5.1 Equation of Lines Using Slope-Intercept
Objectives:
• Learn how to use slope intercept form to write an equation of a line
• Learn how to model a real-life situation with a linear equation
Slope Intercept Form:
y = mx + b
Where m = slope
b = y-intercept
Write an equation when given the slope and y-intercept:
m = –2
m=5
b=4
b = −1
2
y = –2x + 4
y = 5x – 1
2
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5.1 Equation of Lines Using Slope-Intercept
Write an equation from the graph
y = mx + b
y=
2
x
3
y = mx + b
y = mx + b
y = –x –1
y=3
+(–3)
Real-life: the Phone Company charges a flat fee of $0.75 for the first minute of long
distance plus $0.10 per minute after that.
Write an equation to use to figure out each call.
1. Verbal Model:
Total cost = first minute cost +
rate · number of minutes
minute
2. Labels:
y = total cost
c = cost for first minute ($0.75)
x = # of additional minutes
r = rate per additional minute ($0.10)
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5.1 Equation of Lines Using Slope-Intercept
3. Algebraic Model:
Fill in what you know and write in slope-intercept form (y = mx + b)
y = .75 + .10x
4. Solve/Give Answer:
Find values for extra minutes: 0, 5, 10 and graph
x
y
0
.75
y
3
Cost ($)
5
1.25
10
1.75
2
1
1 2 3 4 5 6 7 8 9 10 11
x
minutes
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5.2 Equations of Lines Given the Slope and a Point
Objectives:
• Learn how to use slope and any point to write an equation of the line
• Learn how to model a real-life situation with a linear equation
Slope intercept form:
y = mx + b
Need to know m and b
Given the slope, m = –2 and the point (6, –3), find equation of the line:
y = mx + b
–3 = –2(6) + b
substitute what you know into the slope-intercept form of the
equation, and then solve for b
–3 = –12 + b
+12 +12
9=b
y = –2x + 9
graph check:
10
9
8
7
6
5
4
3
2
1
y
–10–9–8–7–6–5–4–3–2–1 1 2 3 4 5 6 7 8 9 10 x
–2
–3
–4
–5
–6
–7
–8
–9
–10
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5.2 Equations of Lines Given the Slope and a Point
Write the equation for a line that passes through point (6, 7) and has a slope 2
3
y = mx + b
graph:
7 = 6( 2 ) + b
9
8
7
6
5
4
3
2
1
3
7=4+b
–4 –4
3=b
y = 2x + 3
3
–9–8–7–6–5–4–3–2–1
–1
–2
–3
–4
–5
–6
–7
–8
–9
y
1 2 3 4 5 6 7 8 9 x
Real-life: Find an equation for vacation trips y (in millions) in terms of the
year, t. Let t = 0 correspond to 1980.
From 1980 – 1990, vacation trips increased by about 15 million
per year. In 1985, Americans went on 340 million vacation
trips.
Because change is constant, you can model this as a linear equation y = mt + b
Constant is 15 trips per year, so you know slope. In 1985, the year would be t = 0 + 5
and the number of vacation trips (y) would be 340 million.
y = 15t + b
340 = 15(5) + b
340 = 75 + b
–75 –75
265 = b
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5.2 Equations of Lines Given the Slope and a Point
y = 15t + 265
y
600
500
400
300
200
100
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
t
Year ( 0 ↔ 1980)
In 1998, the value of t = 1998 - 1980 = 18
So the number of vacations taken in 1998 would be:
y = 15(18) + 265
y = 270 + 265
y = 535
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5.3 Equations of Lines Given 2 Points
Objectives:
• Learn how to write an equation of a line given 2 points on the line
• Learn how to model a real-life problem with a linear equation
Equation of a line:
y = mx + b
Need slope and y-intercept
Given 2 points, can you find the slope and y-intercept?
Points (3, 5) and (–7, 2)
y − y1
1. Find the slope: m = rise = 2
run x2 − x1
m = 2−5 = 3
−7 − 3 10
y = mx + b
Substitute the value of m back into the
slope-intercept form of the linear equation
y= 3 x+b
10
2. Substitute the x and y values from one point to find the y-intercept (the
value of b):
y = mx + b
(3, 5)
5 = 3 (3) + b
10
5= 9 +b
10
–9 –9
10 10
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5.3 Equations of Lines Given 2 Points
4 1 =b
10
3. Substitute both m and b back into the slope-intercept form of the linear
equation and solve:
y = 3 x + 4 1 This is the equation of the line passing through points
10
10
(3, 5) and (–7, 2)
(–7, 2)
10
9
8
7
6
5
4
3
2
1
y
(3, 5)
–10–9–8–7–6–5–4–3–2–1 1 2 3 4 5 6 7 8 9 10 x
–2
–3
–4
–5
–6
–7
–8
–9
–10
Example: Find linear equation with points (–3, 2) and (5,–2), and then graph
Let’s say (x1, y1) = (–3, 2) and (x2, y2) = (5,–2) (Though it really doesn’t matter)
1. Find the slope, m:
m = −2 − 2 = −4 = − 1
5 − (−3)
8
2
y = mx + b
Substitute m into the slope-intercept form of
the linear equation
y = −1x + b
2
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5.3 Equations of Lines Given 2 Points
2. Substitute the x and y values from one of the two points given in the
question to find the y-intercept (the value of b):
y = mx + b (–3, 2)
2 = − 1 (–3) + b
2
2= 3 +b
2
–3 –3
2 2
1= b
2
3. Substitute both m and b back into the slope-intercept form of the linear
equation and solve:
y = −1x + 1
2
2
10
9
8
7
6
5
4
(–3, 2) 3
2
1
y
–10–9–8–7–6–5–4–3–2–1 1 2 3 4 5 6 7 8 9 10 x
–2
–3
(5, –2)
–4
–5
–6
–7
–8
–9
–10
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5.3 Equations of Lines Given 2 Points
Can we find the equation of a line if we know both the x, and y-intercepts?
Write an equation that has points whose y-intercept is –4 and x-intercept is –6
What are your 2 points? (0, –4) (–6, 0)
m=
y2 − y1 0 − (−4)
=
x2 − x1 −6 − 0
y = mx + b
0 = – 2 (–6) + b
3
0 = +4 + b
–4 –4
= 4
−6
= –2
3
y = mx + b
y = –2x – 4
3
–4 = b
(–6, 0)
8
7
6
5
4
3
2
1
y
–8–7–6–5–4–3–2–1 1 2 3 4 5 6 7 8 x
–2
–3
–4 (0, –4)
–5
–6
–7
–8
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5.3 Equations of Lines Given 2 Points
Real–Life: (see textbook, page 253, Question #29)
The Tunnel (aka, “The Chunnel”) from Calais, France to Dover, England
Write a linear equation of the line formed from A to B
Point A: (0, 60)
m=
Point B: (15, – 70)
y2 − y1 −70 − 60
=
x2 − x1 15,000 − 0
= − 130 = − 13
15000
1500
y = mx + b
60 = − 13 (0) + b
1500
60 = b
y = mx + b
y = – 13 x + 60
1500
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5.4 Exploring Data: Fitting a Line to Data
Objectives:
• Learn how to find a linear equation that approximates a set of data points
• Learn how to use scatter plots to determine positive, negative or no correlation
The winning Olympic times for the 100-meter run from 1928 to 1988 are plotted in the
graph below. Approximate the best-fitting line for these times. Let y represent the
winning time and x the year (x = 0, corresponding to 1928).
y
13
12
11
10
20
40
60 x
Looking at the graph, do you see the trend in the info given?
Can we write a linear equation to match?
Often in life the data collected (no matter how carefully done) to analyze whether a
relationship exists between two variables will not appear as a nice and neat straight
line. However, while all the data may not fall on one line, they may still exhibit a trend
that can best be described as linear. When this happens we draw a single line that
best represents (or approximates) the set of data points. This line is called the line of
best fit.
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5.4 Exploring Data: Fitting a Line to Data
Draw a line of best fit and pick any 2 points along the line (they don’t have to be
actual points plotted on the scatter plot).
Let’s say points (0, 12) and (50, 11)
Use the method described in Chapter 5.3
y − y1 11 − 12
m= rise = 2
=
run x2 − x1 50 − 0
m= − 1
50
y = mx + b
11 = − 1 (50) + b
50
11 = –1 + b
+1 +1
12 = b
y = − 1 x + 12
50
Correlation:
A quantitative assessment of whether a relationship exists between
two variables. Describes the slope of the line of best fit.
y
y
y
x
Positive correlation
x
x
Negative correlation
No correlation
When a line cannot be drawn to represent the set of data points, we say there is no
correlation.
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5.4 Exploring Data: Fitting a Line to Data
Real-life: Make a graph, find the equation of the line of best fit (if possible), and make
a prediction.
Find approximate wing area of a 400 g bird
Bird
Weight (g)
25
47
78
93
143
607
840
2090
Sparrow
Martin
Blackbird
Starling
Dove
Crow
Gull
Blue Heron
Wing Area
(cm2)
87
186
245
190
357
1344
2006
4436
Let x represent weight and y wing area
4800
4400
4000
3600
3200
2800
2400
2000
1600
1200
800
400
y
200 400 600 800 1000 1200 1400 1600 1800 2000 2200
x
Pick any two points:
Let’s say (x1, y1) is (140, 360) and (x2, y2) is (1400, 3200)
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5.4 Exploring Data: Fitting a Line to Data
y = mx + b
m=
y2 − y1 3200 − 360 2840 142
=
=
≈ 2.25
=
x2 − x1 1400 − 140 1260
63
y = mx + b
360 = 2.25 (140) + b
360 = 315 + b
45 = b
y ≈ 2.25x + 45
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5.5 Standard Form of a Linear Equation
Objectives:
• Learn how to transform a linear equation into standard form
• Learn how to model a real-life situation using the standard form of a linear
equation
So far, we have focused on the Slope-Intercept Form of the Linear Equation, though
the Standard Form of the Linear Equation is also commonly used.
Slope-Intercept Form
Standard Form
y = mx + b
Ax + By = C
With the Standard Form of the Linear Equation, notice that the two variables are shown
on the left side of the equation, while a constant is found on the right side. One
advantage of the Standard Form is that it can be used for any type of line, including
vertical lines.
Changing/Transforming Slope-Intercept Form to Standard Form:
y = 3x + 2
4
4y = 3x + 8
–3x –3x
–1 · (–3x + 4y) = 8 · (–1)
3x – 4y = –8
1) Multiply by 4 to clear the fraction
2) Subtract 3x from both sides
3) Standard Form
4) Multiply by –1 to make x positive
Same Standard Form -- acceptable in either form
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5.5 Standard Form of a Linear Equation
Example:
y = –.23x + 5.2
100y = –23x + 520
+23x +23x
23x + 100y = 520
1) multiply by 100 to clear the decimal
2) move the term containing x to the left side
Standard Form - it’s now easy to find both the x- and yintercepts by setting each to zero.
Real-life: You have $6 to buy bananas and apples. Bananas cost $0.49 per pound,
and apples cost $0.34 per pound. Write a linear equation that represents
the different amounts of fruit you can buy.
Let x represent bananas and y represent apples
.49x + .34y = 6
y
22
20
18
16
14
12
10
8
6
4
2
2
4
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10 12 14 16
2
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5.6 Point-Slope Form of the Equation of a Line
Objectives:
• Learn how to use the point-slope form to write an equation of a line
• Learn how to model a real-life situation using point-slope form
Consider a line containing the point (x1, y1) = (2, 5) with a slope of 2 . Let (x, y) be
3
any other point on the line.
Because you know (2, 5) and (x, y) are 2 points on the line, you can find the slope of
the line
y −5
m= rise =
and since we know the slope was given as 2 , we can
run x − 2
3
write:
m=
y −5 2
=
x−2 3
by cross-multiplying and rewriting we can say
y – 5 = 2 (x – 2)
3
Point-Slope Form:
This is called point-slope form
y – y1 = m(x – x1)
When using point-slope form, be sure you see the difference between
(x1, y1) and (x, y).
(x1, y1) = the given point
(x, y) = any point on the line
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5.7 Problem Solving Using Linear Models
Objectives:
• Learn how to create and use Linear Models to solve problems
• Learn how to make linear models that are accurate but simple to use
Linear Models are used to Solve 2 Basic Types of Real-Life Problems:
1. Constant rate of change
2. Two variables such that the sum Ax + By = C is a constant
Rate of Change Example:
The cost of renting a lawn mower is $10 for the first hour and $5 for each additional
hour. Write a linear model that gives the cost of renting a lawn mower in terms of the
number of hours rented, then determine how much it would cost to rent a lawn mower
for 3 hours
1. Verbal Model:
Total cost = # of additional hours · (additional hourly cost) + 1st hour cost
2. Label: identify what you know and assign variables for what you don’t know
y = total cost
1st hour cost = $10
Additional hourly cost = $5/hour
x = the total number of hours needed (including the 1st hour), so . . .
x – 1 = number of additional hours needed
3. Algebraic Model:
y = 5(x – 1) + 10
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5.7 Problem Solving Using Linear Models
4. Solve:
y = 5(x – 1) + 10
when x = 3 hours
y = 5(3 – 1) + 10
y = 5(2) + 10
y = 20
It would cost $20 to rent a lawn mower for 3 hours
Constant Addition Example:
The total income from the sale of raffle tickets was $202.50. If student tickets sold for
$0.75 and adult tickets sold for $1.25, write a linear equation that describes the
relationship between the number of student and adult tickets and construct a graph to
provide a visual model
Let: s = the number of student tickets sold
a = the number of adult tickets sold
Linear Equation:
.75s + 1.25a = 202.50
Graph:
If only adult tickets were sold, s = 0
If only student tickets were sold, a = 0
.75s + 1.25a = 202.50
.75s + 1.25a = 202.50
.75(0) + 1.25a = 202.50
.75s + 1.25(0) = 202.50
1.25a = 202.50
1.25
1.25
.75s = 202.50
.75
.75
a = 162
s = 270
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5.7 Problem Solving Using Linear Models
175
a
(0, 162)
150
125
Adult Tickets 100
75
50
25
(270, 0)
25
50
75
100 125 150 175 200 225 250 275
s
Student Tickets
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