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Lines
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Warm-up: Class Notes Page 25
Today we are starting to learn about lines.
In honor of that, here are several one-liners:
 I wondered why the Frisbee was getting bigger, and then it hit me.
 I had a dream I was eating a giant marshmallow, and when I woke up
my pillow was missing!
 Right now I'm having amnesia and deja vu at the same time! I think
I've forgotten this before?
 How’s that new line I introduced you to? Ohh, she goes on forever.
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Find points on a graph
(−2,3)
4
x coordinate first
y coordinate second
3
2
(2,1)
(x , y)
1
0
-3
-2
-1-1 0
-2
-3
-4
1
2
3
4
5
6
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Finding intercepts on a graph
4
3
2
y-intercept
where line touches y-axis
(0,2)
1
(4,0)
0
-3
-2
-1-1 0
-2
-3
-4
1
2
3
4
5
6
x-intercept
where line touches x-axis
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Find slope from a graph
4
slope 
3
rise

run
2
2
1
Rise
0
-3
-2
-1-1 0
1
2
3
-2
-3
-4
Run
4
5
1
6
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Slope formula
The slope of a line passing through two distinct points
m 
y 2  y1
x 2  x1
(x1, y1) and (x2, y2) is given by the formula____________
Exception: The slope is undefined if x1=x2.
Extra 1: Find the slope of the line passing through the points
(−2, −3) and (−4, 5).
(x1, y1) and (x2, y2)
m
y 2  y1
x 2  x1

( 5 )  ( 3 )
( 4 )  ( 2 )

8
2
 4
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Slope-intercept form
The slope-intercept form of the equation of a line with slope m
y

mx

b
and y-intercept (0, b) is given by the formula _____________
2.3.4 Find the slope-intercept form of the equation of the line
with y-intercept (0, −8) and having slope m=2/5.
y 
2
5
x ( 8 )
Solution
y
2
5
x8
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Extra 2: Slope-intercept form
Find the slope-intercept form of the equation of the line displayed
in the graph below.
5
y-intercept
(0, 3)
(x1, y1)
m
4
(2, 4)
(x2, y2)
3
2
y 2  y1
x 2  x1

m 
1
1
2
3
4
-2
-3
-4
(2)  (0 )
1
2
0
-5 -4 -3 -2 -1-1 0
( 4 )  (3)
Solution
5
y 
1
2
x 3
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2.3.4 Sketch a graph
Sketch a graph of the line y=2/5x+(−8).
3
2
y-intercept: (0, −8)
1
0
2
-1 0 1 2 3 4 5 6 7 8 9 10
-5 -4 -3 -2 -1
Slope: m 
-2
5
-3
-4
-5
-6
Two points define a line.
-7
2
-8
-9
5
-10
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2.3.4 Identifying intercepts
Find the y-intercept and the x-intercept of the line y=2/5x−8.
To find the y-intercept,
set x=0 and solve for y.
Setting x equal to zero
To find the x-intercept,
set y=0 and solve for x.
Setting y equal to zero
y  2 /5 0  8
Solution
y  8
y-intercept: ( 0 ,  8 )
0  2 / 5x  8
Solution
x  20
x-intercept: ( 20 , 0 )
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Point-slope form
The point-slope form of the equation of a line passing through the
point (x1, y1) and having slope m is given by the formula
y

y

m
(
x

x
)
1
1
______________________
Extra 3: Find the point-slope form of the equation of the line
passing through the point (− 3,2) with slope m= −2.
(x1, y1)
y  2   2 ( x  ( 3 ) )
Solution
y  2   2 ( x  3)
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Extra 4: Point-Slope form
Find the point-slope form of the equation of the line passing
through the points (−1, 3) and (1, 4).
(x1, y1) and (x2, y2)
First find the slope.
m
y 2  y1
x 2  x1
y  3  1 2 ( x  ( 1))
Solution y  3 
1
2
( x  1)

( 4 )  (3)
(1)  ( 1)

1
2
y  ( 4 )  1 2 ( x  (1) )
Solution y  4 
1
2
( x  1)
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2.3.2: Point-slope form
Find the point-slope form of the equation of the line displayed
in the graph below.
6
-6
(−4, 5)
(x1, y1)
-5
-4
-3
(−2, −3)
(x2, y2)
-2
5
4
3
2
1
0
-1-1 0
-2
-3
-4
-5
First find the slope.
m
1
( 3 )  ( 5 )
( 2 )  ( 4 )

8
2
 4
2
Solution y  5   4 ( x  4)
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Extra 3: Sketch a graph
Sketch a graph of the line y−2= −2(x+3).
5
4
1
Point
Point: (−3, 2)
3
2
Slope:
21
m  2
0
-5 -4 -3 -2 -1-1 0
-2
-3
-4
-5
1
2
3
4
5
Two points define a line.
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2.3.2 Identifying Intercepts
Find the y-intercept and the x-intercept of the line y−5= −4(x+4).
To find the y-intercept,
set x=0 and solve for y.
Setting x equal to zero
y  5   4(0  4)
Solution
y   11
y-intercept:
( 0 ,  11 )
To find the x-intercept,
set y=0 and solve for x.
Setting y equal to zero
0  5  4( x  4)
Solution
x   11 4
x-intercept: ( 11 4 , 0 )
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Point-slope to Slope-intercept form
2.3.2 Write the line y−5=−4(x+4) in slope-intercept form.
Starting Equation
y  5  4( x  4)
Distribute
right hand side
y  5   4 x  16
Slope-intercept form
5
5
y   4 x  11
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Standard form of a line
The standard from of the equation of a line is given by the formula
Ax

By

C
__________________ , where A and B are not both zero.
2.3.1 Write x/3 − 4y +1 =0 in standard form such that
A, B, C are integers.
Subtract 1 from both sides
Multiply both sides by 3
x
 4 y  1
3
x  12 y   3
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2.3.1 Identifying Intercepts
Find the y-intercept and the x-intercept of the line x−12y= −3.
To find the y-intercept,
set x=0 and solve for y.
Setting x equal to zero
0  12 y   3
Solution
y  1/ 4
y-intercept:
To find the x-intercept,
set y=0 and solve for x.
Setting y equal to zero
x  12  0   3
Solution
x  3
( 0 , 1 4 ) x-intercept: (  3 , 0 )
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2.3.1 Sketch a graph
Sketch a graph of the line x−12y= −3.
2
x-intercept: (−3, 0)
1.5
x-intercept
y-intercept: (0, 1/4)
y-intercept
1
0.5
0
-5 -4 -3 -2 -1
-0.5
-1
-1.5
-2
0
1
2
3
4
5
Two points define a line.
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Standard form
Every line can be written in standard form. This is not true of the
other two forms. In particular, verticals lines can only be written in
standard form, because their slope is undefined.
It is difficult to get the equation of a line directly into standard
form given a graph or “usual” information. Instead, you will need
to be able to get the equations of lines into and out of this form.
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Point-slope to Standard form
To get the equation of a line from point-slope form into standard
form, first convert it into slope-intercept form as we have already
learned. Then proceed as on the next slide to convert the equation
into standard form.
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Slope-intercept to Standard form
2.3.4 Write the line y=2/5x−8 in standard form such that
A, B, and C are integers.
y  2 5x8
Starting Equation
2 5x
2 5x
 2 5 x  y  8
Multiply both sides
by 5
5  2 5 x  y   5  8 
Distribute the 5
Alternative Standard form
 2 x  5 y   40
2 x  5 y  40
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Standard to Slope-intercept to form
2.3.1 Write the line x−12y = −3 in slope-intercept form.
Starting Equation
Subtract x
from both sides
Divide by −12
Slope-intercept form
x  12 y   3
x
x
 12 y   x  3
y 
y 
 x3
 12
1
12
x 
1
4
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Standard to Slope-intercept form
One reason for converting the equation of a line from
Standard form to slope-intercept form is to find the slope
of the line. It also helps if you are trying to graph the line.
2.3.1 Find the slope of the x−12y = −3.
Slope-intercept form
y 
1
12
Slope:
m 
1
12
x 
1
4
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Turtle
Sign language for turtle
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Horizontal lines
A horizontal line passing through the point (a, b) is given
y b
by the equation ________.
m  0
The slope of a horizontal line is _________.
2.3.5 Write the equation of the horizontal line passing
through the point (−9/2, 15/2).
(a , b)
Solution
y 
15
2
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Extra 5: Horizontal lines
Find the equation of the horizontal line displayed below.
5
4
(2, 3)
(a, b)
3
2
Solution
1
0
-5 -4 -3 -2 -1-1 0
-2
-3
-4
-5
1
2
3
4
5
y  3
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2.3.5 Sketch a graph
Sketch a graph of the line y=15/2.
(−9/2, 15/2)
9
Slope:
8
7
6
5
4
3
2
1
0
-5 -4 -3 -2 -1-1 0
-2
1
2
3
4
5
m  0
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2.3.5 Identifying intercepts
Find the y-intercept and the x-intercept of the line y=15/2.
To find the y-intercept,
set x=0 and solve for y.
Setting x equal to zero
y  15 / 2
Solution
y-intercept:
y  15 2
To find the x-intercept,
set y=0 and solve for x.
Setting y equal to zero
0  15 / 2
¡PROBLEMA! 0  15 / 2
( 0 , 15 2 ) There is no x-intercept!
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Vertical lines
A vertical line passing through the point (a, b) is given
xa
by the equation ________.
undefined
The slope of a vertical line is _____________.
2.3.3 Write the equation of the vertical line
with x-intercept (1/3, 0).
(a , b)
Solution
x 
1
3
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Extra 6: Vertical lines
Find the equation of the vertical line displayed below.
5
4
(2, 3)
(a, b)
3
2
Solution
1
0
-5 -4 -3 -2 -1-1 0
-2
-3
-4
-5
1
2
3
4
5
x2
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2.3.3 Sketch a graph
Sketch a graph of the line x=1/3.
5
Slope: undefined
4
3
2
1
(1/3, 0)
0
-2
-1
-1 0
-2
-3
-4
-5
1
2
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2.3.3 Identifying intercepts
Find the y-intercept and the x-intercept of the line x=1/3.
To find the y-intercept,
set x=0 and solve for y.
Setting x equal to zero
To find the x-intercept,
set y=0 and solve for x.
Setting y equal to zero
x  1/ 3
0  1/3
¡PROBLEMA! 0  1 / 3
There is no y-intercept!
Solution
x-intercept:
x  1/ 3
(1 / 3 , 0 )
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Horizontal lines vs. Vertical lines
Horizontal Lines
Verticals lines
Equation
yb
xa
Slope
m  0
undefined
x-intercept
y-intercept
none
Exception: y=0
(a, 0 )
( 0 ,b )
none
Exception: y=0