.6 Parallel and Perpendicular Lines3.7 Graphing Inequalities in Two Variables

33


GRAPHS and APPLICATIONS OF LINEAR EQUATIONS

Plot points associated with ordered pairs of numbers, determine the quadrant in which a point lies.

Find the coordinates of a point on a graph.

Determine whether an ordered pair is a solution of an equation with two variables.

Graph linear equations of the type y = mx + b and Ax + By = C, identifying the y-intercept.

Solve applied problems involving graphs of linear equations.

3.3.11aabb

cc

dd

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Objective

Plot points associated with ordered pairs of numbers, determine the quadrant in which a point lies.

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Points and Ordered Pairs

To graph, or plot, points we use two perpendicular number lines called axes. The point at which the axes cross is called the origin. Arrows on the axes indicate the positive directions.

Consider the pair (2, 3). The numbers in such a pair are called the coordinates. The first coordinate in this case is 2 and the second coordinate is 3.


Points and Ordered Pairs continued

To plot the point (2, 3) we start at the origin.Move 2 units in the horizontal direction.

The second number 3, is positive. We move 3 units in the vertical direction (up).

Make a “dot” and label the point.

(2, 3)(2, 3)


Example A Plot the point (4, 3).

SolutionStarting at the origin, we move 4 units in the negative horizontal direction. The second number, 3, is positive, so we move 3 units in the positive vertical direction (up).

3 units up

4 units left

(4, 3)


The horizontal and vertical axes divide the plane into four regions, or quadrants.

In which quadrant is the point (3, 4) located?

IV

In which quadrant is the point (3, 4) located?

III


Objective

Find the coordinates of a point on a graph.

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Example BFind the coordinates of points A, B, C, D, E, F, and G.

SolutionPoint A is 5 units to the right of the origin and 3 units above the origin. Its coordinates are (5, 3). The other coordinates are as follows: B: (2, 4)C: (3, 4)D: (3, 2)E: (2, 3)F: (3, 0)G: (0, 2)

AB

C

D

E

F

G


Objective

Determine whether an ordered pair is a solution of an equation with two variables.

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Example C Determine whether each of the following pairs is a solution of 4y + 3x = 18: a) (2, 3); b)(1, 5).

Solutiona) We substitute 2 for x and 3 for y.

4y + 3x = 18 4•3 + 3•2 | 18 12 + 6 | 18 18 = 18 Trueb) We substitute 1 for x and 5 for y.

4y + 3x = 18 4•5 + 3•1 | 18 20 + 3 | 18 23 = 18 False

Since 18 = 18 is true, the pair (2, 3) is a solution.

Since 23 = 18 is false, the pair (1, 5) is not a solution.


Objective

Graph linear equations of the type y = mx + b and Ax + By = C, identifying the y-intercept.

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Graph of An EquationThe graph of an equation is a drawing that represents all its solutions.


To Graph a Linear Equation

1. Select a value for one variable and calculate the corresponding value of the other value. Form an ordered pair using alphabetical order as indicated by the variables.

2. Repeat step (1) to obtain at least two other ordered pairs. Two points are essential to determine a straight line. A third ordered point serves as a check.

3. Plot the ordered pairs and draw a straight line passing through the points.


Example D Graph y = 3xSolution Find some ordered pairs that are solutions. We choose any number for x and then determine y by substitution.

y

x y = 3x (x, y)

2

1

0

12

6

3

0

36

(2, 6)

(1, 3)

(0, 0)

(1, 3)(2, 6)

1. Choose x.2. Compute y.

3. Form the ordered pair (x, y).4. Plot the points.

(2, 6)

(1, 3)

(0, 0)


Example E Graph y = 4x + 1

SolutionWe select convenient values for x and compute y, and form an ordered pair.If x = 2, then y = 4(2) + 1 = 7 and (2, 7) is a solution.

If x = 0, then y = 4(0) + 1 = 1 and (0, 1) is a solution.

If x = 2, then y = 4(2) + 1 = 9 and (2, 9) is a solution.


Solution (continued)

Results are often listed in a table.

(1) Choose x.(2) Compute y.(3) Form the pair (x, y).(4) Plot the points.

x y (x, y)

2 7 (2, 7)

0 1 (0, 1)

2 9 (2, 9)


Note that all three points line up. If they didn’t we would know that we had made a mistake.

Finally, use a ruler or other straightedge to draw a line.

Every point on the line represents a solution of y = 4x + 1



y-InterceptThe graph of the equation y = mx + b passes through the y-intercept (0, b).

(0, b)


Example F Graph

Solution Complete a table of values.

13

4y x

x y (x, y)

4 4 (4, 4)

0 3 (0, 3)

4 2 (4, 2)

y-intercept (4, 4)

(4, 2)

We see that (0, 3) is a solution. It is the y-intercept.

43

1y x (0, 3) is the y-intercept.


Example G

Graph and identify the y-intercept.

SolutionTo find an equivalent equation in the form y = mx + b, we solve for y:

2 3 0y x

2 3 0y x 32 33 0xy x x 2 3y x2 3

2 2

y x

3

2

xy


continued

Complete a table of values.

x y (x, y)

0 0 (0, 0)

2 3 (2, 3)

2 3 (2, 3)

y-intercept

3

2

xy

(2, 3)

(2, 3)

y-intercept


Objective

Solve applied problems involving graphs of linear equations.

ee


Example H

The cost c, in dollars, of shipping a FedEx Priority Overnight package weighing 1 lb or more a distance of 1001 to 1400 mi is given by c = 2.8w + 21.05 where w is the package’s weight in pounds. Graph the equation and then use the graph to estimate the cost of shipping a 10 ½-pound package.


Solution

Select values for w and then calculate c.c = 2.8w + 21.05If w = 2, then c = 2.8(2) + 21.05 = 26.65If w = 4, then c = 2.8(4) + 21.05 = 32.25If w = 8, then c = 2.8(8) + 21.05 = 43.45

w c

2 26.65

4 32.25

8 43.45


Weight (in pounds)

Mai

l cos

t (in

dol

lars

)


Plot the points.

To estimate an 10 ½ pound package, we locate the point on the line that is above 10 ½ and then find the value on the c-axis that corresponds to that point.

The cost of shipping an 10 ½ pound package is about $51.00. 10 ½ pounds

28 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-WesleySection 3.1

1. Find the coordinates of point A.

a) (3, 1)

b) (1, 3)

c) (3, 1)

d) (1, 3)


1. Find the coordinates of point A.

a) (3, 1)

b) (1, 3)

c) (3, 1)

d) (1, 3)


2. Graph 4x – y = –4.

a) b)

c) d)


More with Graphing and Intercepts

Find the intercepts of a linear equation, and graph using intercepts.

Graph equations equivalent to those of the type x = a and y = b.

3.3.22aa

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Objective

Find the intercepts of a linear equation, and graph using intercepts.

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InterceptsThe y-intercept is (0, b). To find b, let x = 0 and solve the original equation for y.

The x-intercept is (a, 0). To find a, let y = 0 and solve the original equation for x.

(0, b)

(a, 0)

y-intercept

x-intercept


Example A Consider 5x + 2y = 10. Find the intercepts. Then graph the equation using the intercepts.

SolutionTo find the y-intercept, we let x = 0 and solve for y: 5 • 0 + 2y = 10 2y = 10 y = 5The y-intercept is (0, 5).

To find the x-intercept, we let y = 0 and solve for x.5x + 2• 0 = 10 5x = 10 x = 2The x-intercept is (2, 0).

Replacing x with 0

Replacing y with 0


continued

We plot these points and draw the line, or graph. A third point should be used as a check. We substitute any convenient value for x and solve for y. If we let x = 4, then 5 • 4 + 2y = 10 20 + 2y = 10 2y = 10

y = 55x + 2y = 10

x-intercept (2, 0)

y-intercept (0, 5)

x y

0 5

2 0

4 5


Example B Graph y = 4x

SolutionWe know that (0, 0) is both the x-intercept and y-intercept. We calculate two other points and complete the graph, knowing it passes through the origin.

x y

1 40 0

1 4

x-intercepty-intercept

y = 4x

(1, 4)

(1, 4)

(0, 0)


Objective

Graph equations equivalent to those of the type x = a and y = b.

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Example C Graph y = 2

SolutionWe regard the equation y = 2 as 0 • x + y = 2. No matter what number we choose for x, we find that y must equal 2.

y = 2Choose any number for x. x y (x, y)

0 2 (0, 2)

4 2 (4, 2)

4 2 (4 , 2)

y must be 2.


continued y = 2

SolutionWhen we plot the ordered pairs (0, 2), (4, 2) and (4, 2) and connect the points, we obtain a horizontal line.

Any ordered pair of the form (x, 2) is a solution, so the line is parallel to the x-axis with y-intercept (0, 2).

y = 2

(4, 2)

(0, 2)

(4, 2)


x y (x, y)

2 4 (2, 4)

2 1 (2, 1)

2 4 (2, 4)

x must be 2.

Example D Graph x = 2

SolutionWe regard the equation x = 2 as x + 0 • y = 2. We make up a table with all 2 in the x-column.

x = 2

Any number can be used for y.


continued x = 2

SolutionWhen we plot the ordered pairs (2, 4), (2, 1), and (2, 4) and connect them, we obtain a vertical line.

Any ordered pair of the form (2, y) is a solution. The line is parallel to the y-axis with x-intercept (2, 0).

x = 2

(2, 4)

(2, 4)

(2, 1)


Horizontal and Vertical LinesThe graph of y = b is a horizontal line. The y-intercept is (0, b).The graph of x = a is a vertical line. The x-intercept is (a, 0).


1. Find the x- and y- intercepts of x = 5 + 2y.

a) x-intercept: (0, 5); y-intercept:

b) x-intercept: (5, 0); y-intercept:

c) x-intercept: (5, 0); y-intercept:

d) x-intercept: (5, 0); y-intercept:

5,0

2

50,

2

50,

2

20,

5


2. Write an equation for the graph.

a) y = 3

b) y = 3

c) x = 3

d) x = 3


Slope and Applications

Given the coordinates of two points on a line, find the slope of the line, if it exists.

Find the slope, or rate of change, in an applied problem involving slope.

Find the slope of a line from an equation.

3.3.33aa

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cc


Objective

Given the coordinates of two points on a line, find the slope of the line, if it exists.

aa


We have looked at two forms of a linear equation, Ax + By = C and y = mx + b

We know that the y-intercept of a line is (0, b).

y = mx + b

? The y-intercept is (0, b).

What about the constant m? Does it give certain information about the line?


Look at the following graphs and see if you can make any connections between the constant m and the “slant” of the line.

y

x

21

3y x

y

x

101

3y x

y

x

21

3y x

y

x3

110

y x


SlopeThe slope of the line containing points (x1, y1) and (x2, y2) is given by

2 1

2 1

rise change in .

run change in

y yym

x x x


Example A Graph the line containing the points (4, 5) and (4, 1) and find the slope.

Solutionrise change in

Slope = =run change in

y

x

1 5

4 4 =

( )

6

= 8

6 3 = , or

8 4

rise

run

2 1

2 1

y y

x x


The slope of a line tells how it slants.A line with a positive slope slants up from left to right.The larger the slope, the steeper the slant.

A line with a negative slope slants downward from left to right.


Objective

Find the slope, or rate of change, in an applied problem involving slope.

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Applications of Slope

Some applications use slope to measure the steepness. For examples, numbers like 2%, 3%, and 6% are often used to represent the grade of a road, a measure of a road’s steepness. That is, a 3% grade means that for every horizontal distance of 100 ft, the road rises or drops 3 ft.


Example B Find the slope (or grade) of the treadmill.

0.42 ft5.5 ft

Solution

0.4

.5

2

5m

420

5500

427.6%

550

The grade of the treadmill is 7.6%.

** Reminder: Grade is slope expressed as a percent.


Objective

Find the slope of a line from an equation.

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It is possible to find the slope of a line from its equation.

Determining Slope from the Equation y = mx + bThe slope of the line y = mx + b is m. To find the slope of a nonvertical line, solve the linear equation in x and y for y and get the resulting equation in the form y = mx + b. The coefficient of the x-term, m is the slope of the line.


Example CFind the slope of the line.a. b.

c. y = x + 8 d.

24

3y x

2

3y x

0.25 6.8y x

m = 4 = Slopem = = Slope

m = 1 = Slope m = 0.25 = Slope

2

3


Example D

Find the slope of the line 3x + 5y = 15.SolutionWe solve for y to get the equation in the form y = mx + b.

3x + 5y = 15 5y = –3x + 15

3 15

5

xy

3

53y x The slope is

3.

5


Example E Find the slope of the line y = 3

SolutionConsider the points (3, 3) and (2, 3), which are on the line.

A horizontal line has slope 0.

2 (

3

)

3

3m

0

5

0

(3, 3) (2, 3)


Example F Find the slope of the line x = 2

SolutionConsider the points (2, 4) and (2, 2), which are on the line.

The slope of a vertical line is undefined.

2 2

4 ( 2)m

6 undefined

0

(2, 4)

(2, 2)


Slope 0; Slope Not DefinedThe slope of a horizontal line is 0.The slope of a vertical line is not defined.


1. Find the slope of the line containing (–2 , 6) and (–3 , 10).

a) –4

b)

c)

d) 4

1

4

4

5


2. Find the slope of 8 – 4y = 0, if it exists.

a) Not defined

b) 0

c) 2

d) 4


Equations of Lines

Given an equation in the form y = mx + b, find the slope and y-intercept; find an equation of a line when the slope and the y-intercept are given.

Find an equation of a line when the slope and a point on the line are given.

Find an equation of a line when two points on the line are given.

3.3.44aa

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cc


Objective

Given an equation in the form y = mx + b, find the slope and y-intercept; find an equation of a line when the slope and the y-intercept are given.

aa


The Slope-Intercept Equation:y = mx + bThe equation y= mx + b is called the slope-intercept equation. The slope is m and the y-intercept is (0, b).


Example AFind the slope and the y-intercept of 3x – 4y = 9.

Solution We first solve for y:3x – 4y = 9

–4y = –3x + 9 –4 –4

3 9

4 4y x

9

4

3

4y x

The slope is 3

4. The y-intercept is 0, .

9

4

Subtracting 3x

Dividing by –4


Example B

A line has slope –3.2 and y-intercept (0, 5). Find an equation of the line.

SolutionWe use the slope-intercept equation.Substitute –3.2 for m and 5 for b:

y = mx + by = –3.2x + 5. Substituting


Example C

A line has slope 0 and y-intercept (0, 4). Find an equation of the line.SolutionWe use the slope-intercept equation.Substitute 0 for m and 4 for b:

y = mx + by = 0x + 4. Substituting

y = 4


Example D

A line has slope and y-intercept (0, 0). Find an equation of the line.SolutionWe use the slope-intercept equation.Substitute for m and 0 for b:

y = mx + by = x + 0. Substituting

3

5

3

5

3

5

3

5y x


Objective

Find an equation of a line when the slope and a point on the line are given.

bb


Example EFind an equation of the line with slope 3 that contains the point (2, 7).Solution We know that the slope is 3, so the equation is y = 3x + b.The equation is true for (2, 7). Using the point (2, 7), we substitute 2 for x and 7 for y and solve for b.

y = 3x + b Substituting 3 for m. 7 = 3(2) + b Substituting 2 for x and 7 for y

7 = 6 + b 1 = b Solving for b

The equation is y = 3x + 1.


Objective

Find an equation of a line when two points on the line are given.

cc


Example FFind an equation of the line containing the points (2, 2) and (6, 4).Solution First, we find the slope:

Use either point to find b.

4 2

6 2m

6 3

, or 8 4

3

4y x b

2 23

4b

62

4b

1

2b

3 1

4 2y x

Substituting 2 for x and 2 for y

Solving for b

The equation of the line.


1. Write the equation of the line with slope 3 and y-intercept (0, 5).

a) y = 3x – 5

b) y = 3x + 5

c) y = 5x + 3

d) y = –5x + 3


2. Find an equation of the line that contains the points (4, 5) and (5, 13).

a) y = –5x + 13

b) y = 4x – 5

c) y = 2x + 3

d) y = –2x + 3


Graphing Using the Slope and the y-Intercept

Use the slope and the y-intercept to graph a line.

3.3.55aa


Objective

Use the slope and the y-intercept to graph a line.

aa


Example ADraw a line that has slope y-intercept (0, 1).SolutionWe plot (0, 1).Move up 1 unit (since thenumerator is positive.)Move to the right 3 units(since the denominator ispositive).This locates the point (3, 2).We plot (3, 2) and draw a line passing through both points.

1

3

Up 1

Right 3


Example BDraw a line that has slope y-intercept (0, 4).SolutionWe plot (0, 4).Move down 3 units (since thenumerator is negative.)Move to the right 4 units(since the denominator ispositive).This locates the point (4, 1).We plot (4, 1) and draw a line passing through both points.

3

4

Down 3

Right 4


Example CGraph 3x + 2y = 2 using the slope and y-intercept.SolutionWrite the equation in slope-intercept form.3 2 2x y

2 3 2y x 1 1

2 3 22 2

y x

31

2y x

Plot the y-intercept (0, –1).Move down 3 units and to the right 2 units.


1. Graph 3x – 2y = 6.

a) b)

c) d)


Parallel and Perpendicular Lines

Determine whether the graphs of two linear equations are parallel.

Determine whether the graphs of two linear equations are perpendicular.

3.3.66aa

bb


When we graph a pair of linear equations, there are three possibilities:1. The graphs are the same.2. The graphs intersect at exactly one point.3. The graphs are parallel (they do not intersect).


Objective

Determine whether the graphs of two linear equations are parallel.

aa


The graphs shown below are of the linear equations y = 2x + 5 and y = 2x – 3.

The slope of each line is 2.The y-intercepts are (0, 5) and (0, –3).The lines do not intersect and are parallel.

y = 2x + 5

y = 2x – 3


Parallel Lines Parallel nonvertical lines have the same slope,

m1 = m2, and different y-intercepts, b1 b2. Parallel horizontal lines have equations y = p

and y = q, where p q. Parallel vertical ines have equations x = p and

x = q, where p q.


Example ADetermine whether the graphs of the lines y = –2x – 3 and 8x + 4y = –6 are parallel.SolutionThe graphs are shown above, but they are not necessary in order to determine whether the lines are parallel.Solve each equation for y.

8 4 6x y 4 8 6y x

18 6

4y x

32

2y x

The slope of each line is –2 and the y-intercepts are different. The lines are parallel.


Objective

Determine whether the graphs of two linear equations are perpendicular.

bb


Perpendicular lines in a plane are lines that intersect at a right angle. The measure of a right angle is 90 degrees.

The slopes of the lines are 2 and –1/2. Note that 2(– 1/2) = –1. That is, the product of the slopes is –1.

y = 2x – 3

1 1

2 2y x


Perpendicular Lines Two nonvertical lines are perpendicular if the

product of their slopes is –1, m1 m2. (If one lines has slope m, the slope of the line perpendicular to it is –1/m.)

If one equation in a pair of perpendicular lines is vertical, then the other is horizontal. The equations are of the form x = a and y = b.


Example BDetermine whether the graphs of the lines y = 4x + 1 and x + 4y = 4 are perpendicular.SolutionThe graphs are shown above, but they are not necessary in order to determine whether the lines are parallel.Solve each equation for y.

4 4x y 4 4y x

14

4y x

11

4y x

The slopes are 4 and –1/4. The product of the slopes is –1. The lines are perpendicular.


1. Write an equation of the line parallel to the y-axis and 3 units to the right of it.

a) x = –3

b) y = –3

c) x = 3

d) y = 3


2. Given the line 2x + 3y = –8, find the slope of a line parallel to it.

a)

b)

c)

d)

2

38

3

2

3

3

2


Graphing Inequalities in Two Variables

Determine whether an ordered pair of numbers is a solution of an inequality in two variables.

Graph linear inequalities.

3.3.77aa

bb


Objective

Determine whether an ordered pair of numbers is a solution of an inequality in two variables.

aa


The solutions of inequalities in two variables are ordered pairs.EXAMPLE ADetermine whether (1, 2) is a solution of 3x + 4y < 15.We use alphabetical order to replace x with 1 and y with 3.

3x + 4y < 153(1) + 4(2) ? 15 3 + 8 11 True

Since 11 < 15 is true, (1, 2) is a solution.


Objective

Graph linear inequalities.

bb


Example B Graph y < 2xSolution We first graph the line y = 2x. Every solution of y = 2x is an ordered pair like (1, 2).We draw the line dashed because its points are not solutions.Select a point on one side of the half-plane and check in the inequality.Try (2, 0)y < 2x0 < 2(2)0 < 4 TRUE Shade the half-plane containingthe point.


To graph an inequality in two variables:1. Replace the inequality symbol with an equals sign and

graph this related equation.2. If the inequality symbol is < or >, draw the line dashed.

If the inequality symbol is or , draw the line solid.3. The graph consists of a half-plane, either above or

below or left or right of the line, and, if the line is solid, the line as well. To determine which half-plane to shade, choose a point not on the line as a test point. Substitute to find whether that point is a solution of the inequality. If it is, shade the half-plane containing the point. If it is not, shade the half-plane on the opposite side of the line.


Example C Graph: x – 2y 61. First, we graph the line. The

intercepts are (0, –3) and (6, 0).

2. Since the inequality contains the symbol, we draw the line solid to indicate that any pair on the line is a solution.

3. Next, we choose a test point. (0, 0) x – 2y 60 – 2(0) 6

0 6 TRUE

Shade the half-plane containing the point.


Example D Graph x > 1 1. First, we graph the line. x = 12. Since the inequality symbol

is >, we use a dashed line. 3. Next, we choose a test point.

(0, 0) x + 0y > 1 0 – 0(0) > 1

0 > 1 FALSEShade on the other half-plane.We see from the graph that the

solutions are all ordered pairs with first coordinates > 1.


1. Graph y > –x – 2.

a) b)

c) d)


2. Graph x > 4 + y.

a) b)

c) d)

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