Chapter11

11-1

Graphing Linear Equations

Course 3

Warm UpSolve each equation for y.

1. 6y – 12x = 24

2. –2y – 4x = 20

3. 2y – 5x = 16

4. 3y + 6x = 18

y = 2x + 4

y = –2x – 10

Course 3

11-1 Graphing Linear Equations

y = –2x + 6

y = x + 852

Learn to identify and graph linear equations.

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WHAT IS A LINEAR EQUATION?

A linear equation is an equation whose solutions fall on a line on the coordinate plane. All solutions of a particular linear equation fall on the line, and all the points on the line are solutions of the equation.

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HOW DO I KNOW IF AN EQUATION IS LINEAR?

If an equation is linear, a constant change in the x-value corresponds to a constant change in the y-value.

3

3

3

2

2

2

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The graph shows an example where each time the x-value increases by 3, the y-value increases by 2.

Make a table, graph the equation, and tell whether it is linear.

A. y = 3x – 1

Graphing Equations

x 3x – 1 y (x, y)

–2

–1

0

1

2

–73(–2) – 13(–1) – 1

3(0) – 13(1) – 1

3(2) – 1

–4

–1

2

5

(–2, –7)

(–1, –4)(0, –1)

(1, 2)(2, 5)

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Each time x increases by 1 unit, y increases by 3 units.

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(x, y)

(-2, -7)

(-1, -4)

(0, -1)

(1, 2)

(2, 5)

y = 3x – 1

Make a table, graph the equation, and tell whether it is linear.

B. y = x3

Graphing Equations

x x3 y (x, y)

–2

–1

0

1

2

–8(–2)3

(–1)3

(0)3

(1)3

(2)3

–1

0

1

8

(–2, –8)

(–1, –1)(0, 0)

(1, 1)(2, 8)

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The equation y = x3 is not a linear equation because its graph is not a straight line. Also notice that as x increases by a constant of 1 unit, the change in y is not constant.

x –2 –1 0 1 2

y –8 –1 0 1 8

+7 +1 +1 +7Course 3


Fill in the table below. Then, graph the equation and tell whether it is linear.

A. y = 2x + 1

Try This

x 2x + 1 y (x, y)

–2

–1

0

1

2

–32(–2) + 12(–1) + 1

2(0) + 12(1) + 1

2(2) + 1

–1

1

3

5

(–3, –3)

(–2, –1)(–1, 1)

(0, 3)(2, 5)

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The equation y = 2x + 1is a linear equation.

Each time x increase by 1 unit, y increases by 2 units.

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Complete the first two problems on page 85

REVIEW• An equation that forms a line on a coordinate

grid is called a linear equation. All of the points on the line are solutions.

• In order to be linear, there must be a constant pattern found in the y-vaules.

• To solve, first create a table and then graph the points.

• Watch this:

http://www.brainpop.com/math/algebra/graphinglinearequations

More Graphing EquationsGraph the equation and tell whether it is linear.

C. y = – 3x4

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Additional Example 1 Continued

The equation y = –

is a linear equation.

3x4

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Graph the equation and tell whether it is linear.

D. y = 2

More Graphing Equations

For any value of x, y = 2.

x 2 y (x, y)

–2

–1

0

1

2

222

2

2

2

2

2

2

2

(–2, 2)

(–1, 2)(0, 2)

(1, 2)(2, 2)

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Additional Example 1D Continued

The equation y = 2 is a linear equation because the points form a straight line.

As the value of x increases, the value of y has a constant change of 0.

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Try ThisGraph the equation and tell whether it is linear.

C. y = x

x y (x, y)

–8

–6

0

4

8

–8

–6

0

4

8

(–8, –8)

(–6, –6)(0, 0)

(4, 4)(8, 8)

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Try This: Example 1C Continued

The equation y = x is a linear equation because the points form a straight line.

Each time the value of x increases by 1, the value of y increases by 1.

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ApplicationIn an amusement park ride, a car travels according to the equation D = 1250t where t is time in minutes and D is the distance in feet the car travels.

Graph the relationship between time and distance. How far has each person traveled?

Rider Time

Ryan 1 min

Greg 2 min

Colette 3 min

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Continued

t D =1250t D (t, D)

1 1250(1) 1250 (1, 1250)

2 1250(2) 2500 (2, 2500)

3 1250(3) 3750 (3, 3750)

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The distances are: Ryan, 1250 ft; Greg, 2500 ft; and Collette, 3750 ft.

Continued

x

y

This is a linear equation because when t increases by 1 unit, D increases by 1250 units.

1250

2500

1 2

3750

5000

3 4Time (min)

Dis

tan

ce (

ft)

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Sports Application

A lift on a ski slope rises according to the equation a = 130t + 6250, where a is the altitude in feet and t is the number of minutes that a skier has been on the lift.

Five friends are on the lift. What is the altitude of each person if they have been on the ski lift for the times listed in the table?

Draw a graph that represents the relationship between the time on the lift and the altitude.

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The altitudes are: Anna, 6770 feet; Tracy, 6640 feet; Kwani, 6510 feet; Tony, 6445 feet; George, 6380 feet. This is a linear equation because when t increases by 1 unit, a increases by 130 units. Note that a skier with 0 time on the lift implies that the bottom of the lift is at an altitude of 6250 feet.


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Sit down and check your homework!

12. (-1, -4) (0,0) (1, 4)

13. (-1, 3) (0, 5) (1, 7)

14. (-1,-9) (0, -3) (1, 3)

15. (-1, -11) (0, -10) (1,-9)

16. (-1, -6) (0, -2) (1, 2)

17. (-1, -1) (0, 3) (1, 7)

18. (-1, -6) (0, -4) (1, -2)

19. (-1, 6) (0, 7) (1, 8)

20. (-1, -0.5) (0, 2.5) (1, 5.5)

22. $43.20 $46.40 $49.60

$52.80 $56.00 $59.20

11-2

Slope of a Line

Course 3

Warm UpEvaluate each equation for x = –1, 0, and 1.

1. y = 3x

2. y = x – 7

3. y = 2x + 5

4. y = 6x – 2

–3, 0, 3

–8, –7, –6

3, 5, 7

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11-2 Slope of a Line

–8, –2, 4

Learn to find the slope of a line and use slope to understand and draw graphs.

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What is slope?

vertical change horizontal change

change in y change in x=

This ratio is often referred to as , or “rise

over run,” where rise indicates the number of units moved up or down and run indicates the number of units moved to the left or right. Slope can be positive, negative, zero, or undefined.

rise run

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GET ON

YOUR FEET!

Finding Slope from a Graph

Choose two points on the line: (0, 1) and (3, –4).

Guess by looking at the graph:

riserun = –5

3 = – 5 3

–5

3

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How to find the slope of a line (without a visual)…

If you have two points (x1, y1) & (x2, y2)

Use the following formula:

yy22 –– yy11

xx22 –– xx11

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Find the slope of the line that passes through (–2, –3) and (4, 6).

Finding Slope, Given Two Points

Let (x1, y1) be (–2, –3) and (x2, y2) be (4, 6).

6 – (–3)4 – (–2)

Substitute 6 for y2, –3 for y1, 4 for x2, and –2 for x1.

96=

The slope of the line that passes through (–

2, –3) and (4, 6) is . 32

=y2 – y1

x2 – x1

32=

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Find the slope of the line that passes through (–4, –6) and (2, 3).

Try This: Example 1

Let (x1, y1) be (–4, –6) and (x2, y2) be (2, 3).

3 – (–6)2 – (–4)

Substitute 3 for y2, –6 for y1, 2 for x2, and –4 for x1.

96=

The slope of the line that passes through (–

4, –6) and (2, 3) is . 32

=y2 – y1

x2 – x1

32=

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Use the graph of the line to determine its slope.

Try This: Example 2

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Try This: Example 2 Continued

Choose two points on the line: (1, 1) and (0, –1).

Guess by looking at the graph:

riserun = 2

1 = 2

Use the slope formula.

Let (1, 1) be (x1, y1) and (0, –1) be (x2, y2).

=y2 – y1

x2 – x1

–2–1=

–1 – 1 0 – 1

= 2

12

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Class Assignment/Homework

Workbook 11-2Complete #1-6

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REVIEW• A slope is a ratio that tells how steep or flat a

line is. It is read as “rise over run”.

• Slope can be used and seen in constructions, sports, hills, roller coasters, etc.

• There are four types of slopes: positive, negative, zero, and undefined (no slope).

• To find the slope without a graph, write the difference in y over the difference in x.

Something you need to remember:

Parallel lines have the same slope.

The slopes of two perpendicular lines are negative reciprocals of each other.

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Parallel and Perpendicular Lines:On the following slides, you need to figure out one thing…

Are the lines passing through the given points parallel or perpendicular?

1. Find the slope of the first line.

2. Find the slope of the second line.

3. Compare.

~ If they are the same Parallel

~ If they are opposite reciprocals Perpendicular

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Parallel and Perpendicular Lines by Slope

line 1: (–6, 4) and (2, –5)

line 2: (–1, –4) and (8, 4)

slope of line 1:

slope of line 2:

Line 1 has a slope equal to – and line 2 has a slope

equal to , – and are negative reciprocals of each

other, so the lines are perpendicular.

98

89

89

98

=y2 – y1

x2 – x1

–9 8= –5 – 4

2 – (–6)

4 – (–4)8 – (–1)=

y2 – y1

x2 – x1

8 9=

9 8= –

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Parallel and Perpendicular Lines by Slope

line 1: (0, 5) and (6, –2)

line 2: (–1, 3) and (5, –4)

Both lines have a slope equal to – , so the lines are parallel.

76

slope of line 1:

slope of line 2:

=y2 – y1

x2 – x1

–7 6= –2 – 5

6 – 0

=y2 – y1

x2 – x1

7 6= –

–7 6= 7

6= – –4 – 35 – (–1)

Course 3


Try This

line 1: (1, 1) and (2, 2)

line 2: (1, –2) and (2, -1)

Line 1 has a slope equal to 1 and line 2 has a slope equal to –1. 1 and –1 are negative reciprocals of each other, so the lines are perpendicular.

slope of line 1:

slope of line 2:

=y2 – y1

x2 – x1

1 1= 2 – 1

2 – 1

=y2 – y1

x2 – x1

–1 1= –1 – (–2)

2 – (1)

= 1

= –1

Course 3


Try This

line 1: (–8, 2) and (0, –7)line 2: (–3, –6) and (6, 2)

slope of line 1:

slope of line 2:

Line 1 has a slope equal to – and line 2 has a slope

equal to , – and are negative reciprocals of each

other, so the lines are perpendicular.

98

89

89

98

=y2 – y1

x2 – x1

–9 8= –7 – 2

0 – (–8)

2 – (–6)6 – (–3)=

y2 – y1

x2 – x1

8 9=

9 8= –

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Graphing a Line Using a Point and the Slope

2. Use the slope to count your units. Place another point once you get to the end.

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3. Continue until you have enough point and then draw a line.

1. Plot the point given to you.

Graphing a Line Using a Point and the Slope

Graph the line passing through (3, 1) with slope 2.

Plot the point (3, 1). Then move 2 units up and right 1 unit and plot the point (4, 3). Use a straightedge to connect the two points.

The slope is 2, or . So for every 2 units up, you will move right 1 unit, and for every 2 units down, you will move left 1 unit.

21

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1

2(3, 1)

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Try This: Example 4

Graph the line passing through (1, 1) with slope 2.

Plot the point (1, 1). Then move 2 units up and right 1 unit and plot the point (2, 3). Use a straightedge to connect the two points.

The slope is 2, or . So for every 2 units up, you will move right 1 unit, and for every 2 units down, you will move left 1 unit.

21

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1

2(1, 1)

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Sit down and check your homework!10. - 3/2

11. - 1/2

12. - 3

13. - 2/3

14. 7/4

15. - 5/6

16. - 3/2

17. 0

18. parallel

19. perpendicular

Warm UpFind the slope of the line that passes through each pair of points.

1. (3, 6) and (-1, 4)

2. (1, 2) and (6, 1)

3. (4, 6) and (2, -1)

4. (-3, 0) and (-1, 1)

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11-3 Using Slopes and Intercepts

12

- 157212

11-3

Using Slopes and Intercepts

Course 3

Learn to use slopes and intercepts to graph linear equations.

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As you watch the video, take notes on your handout.

Insert Lesson Title Here

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11-3

http://www.brainpop.com/math/algebra/slopeandintercept

One way to graph a linear equation easily is by finding the x-intercept and the y-intercept.

The x-intercept is the value of x where the line crosses the x-axis (y = 0).

The y-intercept is the value of y where the line crosses the y-axis (x = 0).

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11-3 Using Intercepts

Find the x-intercept and y-intercept of the line 4x – 3y = 12. Use the intercepts to graph the equation.

Example 1

Find the x-intercept (y = 0).

4x – 3y = 12

4x – 3(0) = 12

4x = 124x4

124=

x = 3The x-intercept is 3.

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Example 1 Continued

Find the y-intercept (x = 0).

4x – 3y = 12

4(0) – 3y = 12

–3y = 12

-3y-3

12-3 =

y = –4

The y-intercept is –4.

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4x – 3y = 12

Crosses the x-axis at the point (3, 0)

Crosses the y-axis at the point (0, –4)

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Find the x-intercept and y-intercept of the line 8x – 6y = 48. Use the intercepts to graph the equation.

Try This

Find the x-intercept (y = 0).

8x – 6y = 48

8x – 6(0) = 48

8x = 488x8

488=

x = 6The x-intercept is 6 so the point is (6, 0).

Course 3


Try This

Find the y-intercept (x = 0).

8x – 6y = 48

8(0) – 6y = 48

–6y = 48

-6y-6

48-6 =

y = –8

The y-intercept is –6 so the point is (0, -8).

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The graph of 8x – 6y = 48 is the line that crosses the x-axis at the point (6, 0) and the y-axis at the point (0, –8).

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In an equation written in slope-intercept form, y = mx + b, m is the slope and b is the y-intercept.

y = mx + b

Slope y-intercept

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For an equation such as y = x – 6, write it as y = x + (–6) to read the y-intercept, –6. The point would be (0,-6)

Helpful Hint

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Using the Slope-Intercept Form

1. Isolate the y so that you equation is in y = mx+b form.

2. Slope will always be “m” (the number in front of x).

3. The y-intercept will always be “b” (the number by itself). Write this point as (0,b)

Example 2

Write each equation in slope-intercept form, and then find the slope and y-intercept.

A. 2x + y = 32x + y = 3–2x –2x Subtract 2x from both sides.

y = 3 – 2x

y = –2x + 3 The equation is in slope-intercept form.

m = –2 b = 3The slope of the line is –2, and the y-intercept is 3.

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More Examples

B. 5y = 3x

5y = 3x

Divide both sides by 5 to solve for y.

The equation is in slope-intercept form.

b = 0

= x35

5y5

y = x + 035

m =35

The slope of the line is , and they-intercept is 0.

35

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More Examples

C. 4x + 3y = 9

4x + 3y = 9Subtract 4x from both sides.

b = 3

y =- x + 343

m =- 43

The slope of the line 4x+ 3y = 9 is – , and the y-intercept is 3.4

3

–4x –4x

3y = –4x + 9

= + –4x 3

3y3

93 Divide both sides by 3.


Course 3


Try This

Write each equation in slope-intercept form, and then find the slope and y-intercept.

A. 4x + y = 4–4x –4x Subtract 4x from both sides.

y = 4 – 4xRewrite to match slope-intercept form.

y = –4x + 4 The equation is in slope-intercept form.

m = –4 b = 4The slope of the line 4x + y = 4 is –4, and the y-intercept is 4.

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Try This

B. 7y = 2x

7y = 2x

Divide both sides by 7 to solve for y.


b = 0

= x27

7y7

y = x + 027

m =27

The slope of the line 7y = 2x is , and they-intercept is 0.

27

Course 3


Try This

C. 5x + 4y = 8

5x + 4y = 8Subtract 5x from both sides.

Rewrite to match slope-intercept form.

b = 2

y =- x + 254

The slope of the line 5x + 4y = 8 is – , and the y-intercept is 2.5

4

–5x –5x

4y = 8 – 5x

5x + 4y = 8

= + –5x 4

4y4

84 Divide both sides by 4.


m =- 54

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Additional Example 3: Entertainment Application

A video club charges $8 to join, and $1.25 for each DVD that is rented. The linear equation y = 1.25x + 8 represents the amount of money y spent after renting x DVDs. Graph the equation by first identifying the slope and y-intercept.

y = 1.25x + 8The equation is in slope-intercept form.

b = 8m =1.25

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The slope of the line is 1.25, and the y-intercept is 8. The line crosses the y-axis at the point (0, 8) and moves up 1.25 units for every 1 unit it moves to the right.

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Try This: Example 3

A salesperson receives a weekly salary of $500 plus a commission of 5% for each sale. Total weekly pay is given by the equation S = 0.05c + 500. Graph the equation using the slope and y-intercept.

y = 0.05x + 500 The equation is in slope-intercept form.

b = 500m =0.05

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The slope of the line is 0.05, and the y-intercept is 500. The line crosses the y-axis at the point (0, 500) and moves up 0.05 units for every 1 unit it moves to the right.

x

y

500

1000

1500

2000

10,0005000 15,000

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Additional Example 4: Writing Slope-Intercept Form

Write the equation of the line that passes through (3, –4) and (–1, 4) in slope-intercept form.

Find the slope.

The slope is –2.

Choose either point and substitute it along with the slope into the slope-intercept form.

y = mx + b

4 = –2(–1) + b

4 = 2 + b

Substitute –1 for x, 4 for y, and –2 for m.

Simplify.

4 – (–4) –1 – 3

=y2 – y1

x2 – x1

8–4= = –2

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Solve for b.

Subtract 2 from both sides.

Write the equation of the line, using –2 for m and 2 for b.

4 = 2 + b–2 –2

2 = b

y = –2x + 2

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Try This: Example 4

Write the equation of the line that passes through (1, 2) and (2, 6) in slope-intercept form.

Find the slope.

The slope is 4.

Choose either point and substitute it along with the slope into the slope-intercept form.

y = mx + b

2 = 4(1) + b

2 = 4 + b

Substitute 1 for x, 2 for y, and 4 for m.

Simplify.

6 – 2 2 – 1

=y2 – y1

x2 – x1

4 1= = 4

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Solve for b.

Subtract 4 from both sides.

Write the equation of the line, using 4 for m and –2 for b.

2 = 4 + b–4 –4

–2 = b

y = 4x – 2

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Clear your desk except for a pencil and your calculator.

11-4 Point-Slope Form

Course 3

Warm UpWarm Up

Problem of the DayProblem of the Day

Lesson PresentationLesson Presentation

Warm UpWrite the equation of the line that passes through each pair of points in slope-intercept form.

1. (0, –3) and (2, –3)

2. (5, –3) and (5, 1)

3. (–6, 0) and (0, –2)

4. (4, 6) and (–2, 0)

y = –3

x = 5

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y = x + 2

y = – x – 213

Problem of the Day

Without using equations for horizontal or vertical lines, write the equations of four lines that form a square.

Possible answer: y = x + 2, y = x – 2, y = –x + 2, y = –x – 2

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Learn to find the equation of a line given one point and the slope.

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Vocabulary

point-slope form


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Point on the line((xx11, , yy11))

Point-slope formyy – – yy11 = = mm ( (xx – – xx11))

slopeslope

The point-slope of an equation of a line with slope m passing through (x1, y1) is y – y1 = m(x – x1).

Course 3


Use the point-slope form of each equation to identify a point the line passes through and the slope of the line.

A. y – 7 = 3(x – 4)

Additional Example 1: Using Point-Slope Form to Identify Information About a Line

y – y1 = m(x – x1)

y – 7 = 3(x – 4)

m = 3

(x1, y1) = (4, 7)

The line defined by y – 7 = 3(x – 4) has slope 3, and passes through the point (4, 7).

The equation is in point-slope form. Read the value of m from the equation. Read the point from the equation.

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B. y – 1 = (x + 6)

Additional Example 1B: Using Point-Slope Form to Identify Information About a Line

y – y1 = m(x – x1)

(x1, y1) = (–6, 1)

Rewrite using subtraction instead of addition.

13

13

y – 1 = (x + 6)

y – 1 = [x – (–6)]13

m =13

The line defined by y – 1 = (x + 6) has slope , and

passes through the point (–6, 1).

13

13

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Use the point-slope form of each equation to identify a point the line passes through and the slope of the line.

A. y – 5 = 2 (x – 2)

Try This: Example 1

y – y1 = m(x – x1)

y – 5 = 2(x – 2)

m = 2

(x1, y1) = (2, 5)

The line defined by y – 5 = 2(x – 2) has slope 2, and passes through the point (2, 5).

The equation is in point-slope form. Read the value of m from the equation. Read the point from the equation.

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B. y – 2 = (x + 3)Try This: Example 1B

23

(x1, y1) = (–3, 2)

Rewrite using subtraction instead of addition.

23

y – 2 = (x + 3)

y – 2 = [x – (–3)]23

m =23

The line defined by y – 2 = (x + 3) has slope , and

passes through the point (–3, 2).

23

23

y – y1 = m(x – x1)

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Write the point-slope form of the equation with the given slope that passes through the indicated point.

A. the line with slope 4 passing through (5, -2)

Additional Example 2: Writing the Point-Slope Form of an Equation

y – y1 = m(x – x1)

The equation of the line with slope 4 that passes through (5, –2) in point-slope form is y + 2 = 4(x – 5).

Substitute 5 for x1, –2 for y1, and 4 for m.

[y – (–2)] = 4(x – 5)

y + 2 = 4(x – 5)

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B. the line with slope –5 passing through (–3, 7)

Additional Example 2: Writing the Point-Slope Form of an Equation

y – y1 = m(x – x1)

The equation of the line with slope –5 that passes through (–3, 7) in point-slope form is y – 7 = –5(x + 3).

Substitute –3 for x1, 7 for y1, and –5 for m.

y – 7 = -5[x – (–3)]

y – 7 = –5(x + 3)

Course 3


Write the point-slope form of the equation with the given slope that passes through the indicated point.

A. the line with slope 2 passing through (2, –2)

Try This: Example 2A

y – y1 = m(x – x1)

The equation of the line with slope 2 that passes through (2, –2) in point-slope form is y + 2 = 2(x – 2).

Substitute 2 for x1, –2 for y1, and 2 for m.

[y – (–2)] = 2(x – 2)

y + 2 = 2(x – 2)

Course 3


B. the line with slope -4 passing through (-2, 5)

Try This: Example 2B

y – y1 = m(x – x1)

The equation of the line with slope –4 that passes through (–2, 5) in point-slope form is y – 5 = –4(x + 2).

Substitute –2 for x1, 5 for y1, and –4 for m.

y – 5 = –4[x – (–2)]

y – 5 = –4(x + 2)

Course 3


A roller coaster starts by ascending 20 feet for every 30 feet it moves forward. The coaster starts at a point 18 feet above the ground. Write the equation of the line that the roller coaster travels along in point-slope form, and use it to determine the height of the coaster after traveling 150 feet forward. Assume that the roller coaster travels in a straight line for the first 150 feet.

Additional Example 3: Entertainment Application

As x increases by 30, y increases by 20, so the slope

of the line is or . The line passes through the point (0, 18).

2030

23

Course 3



y – y1 = m(x – x1) Substitute 0 for x1, 18 for y1,

and for m.23

The equation of the line the roller coaster travels along, in point-slope form, is y – 18 = x. Substitute 150 for x to find the value of y.

23

y – 18 = (150)23

y – 18 = 100

y – 18 = (x – 0)23

y = 118

The value of y is 118, so the roller coaster will be at a height of 118 feet after traveling 150 feet forward.

Course 3


Try This: Example 3A roller coaster starts by ascending 15 feet for every 45 feet it moves forward. The coaster starts at a point 15 feet above the ground. Write the equation of the line that the roller coaster travels along in point-slope form, and use it to determine the height of the coaster after traveling 300 feet forward. Assume that the roller coaster travels in a straight line for the first 300 feet.

As x increases by 45, y increases by 15, so the slope

of the line is or . The line passes through the point (0, 15).

1545

13

Course 3



y – y1 = m(x – x1) Substitute 0 for x1, 15 for y1,

and for m.13

The equation of the line the roller coaster travels along, in point-slope form, is y – 15 = x. Substitute 300 for x to find the value of y.

13

y – 15 = (300)13

y – 15 = 100

y – 15 = (x – 0)13

y = 115

The value of y is 115, so the roller coaster will be at a height of 115 feet after traveling 300 feet forward.

Course 3


11-5 Direct Variation

Course 3

Warm UpWarm Up



Warm UpUse the point-slope form of each equation to identify a point the line passes through and the slope of the line.

1. y – 3 = – (x – 9)

2. y + 2 = (x – 5)

3. y – 9 = –2(x + 4)

4. y – 5 = – (x + 7)

(–4, 9), –2

Course 3


17

23

14

(9, 3), – 17

(5, –2), 23

(–7, 5), – 14

Problem of the Day

Where do the lines defined by the equations y = –5x + 20 and y = 5x – 20 intersect?(4, 0)

Course 3


Learn to recognize direct variation by graphing tables of data and checking for constant ratios.

Course 3


Vocabulary

direct variationconstant of proportionality


Course 3


Course 3


Course 3


The graph of a direct-variation equation is always linear and always contains the point (0, 0). The variables x and y either increase together or decrease together.

Helpful Hint

Determine whether the data set shows direct variation.

A.

Additional Example 1A: Determining Whether a Data Set Varies Directly

Course 3


Make a graph that shows the relationship between Adam’s age and his length.

Additional Example 1A Continued

Course 3


You can also compare ratios to see if a direct variation occurs.

223

2712=

?81

264

81 ≠ 264

The ratios are not proportional.

The relationship of the data is not a direct variation.


Course 3



B.

Additional Example 1B: Determining Whether a Data Set Varies Directly

Course 3


Make a graph that shows the relationship between the number of minutes and the distance the train travels.

Additional Example 1B Continued

Plot the points.

The points lie in a straight line.

Course 3


(0, 0) is included.


The ratios are proportional. The relationship is a direct variation.

2510

5020

7530

10040= = = Compare ratios.


Course 3



A.


Kyle's Basketball Shots

Distance (ft) 20 30 40

Number of Baskets 5 3 0

Course 3


Make a graph that shows the relationship between number of baskets and distance.

Try This: Example 1A Continued

Num

ber

of

Bask

ets

Distance (ft)

2

3

4

20 30 40

1

5

Course 3




520

330=

?60

150

150 60.

The ratios are not proportional.

The relationship of the data is not a direct variation.

Course 3



B.


Ounces in a Cup

Ounces (oz) 8 16 24 32

Cup (c) 1 2 3 4

Course 3


Make a graph that shows the relationship between ounces and cups.

Try This: Example 1B Continued

Num

ber

of

Cup

s

Number of Ounces

2

3

4

8 16 24

1

32

Course 3


Plot the points.

The points lie in a straight line.

(0, 0) is included.



Course 3


The ratios are proportional. The relationship is a direct variation.

Compare ratios. = 1 8 = =2

163

24 432

Find each equation of direct variation, given that y varies directly with x.

A. y is 54 when x is 6

Additional Example 2A: Finding Equations of Direct Variation

y = kx

54 = k 6

9 = k

y = 9x

y varies directly with x.

Substitute for x and y.

Solve for k.

Substitute 9 for k in the original equation.

Course 3


B. x is 12 when y is 15

Additional Example 2B: Finding Equations of Direct Variation

y = kx

15 = k 12



Solve for k. = k54

Substitute for k in the original equation.

54y = k5

4

Course 3


C. y is 8 when x is 5

Additional Example 2C: Finding Equations of Direct Variation

y = kx

8 = k 5



Solve for k. = k85


85y = k8

5

Course 3


Find each equation of direct variation, given that y varies directly with x.

A. y is 24 when x is 4


y = kx

24 = k 4

6 = k

y = 6x



Solve for k.

Substitute 6 for k in the original equation.

Course 3


B. x is 28 when y is 14


y = kx

14 = k 28



Solve for k. = k12


12y = k1

2

Course 3


C. y is 7 when x is 3

Try This: Example 2C

y = kx

7 = k 3



Solve for k. = k73


73y = k7

3

Course 3


Mrs. Perez has $4000 in a CD and $4000 in a money market account. The amount of interest she has earned since the beginning of the year is organized in the following table. Determine whether there is a direct variation between either of the data sets and time. If so, find the equation of direct variation.

Additional Example 3: Money Application

Course 3



A. interest from CD and time

interest from CDtime = 17

1interest from CD

time = = 17342

The second and third pairs of data result in a common ratio. In fact, all of the nonzero interest from CD to time ratios are equivalent to 17.

The variables are related by a constant ratio of 17 to 1, and (0, 0) is included. The equation of direct variation is y = 17x, where x is the time, y is the interest from the CD, and 17 is the constant of proportionality.

= = = 17interest from CDtime = = 17

1342

513

684

Course 3



B. interest from money market and time

interest from money markettime = = 19

191

interest from money markettime = =18.5

372

19 ≠ 18.5

If any of the ratios are not equal, then there is no direct variation. It is not necessary to compute additional ratios or to determine whether (0, 0) is included.

Course 3


Mr. Ortega has $2000 in a CD and $2000 in a money market account. The amount of interest he has earned since the beginning of the year is organized in the following table. Determine whether there is a direct variation between either of the data sets and time. If so, find the equation of direct variation.

Try This: Example 3

Course 3



Interest Interest from

Time (mo) from CD ($) Money Market ($)

0 0 0

1 12 15

2 30 40

3 40 45

4 50 50

Course 3



interest from CDtime = 12

1interest from CD

time = = 15302

The second and third pairs of data do not result in a common ratio.


A. interest from CD and time

Course 3



B. interest from money market and time

interest from money markettime = = 1515

1interest from money market

time = =20 402

15 ≠ 20


Course 3


11-6 Graphing Inequalities in Two Variables

Course 3

Warm UpWarm Up



Warm UpFind each equation of direct variation, given that y varies directly with x.

1. y is 18 when x is 3.

2. x is 60 when y is 12.

3. y is 126 when x is 18.

4. x is 4 when y is 20.

y = 6x

y = 7x

Course 3


y = 5x

y = x15

Problem of the Day

The circumference of a pizza varies directly with its diameter. If you graph that direct variation, what will the slope be?

Course 3


Learn to graph inequalities on the coordinate plane.

Course 3


Vocabulary

boundary linelinear inequality


Course 3


A graph of a linear equation separates the coordinate plane into three parts: the points on one side of the line, the points on the boundary line, and the points on the other side of the line.

Course 3


Course 3


When the equality symbol is replaced in a linear equation by an inequality symbol, the statement is a linear inequality. Any ordered pair that makes the linear inequality true is a solution.

Course 3


Graph each inequality.

A. y < x – 1

Additional Example 1A: Graphing Inequalities

First graph the boundary line y = x – 1. Since no points that are on the line are solutions of y < x – 1, make the line dashed. Then determine on which side of the line the solutions lie.

(0, 0)

y < x – 1

Test a point not on the line.

Substitute 0 for x and 0 for y.0 < 0 – 1?

0 < –1?

Course 3



Since 0 < –1 is not true, (0, 0) is not a solution of y < x – 1. Shade the side of the line that does not include (0, 0).

Course 3


B. y 2x + 1

Additional Example 1B: Graphing Inequalities

First graph the boundary line y = 2x + 1. Since points that are on the line are solutions of y 2x + 1, make the line solid. Then shade the part of the coordinate plane in which the rest of the solutions of y 2x + 1 lie.

(0, 4) Choose any point not on the line.

Substitute 0 for x and 4 for y.

y ≥ 2x + 1

4 ≥ 0 + 1?

Course 3



Since 4 1 is true, (0, 4) is a solution of y 2x + 1. Shade the side of the line that includes (0, 4).

Course 3


C. 2y + 5x < 6

Additional Example 1C: Graphing Inequalities

First write the equation in slope-intercept form.

2y < –5x + 6

2y + 5x < 6

y < – x + 352

Then graph the line y = – x + 3. Since points that

are on the line are not solutions of y < – x + 3,

make the line dashed. Then determine on which

side of the line the solutions lie.

52 5

2

Subtract 5x from both sides.

Divide both sides by 2.

Course 3


Additional Example 1C Continued

Since 0 < 3 is true, (0, 0) is a

solution of y < – x + 3.

Shade the side of the line

that includes (0, 0).

52


y < – x + 352

0 < 0 + 3?

0 < 3?

Course 3


Graph each inequality.

A. y < x – 4


First graph the boundary line y = x – 4. Since no points that are on the line are solutions of y < x – 4, make the line dashed. Then determine on which side of the line the solutions lie.

(0, 0)

y < x – 4

Test a point not on the line.

Substitute 0 for x and 0 for y.0 < 0 – 4?

0 < –4?

Course 3


Try This: Example 1A Continued

Since 0 < –4 is not true, (0, 0) is not a solution of y < x – 4. Shade the side of the line that does not include (0, 0).

Course 3


B. y > 4x + 4


First graph the boundary line y = 4x + 4. Since points that are on the line are solutions of y 4x + 4, make the line solid. Then shade the part of the coordinate plane in which the rest of the solutions of y 4x + 4 lie.



y ≥ 4x + 4

3 ≥ 8 + 4?

Course 3



Since 3 12 is not true, (2, 3) is not a solution of y 4x + 4. Shade the side of the line that does not include (2, 3).

Course 3


C. 3y + 4x 9

Try This: Example 1C

First write the equation in slope-intercept form.

3y –4x + 9

3y + 4x 9

y – x + 343

43Then graph the line y = – x + 3. Since points that

are on the line are solutions of y – x + 3, make

the line solid. Then determine on which side of the

line the solutions lie.

43

Subtract 4x from both sides.

Divide both sides by 3.

Course 3


Try This: Example 1C Continued

Since 0 3 is not true, (0, 0) is

not a solution of y – x + 3.

Shade the side of the line that

does not include (0, 0).

43


y – x + 343

0 0 + 3?

0 3?

Course 3


A successful screenwriter can write no more than seven and a half pages of dialogue each day. Graph the relationship between the number of pages the writer can write and the number of days. At this rate, would the writer be able to write a 200-page screenplay in 30 days?

Additional Example 2: Career Application

First find the equation of the line that corresponds to the inequality.

In 0 days the writer writes 0 pages.

point (0, 0)

point (1, 7.5)In 1 day the writer writes no more than 7 pages.1

2Course 3



With two known points, find the slope.

y 7.5 x + 0 The y-intercept is 0. No more than means .

Graph the boundary line y = 7.5x. Since points on

the line are solutions of y 7.5x make the line solid.

Shade the part of the coordinate plane in which the

rest of the solutions of y 7.5x lie.

Course 3


m = 7.5 – 01 – 0

7.5 1

= = 7.5


y 7.5x


Since 2 15 is true, (2, 2) is a solution of y 7.5x. Shade the side of the line that includes point (2, 2).


2 7.5 2?

2 15?

Course 3


The point (30, 200) is included in the shaded area, so the writer should be able to complete the 200 page screenplay in 30 days.


Course 3


A certain author can write no more than 20 pages every 5 days. Graph the relationship between the number of pages the writer can write and the number of days. At this rate, would the writer be able to write 140 pages in 20 days?

Try This: Example 2

First find the equation of the line that corresponds to the inequality.

In 0 days the writer writes 0 pages. point (0, 0)

point (5, 20)In 5 days the writer writes no more than 20 pages.

Course 3



20 - 0 5 - 0m = = 20

5 = 4 With two known points, find the slope.

y 4x + 0 The y-intercept is 0. No more than means .

Graph the boundary line y = 4x. Since points on the line are solutions of y 4x make the line solid. Shade the part of the coordinate plane in which the rest of the solutions of y 4x lie.

Course 3



y 4x


Since 60 20 is not true, (5, 60) is not a solution of y 4x. Shade the side of the line that does not include (5, 60).


60 4 5?

60 20?

Course 3


The point (20, 140) is not included in the shaded area, so the writer will not be able to write 140 pages in 20 days.


x

y200

180

160

140120

100

80

60

40\

20

Pag

es

5 10 15 20 25 30 35 40 45 50

Days

Course 3


11-7 Lines of Best Fit

Course 3

Warm UpWarm Up



Warm UpAnswer the questions about the inequality 5x + 10y > 30.

1. Would you use a solid or dashed boundary line?

2. Would you shade above or below the boundary line?

3. What are the intercepts of the graph?

dashed

above

(0, 3) and (6, 0)

Course 3


Problem of the Day

Write an inequality whose positive solutions form a triangular region with an area of 8 square units. (Hint: Sketch such a region on a coordinate plane.)

Possible answer: y < –x + 4

Course 3


Learn to recognize relationships in data and find the equation of a line of best fit.

Course 3


Course 3


When data show a correlation, you can estimate and draw a line of best fit that approximates a trend for a set of data and use it to make predictions.

To estimate the equation of a line of best fit:

• calculate the means of the x-coordinates and y-coordinates: (xm, ym)

• draw the line through (xm, ym) that appears to best fit the data.

• estimate the coordinates of another point on the line.

• find the equation of the line.

Plot the data and find a line of best fit.

Additional Example 1: Finding a Line of Best Fit

Course 3


Plot the data points and find the mean of the x- and y-coordinates.

xm = = 6 4 + 7 + 3 + 8 + 8 + 66

ym = = 4 4 + 5 + 2 + 6 + 7 + 46

23

x 4 7 3 8 8 6

y 4 5 2 6 7 4

23(xm, ym)= 6, 4

Course 3


The line of best fit is the line that comes closest to all the points on a scatter plot. Try to draw the line so that about the same number of points are above the line as below the line.

Remember!


Course 3


Draw a line through 6, 4 that best represents the data. Estimate and plot the coordinates of another point on that line, such as (8, 6). Find the equation of the line.

23

Course 3


Find the slope.

y – y1 = m(x – x1) Use point-slope form.

y – 4 = (x – 6)23

23 Substitute.

y – 4 = x – 423

23

23y = x +2

3

The equation of a line of best fit is .23y = x +23

Additional Example 1 Continued23

13m = = =

6 – 4

8 – 6

1

223

Plot the data and find a line of best fit.

Try This: Example 1

Course 3


Plot the data points and find the mean of the x- and y-coordinates.

xm = = 2 –1 + 0 + 2 + 6 + –3 + 8 6

ym = = 1 –1 + 0 + 3 + 7 + –7 + 46

x –1 0 2 6 –3 8

y –1 0 3 7 –7 4

(xm, ym) = (2, 1)


Course 3


Draw a line through (2, 1) that best represents the data. Estimate and plot the coordinates of another point on that line, such as (10, 10). Find the equation of the line.

Course 3


Find the slope.


y – 1 = (x – 2)98 Substitute.

y – 1 = x –98

94

The equation of a line of best fit is . y = x –98

54


m = = 10 – 1 10 – 2

98

y = x –98

54

Find a line of best fit for the Main Street Elementary annual softball toss. Use the equation of the line to predict the winning distance in 2006.

Additional Example 2: Sports Application

Course 3


Let 1990 represent year 0. The first point is then (0, 98), and the last point is (12, 107).

xm = = 50 + 2 + 4 + 7 + 125

Year 1990 1992 1994 1997 2002

Distance (ft) 98 101 103 106 107

ym = = 103 98 + 101 + 103 + 106 + 107 5

(xm, ym) = (5, 103)


Course 3


Draw a line through (5, 103) that best represents the data. Estimate and plot the coordinates of another point on that line, such as (10, 107). Find the equation of the line.

Course 3


m = = 0.8 107 - 10310 - 5 Find the slope.


y – 103 = 0.8(x – 5) Substitute.

y – 103 = 0.8x – 4

y = 0.8x + 99

The equation of a line of best fit is y = 0.8x + 99. Since 1990 represents year 0, 2006 represents year 16.


Course 3


Substitute.

y = 12.8 + 99

y = 0.8(16) + 99

The equation predicts a winning distance of about 112 feet for the year 2006.

y = 111.8


Predict the winning weight lift in 2010.

Try This: Example 2

Course 3


Let 1990 represent year 0. The first point is then (0, 100), and the last point is (10, 170).

xm = = 60 + 5 + 7 + 8 + 105

ym = = 132100 + 120 + 130 + 140 + 170 5

Year 1990 1995 1997 1998 2000

Lift (lb) 100 120 130 140 170

(xm, ym) = (6, 132)


Course 3


Draw a line through (5, 132) the best represents the data. Estimate and plot the coordinates of another point on that line, such as (7, 140). Find the equation of the line.

Years since 1990

weig

ht

(lb)

0100

120

140

160

180

2 4 6 8 10

200

Course 3


m = = 4 140 – 132 7 – 5 Find the slope.


y – 132 = 4(x – 5) Substitute.

y – 132 = 4x – 20

y = 4x + 112

The equation of a line of best fit is y = 4x + 112. Since 1990 represents year 0, 2010 represents year 20.


Course 3


Substitute.

y = 192

y = 4(20) + 112

The equation predicts a winning lift of about 192 lb for the year 2010.


Chapter11

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