5.1 Equations of Lines

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Equations of the form ax + by = c are called linear equations in two variables.

The point (0,4) is the y-intercept.

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

x

y

2-2

This is the graph of the equation 2x + 3y = 12.

(0,4)

(6,0)

5.1 Equations of Lines

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The slope of a line is a number, m, and is defined.

m = 0

m is undefined

negative

risesloperun

If a line goes up from left to right, then it has a positive slope.

If a line goes down from left to right, then it has a negative slope.

positive


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y

x2-2

The slope of a line is a number, m, which measures its steepness.

m = 0

m = 2m is undefined

m = 12

m = - 14


4

y

x1-1

4 22

m


(-4,0)(0,-2)

5

x

y

x2 – x1

y2 – y1

change in y

change in x

The slope of the line passing through the two points (x1, y1) and (x2, y2) is given by the formula The slope is the change in y divided by the change in x as we move along the line from (x1, y1) to (x2, y2).

y2 – y1

x2 – x1

m = , (x1 ≠ x2 ).

(x1, y1)

(x2, y2)


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Example: Find the slope of the line passing through the points (2, 3) and (4, 5).

Use the slope formula with x1= 2, y1 = 3, x2 = 4, and y2 = 5.

y2 – y1

x2 – x1

m = 5 – 3 4 – 2

= = 22

= 1

2

2(2, 3)

(4, 5)

x

y


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y = -2x + 3 x + y = 10 2x = 4 y = 5 x + 2y – 5 = 0


State the slope and the y-intercept

m = -2 b = 3 m =-1 b = 10 none m = 0 b = 5 m = - 1/2 b = 5

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A linear equation written in the form y = mx + b is in slope-intercept form.

To graph an equation in slope-intercept form:

1. Write the equation in the form y = mx + b. Identify m and b.

The slope is m and the y-intercept is (0, b).

2. Plot the y-intercept (0, b).

3. Starting at the y-intercept, find another point on the line using the slope.

4. Draw the line through (0, b) and the point located using the slope.


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1

Example: Graph the line y = 2x – 4.

2. Plot the y-intercept, (0, - 4).

1. The equation y = 2x – 4 is in the slope-intercept form. So, m = 2 and b = - 4.

3. The slope is 2.

The point (1, -2) is also on the line.

1= change in y

change in xm = 2

4. Start at the point (0, 4). Count 1 unit to the right and 2 units up to locate a second point on the line.

2

x

y

5. Draw the line through (0, 4) and (1, -2).

(0, - 4)

(1, -2)


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In 2000, there were 200 Twins fans at McDonell. The number will grow by 20 each year. Write a linear model and sketch the graph.


N 20t 200

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N 20t 200How many fans will there be in 5 years?

300200)5(20 NWhen will there be360 fans?360 = 20t + 200

160 = 20t t = 8 years

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5.2 Equations of LinesPoint Slope

Write an equation of the line that passes through the point (2,-1) with a slope of 2.

A(2,-1)

bmxy b )2(21

b 41b 5

52 xy

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Write an equation of the line that passes through the point (-1,-1) with a slope of -3

A(-1,-1)

bmxy b )1(31

b 31b 4

43 xy

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Write an equation of the line that passes through the point (1,0) with a slope of 1/2

A(1,0)

bmxy

b )1(210

b210

b21

21

21

xy

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5.3 Equations of LinesTwo Points

Write an equation of the line that passes through the point (2,1) and (3,-3)

A(2,1)

12

12

xxyym

b )2(41

414

2313

m

bmxy

b9

94 xy

B(3,-3)

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Write an equation of the line that passes through the point (1,4) and (0, 3)

A(1,4)

12

12

xxyym

b )1(14

111

0134

m

bmxy

b3

3xy B(0,3)

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Write an equation of the line that passes through the point (-1,6) and (3, -2)

A(-1,6)12

12

xxyym

b )1(26

24

831)2(6

m

bmxy

b4

42 xy

B(3,-2)

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Write an equation of the line that passes through the point (1,-3) and (3, -2)

A(-1,-3)

12

12

xxyym

b )1(213

21

21

31)2(3

m

bmxy

27

213 b

27

21

xy

B(3,-2)

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5.4 Exploring Data: Fitting a Line to Data

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Find the equation of the line using 1928 and 1988

12

12

xxyym

19201988

2.125.10

028.60

7.1

m

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Find the equation of the line using 1928 and 1988

028.60

7.1

m

bmxy

b )1988)(028.(5.10

b 3.565.10

b8.66

8.66028. xy

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What would the time be in the year2000?

sec8.10y

8.66)2000(028. y

23


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Year Time1928 12.21932 11.91936 11.51948 11.91952 11.51956 11.51960 11.01964 11.41968 11.01972 11.11976 11.01980 11.61984 10.91988 10.5

Times vs Yearsy = -0.0197x +

49.953R2 = 0.6676

10.010.511.011.512.012.5

1900 1950 2000

Years

Tim

e

Series1

Linear(Series1)

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Correlation: The agreement between two sets of data

-1 < r < 1r is called the correlationcoefficient

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5.5 Standard Form of a Linear Equation

Form Standard FormIntercept Slope

cby axbmxy

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Form Standard to43Transform xy

43 xy Subtract y from both sides

430 yx Add 4 to both sides

yx 34 Switch

43 yx

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Form Standard to121Transform xy

121

xy Multiply both sides by 2

22 xy Subtract 2y from both sides


22 yx

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Form Standard to41

32Transform xy

41

32




3128 yx

30


Form Standard to21

53Transform xy

21

53




5106 yx

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A linear equation written in the form y – y1 = m(x – x1) is in point-slope form. It is used mainly to write the equation of a line. The graph of this equation is a line with slope m passing through the point (x1, y1).

The graph of the equation

y – 3 = - (x – 4) is a line

of slope m = - passing

through the point (4, 3).

12 1

2

(4, 3)

m = - 12

x

y

4

4

8

8

5.6 Point-Slope Form of a Line

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Example: Write the slope-intercept form for the equation of the line through the point (-2, 5) with a slope of 3.

Use the point-slope form, y – y1 = m(x – x1), with m = 3 and (x1, y1) = (-2, 5).

y – y1 = m(x – x1) Point-slope form

y – y1 = 3(x – x1) Let m = 3.

y – 5 = 3(x – (-2)) Let (x1, y1) = (-2, 5).

y – 5 = 3(x + 2) Simplify.

y = 3x + 11 Slope-intercept form


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Example: Write the slope-intercept form for the equation of the line through the points (4, 3) and (-2, 5).

y – y1 = m(x – x1) Point-slope form

Slope-intercept formy = - x + 133

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2 1 5 – 3 -2 – 4

= - 6

= - 3

Calculate the slope.m =

Use m = - and the point (4, 3).y – 3 = - (x – 4)13 3

1


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Summary of Equations of Lines


Slope of a line through two points:

Vertical line (undefined slope):

Horizontal line (zero slope):

Slope-intercept form:

Point-slope form:

Standard form:

12

12

xxyym

x = a

y = by = mx + by2 – y1 = m(x2 – x1)

Ax + By = C

5.1 Equations of Lines

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