Unit 16

THE CARTESIAN COORDINATE SYSTEM AND GRAPHS OF

LINEAR EQUATIONS

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GRAPHING A LINEAR EQUATION The graph of a straight line (linear) equation has an

infinite number of points that satisfy the equation Procedure for graphing an equation:

1. Write the equation in terms of either variable (x or y)

2. a. Make a table of values. Head one column x and the other column yb. Select a least three convenient values for one variable and find their corresponding values for the second variable by substituting the first variable values in the equation

3. Draw a label the x axis and the y axis on the graph

4. Plot the coordinates on the graph5. Connect the points. The straight line is the graph

of the equation

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GRAPHING EXAMPLE

Graph the equation: 2x – y = 5

– Write the equation in terms of x or y. We will use y

– Make a table of values that satisfy the equation by choosing a value for x and finding the corresponding y value

– Graphing is shown on the next slide

y = 2x – 5

1 3

-3 1

-5 0

y

x

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EXAMPLE GRAPH

• Graph of the equation 2x – y = 5

• Draw and label the x and y axes.

• Plot the points on the graph.

• Connect the points

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SLOPE OF A LINEAR EQUATION

The slope of a linear equation (a straight line) refers to its steepness or inclination

Slope is defined as follows:

slope rise

run

change in y

change in x

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SLOPE OF A LINEAR EQUATION

If a line rises moving from left to right, it has a positive slope

If a line falls moving from left to right, it

has a negative slope

A horizontal line has a zero slope

A vertical line has an undefined slope

Positive slope

Negative slope

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Determine the slope of the line that passes through the points (1,3) and (–1,2):

Determine the slope of the line that passes through the points (–2,4) and (3,–3):

DETERMINING SLOPE

slope change in y

change in x

3 2

1 1

1

2Ans.

slope change in y

change in x

4 3

2 3

7

5

7

5Ans.

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SLOPE-INTERCEPT FORM

The equation of the line can be determined if the slope and the y-intercept of a straight line are known

The y-intercept is the y-coordinate where the line intercepts (crosses) the y axis

The value of the x at this point is 0 The general equation for the slope-intercept

of a straight line is: y = mx + b

where m is the slope and b is the y-intercept

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SLOPE-INTERCEPT FORM

Determine the equation of a line when the slope is 2 and the y intercept is –5 Given: m = 2 and b = –5 Substituting these values into the slope-intercept

form: y = mx + b Becomes y = 2x – 5 Ans

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DETERMINING THE EQUATION OF A LINE GIVEN TWO POINTS

If two points on a straight line are known, an equation can be determined Step 1: Find the slope of the line Step 2: Find the y-intercept

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DETERMINING THE EQUATION OF THE LINE

Find the equation of a line that passes through (1,9) and (–2,3). Find the slope:

Find the y-intercept. Use either point since they both lie on the line

Substitute the point (1,9) (can use either point) and m = 2, into the slope-intercept form

Solve for b

Substitute the slope and y-intercept into the slope-intercept form:

slope 9 3

1 2

6

32

y = mx + b becomes 9 = 2(1) + b; b = 7

y = 2x + 7 Ans

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POINT-SLOPE FORM

The equation of the line can be determined if the slope and one point on the line are known

The general equation for the point-slope equation of a line:

Where m is the slope and (x1,y1) is a point on the line.

)( 11 xxmyy

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POINT-SLOPE FORM

Determine the equation of a line when its slope is 2 and it passes through the point (2,–5)

m = 2, x1 = 2, and y1 = –5

Ansyx

yx

yx

122

1022

)5(22

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Application

You rent a car from Rentus Car Co. They tell you that the fee is $30 and then $.05 per mile you drive during the rental. What is the equation for the car rental?

$30 is a flat fee paid once…like a y-intercept….it is where your bill will start if you never drive the car

$.05 is how much more you will pay for every mile you drive…..like a slope….climbing

Y = .05x + 30

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Application

You rent a car from Rentus Car Co. They tell you that the fee is $30 and then $.05 per mile you drive during the rental. How much will it cost to drive 300 miles?

You know that x is the miles driven and y is the price from the original equation.

So $45 for the car.

45

30)300(05.

y

y

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Application

You rent a car from Rentus Car Co. They tell you that the fee is $30 and then $.05 per mile you drive during the rental. How much can you drive for $65?

You know that x is the miles driven and y is the price from the original equation.

So 700 miles

x

x

x

700

05.35

3005.65

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Application

You have found another company, Rentalot, that you can rent from and they charge $20 and then $.10 per mile. What is the equation?

What will it cost to drive 300 miles? $50

So you can see this is more than Rentus was. So where is the point that you would want to change companies?

201.0 xy

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Application

When you are looking for the point where a deal switches companies, you are looking for an intersection point. So where will that cost be the same? Remember cost is y…so when will y be the same? We don’t care what that number is right now but it must be

the same…..so we have two equations.

So if we want y to be the same in both equation we can substitute.

201.03005. xyandxy

x

x

xx

200

05.10

201.03005.0

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Application

So at 200 miles is where the companies are even in price. From the examples before we know that less than 200 we will rent from Rentalot and for more than 200 miles Rentus to always get the best deal.

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PRACTICE PROBLEMS

Graph the linear equations for problems 1 and 2 by plotting points:

1. 2x – y = 62. 6x + 3y = 15

Determine the slope of the lines that pass through the points given in problems 3 and 4:

3. (–2, –3) and (4,5)4. (5, –7) and (–3, –2)

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PRACTICE PROBLEMS (Cont)

Write the equation of the line for problems 5 and 6 given the slope (m) and the y intercept (b):

5. m = –2; b = 16. m = 2/3; b = –4

Find the equation for each of the lines passing through the points given in problems 7 and 8:

7. (–1,4) and (3,6)8. (0, –6) and (10, –2)

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PRACTICE PROBLEMS (Cont)

Two cell companies have plans that you are debating over. The first is $50 for the phone and $.01 per minute you talk a month. The second is $40 per month and $.05 per minute.

9. What is the equation for each company?

10. How long can I talk on each plan if I only can afford $60 per month?

11. How many minutes can I talk on both plans and pay the same amount?

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PROBLEM ANSWER KEY

1. 2.

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PROBLEM ANSWER KEY

3. 4/3

4. –5/8

5. y = –2x + 1

6. y = 2/3x – 4

7. y = –x + 9/2

8. y = 2/5x – 6

9. y = 0.05x + 40, y = 0.01x + 50

10. 400 and 1000 minutes

11. 250 minutes