Transcript eq of lines

2
Graphs and
Functions
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Equations of Lines; Curve Fitting
2.5
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Point-Slope Form
Slope-Intercept Form
Vertical and Horizontal Lines
Parallel and Perpendicular Lines
Modeling Data
Solving Linear Equations in One Variable by
Graphing
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Point-Slope Form
The point–slope form of the equation of
the line with slope m passing through the
point (x1, y1) is
y  y 1  m ( x  x1 )
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Example 1
USING THE POINT-SLOPE FORM
(GIVEN A POINT AND THE SLOPE)
Write an equation of the line through (–4, 1)
having slope –3.
Solution H e re x 1   4, y 1  1, a n d m   3 .
y  y 1  m  x  x1 
y  1   3  x  (  4 )
Point-slope form
x 1   4, y 1  1, m   3
y  1  3  x  4 
Be careful
with signs.
y  1  3 x  12
Distributive property
y  3 x  11
Add 1.
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Example 2
USING THE POINT-SLOPE FORM
(GIVEN TWO POINTS)
Write an equation of the line through (–3, 2)
and (2, –4). Write the result in standard form
Ax + By = C.
Solution Find the slope first.
m 
4  2
2  (3)
T h e slo p e m is 
 
6
5
6
Definition of slope
. E ith e r (  3, 2 ) o r (2,  4 )
5
ca n b e u se d fo r
 x 1, y 1  .
W e ch o o se   3, 2  .
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Example 2
USING THE POINT-SLOPE FORM
(GIVEN TWO POINTS)
Write an equation of the line through (–3, 2)
and (2, –4). Write the result in standard form
Ax + By = C.
Solution
y  y 1  m  x  x1 
Point-slope form
6
y  2    x  (  3 )  x1 = –3, y1 = 2, m = – 6/5
5
Multiply by 5.
5  y  2   6  x  3 
5 y  10  6 x  18
Distributive property.
6 x  5 y  8
Standard form
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Slope-Intercept Form
As a special case,
suppose that a line
passes through the
point (0, b), so the line
has y-intercept b. If
the line has slope m,
then using the pointslope form with x1 = 0
and y1 = b gives the
following.
y  y 1  m  x  x1 
y  b  m x  0
y  mx  b
Slope
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y-intercept
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Slope-Intercept Form
The slope-intercept form of the
equation of the line with slope m and
y-intercept b is
y  m x  b.
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FINDING THE SLOPE AND y-INTERCEPT
FROM AN EQUATION OF A LINE
Example 3
Find the slope and y-intercept of the line with
equation 4x + 5y = –10.
Solution Write the equation in slopeintercept form.
4 x  5 y  10
5 y   4 x  10
4
y   x2
5
m
The slope is

4
5
Subtract 4x.
Divide by 5.
b
and the y-intercept is –2.
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USING THE SLOPE-INTERCEPT
FORM (GIVEN TWO POINTS)
Example 4
Write an equation of a line through (1,1) and
(2,4). Then graph the line using the slopeintercept form.
Solution Use the slope intercept form.
First, find the slope.
m 
4 1
2 1

3
3
Definition of slope.
1
Substitute 3 for m in y = mx + b and choose
one of the given points, say (1,1), to find b.
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Example 4
USING THE SLOPE-INTERCEPT
FORM (GIVEN TWO POINTS)
Solution y  m x  b
y-intercept
Slope-intercept form
1  3  1  b
m = 3, x = 1, y = 1
b  2
Solve for b.
The slope intercept form
is y  3 x  2 .
Plot (0,–2) and then use
the definition of slope to
arrive at (1,1).
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FINDING AN EQUATION FROM A
GRAPH
Example 5
Use the graph of the linear function  shown
to complete the following.
(a) Find the slope, y-intercept,
and x-intercept.
Solution The line
falls 1 unit each time
the x-value
increases by 3 units.
S lo p e 
1
3
 
1
–3
–1
y = (x)
.
3
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Example 5
FINDING AN EQUATION FROM A
GRAPH
Solution
The graph intersects
the y-axis at the
point (0,–1) and
intersects the
x-axis at the point
(–3,0). The
y-intercept is –1 and
the x-intercept is –3.
–3
–1
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y = (x)
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FINDING AN EQUATION FROM A
GRAPH
Example 5
Use the graph of the linear function  shown
to complete the following.
(b) Write the equation that
defines .
Solution
1
The slope is m = 
3
and the y-intercept
is b = –1.
( x )  
1
3
–3
–1
y = (x)
x 1
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Equations of Vertical and
Horizontal lines
An equation of the vertical line
through the point (a, b) is x = a.
An equation of the horizontal line
through the point (a, b) is y = b.
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Parallel Lines
Two distinct nonvertical lines are
parallel if and only if they have the
same slope.
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Perpendicular Lines
Two lines, neither of which is vertical,
are perpendicular if and only if their
slopes have a product of –1. Thus, the
slopes of perpendicular lines, neither
of which are vertical, are negative
reciprocals.
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Example 6
FINDING EQUATIONS OF PARALLEL
AND PERPENDICULAR LINES
Write the equation in both slope-intercept
and standard form of the line that passes
through the point (3, 5) and satisfies the
given condition.
(a) parallel to the line 2x + 5y = 4
Solution The point (3, 5) is on the line, so
we need only to find the slope to use the
point-slope form. We find the slope by
writing the equation of the given line in slopeintercept form. (That is, we solve for y.)
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Example 6
FINDING EQUATIONS OF PARALLEL
AND PERPENDICULAR LINES
Write the equation in both slope-intercept
and standard form of the line that passes
through the point (3, 5) and satisfies the
given condition.
(a) parallel to the line 2x + 5y = 4
Solution
2x  5y  4
5 y  2 x  4
2
4
y   x 
5
5
Subtract 2x.
Divide by 5.
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Example 6
FINDING EQUATIONS OF PARALLEL
AND PERPENDICULAR LINES
Write the equation in both slope-intercept
and standard form of the line that passes
through the point (3, 5) and satisfies the
given condition.
(a) parallel to the line 2x + 5y = 4
2
4
Solution
y   x 
5
5
The slope is –2/5. Since the lines are
parallel, –2/5 is also the slope of the line
whose equation is to be found.
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Example 6
FINDING EQUATIONS OF PARALLEL
AND PERPENDICULAR LINES
Write the equation in both slope-intercept
and standard form of the line that passes
through the point (3, 5) and satisfies the
given condition.
(a) parallel to the line 2x + 5y = 4
Solution y  y 1  m  x  x 1  Point-slope form
y 5  
y 5  
2
5
2
5
x  3
x 
6
5
m = – 2/5, x1 = 3,
y1 = 5
Distributive property
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Example 6
FINDING EQUATIONS OF PARALLEL
AND PERPENDICULAR LINES
Write the equation in both slope-intercept
and standard form of the line that passes
through the point (3, 5) and satisfies the
given condition.
(a) parallel to the line 2x + 5y = 4
Solution
slope-intercept
form
y  
2
x 
31
5
5
5 y  2 x  31
standard
2 x  5 y  31
form
Add 5 = 25/5.
Multiply by 5.
Add 2x.
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FINDING EQUATIONS OF PARALLEL
AND PERPENDICULAR LINES
Example 6
Write the equation in both slope-intercept and
standard form of the line that passes through
the point (3, 5) and satisfies the given condition.
(b) perpendicular to the line 2x + 5y = 4
Solution In part (a) we found that the slope of
the line 2x + 5y = 4 is –2/5. The slope of any
line perpendicular to it is 5/2.
y  y 1  m  x  x1 
y 5 
5
2
x  3
m 
5
2
, x 1  3, y 1  5
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Example 6
FINDING EQUATIONS OF PARALLEL
AND PERPENDICULAR LINES
Write the equation in both slope-intercept and
standard form of the line that passes through
the point (3, 5) and satisfies the given condition.
(b) perpendicular to the line 2x + 5y = 4
Solution
y 5 
slope-intercept
form
standard
form
y 
5
2
5
x 
x 
15
2
5
2
2
2y  5 x  5
Distributive property
Add 5 = 10/2.
Multiply by 2.
5 x  2 y  5 Subtract 2y, add 5, and rewrite.
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Equation
y = mx +b
Description
Slope-Intercept Form
Slope is m.
y-intercept is b.
y – y1 = m(x – x1) Point-Slope Form
Slope is m.
Line passes through (x1, y1)
When to Use
Slope and y-intercept
easily identified and
used to quickly graph
the equation. Also
used to find the
equation of a line
given a point and the
slope.
Ideal for finding the
equation of a line if
the slope and a point
on the line or two
points on the line are
known.
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Equation
Ax + By = C
Description
Standard Form
(If the coefficients and
constant are rational, then A,
B, and C are expressed as
relatively prime integers, with
A ≥ 0).
S lo p e is 
A
B
x -in te rce p t is
y -in te rce p t is
When to Use
The x- and yintercepts can be
found quickly and
used to graph the
equation. The slope
must be calculated.
( B  0 ).
C
A
C
B
( A  0 ).
( B  0 ).
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Equation
Description
When to Use
y=b
Horizontal Line
Slope is 0.
y-intercept is b.
If the graph intersects
only the y-axis, then y
is the only variable in
the equation.
x=a
Vertical Line
Slope is undefined.
x-intercept is a.
If the graph intersects
only the x-axis, then x
is the only variable in
the equation.
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Example 7
FINDING AN EQUATION OF A LINE
THAT MODELS DATA
Average annual tuition and fees for in-state
students at public four-year colleges are
shown in the table for selected years and
graphed as ordered pairs of points where
x = 0 represents 2005, x = 1 represents 2006,
and so on, and y represents the cost in
dollars. This graph of ordered pairs of data is
called a scatter diagram.
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FINDING AN EQUATION OF A LINE
THAT MODELS DATA
Example 7
Year
2005
Cost
(in dollars)
5492
2006
5804
2007
6191
2008
6591
2009
7050
2010
7605
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FINDING AN EQUATION OF A LINE
THAT MODELS DATA
(a) Find an equation that
models the data.
Solution The points
lie approximately on a
straight line. Write a
linear equation that
models the
relationship between
year x and cost y.
Choose two data
points, (0, 5492) and
(5, 7605) to find the
slope of the line.
Example 7
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FINDING AN EQUATION OF A LINE
THAT MODELS DATA
(a) Find an equation that
models the data.
Example 7
Solution
m 

7605  5492
50
2113
5
 4 2 2 .6
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FINDING AN EQUATION OF A LINE
THAT MODELS DATA
(a) Find an equation that
models the data.
Example 7
Solution The slope
422.6 indicates that
the cost of tuition and
fees increased by
about $423 per year.
Use this slope, the
y-intercept 5492, and
the slope-intercept
form to write the
equation of the line.
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FINDING AN EQUATION OF A LINE
THAT MODELS DATA
(a) Find an equation that
models the data.
Example 7
Solution
y  mx  b
y  4 2 2 .6 x  5 4 9 2
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FINDING AN EQUATION OF A LINE
THAT MODELS DATA
(b) Use the equation
from part (a) to
predict the cost of
tuition and fees at
public 4-year colleges
in 2012.
Solution The value
x = 7 corresponds to
the year 2012, so
we substitute 7 for
x.
Example 7
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FINDING AN EQUATION OF A LINE
THAT MODELS DATA
(b) Use the equation
from part (a) to
predict the cost of
tuition and fees at
public 4-year colleges
in 2012.
Solution
y  4 2 2 .6 x  5 4 9 2
Example 7
Model from part (a)
y  4 2 2 .6  7   5 4 9 2
y  8 4 5 0 .2
Let x = 7.
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FINDING AN EQUATION OF A LINE
THAT MODELS DATA
(b) Use the equation
from part (a) to
predict the cost of
tuition and fees at
public 4-year colleges
in 2012.
Solution The model
predicts that average
tuition and fees for
in-state students at
public four-year
colleges in 2012 would
be about $8450.
Example 7
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Guidelines for Modeling
Step 1 Make a scatter diagram of the data.
Step 2 Find an equation that models the
data. For a line, this involves
selecting two data points and
finding the equation of the line
through them.
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Example 8
SOLVING AN EQUATION WITH A
GRAPHING CALCULATOR
Use a graphing calculator to solve
2 x  4 2  x   3 x  4.
Solution
We write an equivalent equation with 0 on
one side.
2 x  4 2  x   3 x  4  0
Subtract 3x and 4.
Then we graph Y = –2X – 4(2 – X) – 3X – 4
to find the x-intercept. The standard viewing
window cannot be used because the
x-intercept does not lie in the interval [–10,10].
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Example 8
SOLVING AN EQUATION WITH A
GRAPHING CALCULATOR
Use a graphing calculator to solve
2 x  4 2  x   3 x  4.
Solution
As seen in the figure,
the x-intercept of the
graph is –12, and thus
the solution of the
equation is –12. (–12
is the zero of the
function Y.) The
solution set is {–12}.
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