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Section 1.4
Lines
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Find the slope of the line containing the points (-1, 4) and (2, -3).
3 4 72 1 3
m − −= = −
+4 3 71 2 3
m += = −− −
The average rate of change of y with respect to x is 73
−
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Compute the slopes of the lines L1, L2, L3, and L4 containing the following pairs of points. Graph all four lines on the same set of coordinate axes.
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Graph the equation: x = –2
−4 −3 −2 −1 1 2
−1
1
2
3
4
x
y
−4 −3 −2 −1 1 2
−1
1
2
3
4
x
y
(-2,4)
(-2,2)
(-2,0)
(-2,1)
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Find the equation of a line with slope –3 and containing the point (–1, 4).
( )( )4 3 1y x− = − − −
4 3 3y x− = − −
3 1y x= − +
−4 −3 −2 −1 1 2
−1
1
2
3
4
x
y
Run = 1
Rise = -3
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Find the equation of a horizontal line containing the point (2, – 4).
( ) ( )4 0 2y x− − = ⋅ −
4 0y + =
4y = −
−3 −2 −1 1 2 3
−4
−3
−2
−1
1
x
y
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Find an equation of the line L containing the points (–1, 4) and (3, –1). Graph the line L.
−5 −4 −3 −2 −1 1 2 3 4 5
−5
−4
−3
−2
−1
1
2
3
4
5
(-1, 4)
(3, -1)
( )1 4
3 1m − −=
− −54
= −( )( )54 1
4y x− = − − −
( )54 14
y x− = − +
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−5 −4 −3 −2 −1 1 2 3 4 5
−5
−4
−3
−2
−1
1
2
3
4
5
(0, -3)
(2, 0)
Find the slope m and y-intercept b of the equation 3x – 2y = 6. Graph the equation.
3 32
y x= −
3
2
3x – 2y = 6
– 2y = –3x+6
3 32
y x= −
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Graph the linear equation 3x + 2y = 6 by finding its intercepts.
The x-intercept is at the point (2, 0)
The y-intercept is at the point (0, 3) −5 −4 −3 −2 −1 1 2 3 4 5
−5
−4
−3
−2
−1
1
2
3
4
5
(0, 3)
(2, 0)
3x + 2(0) = 6
3x = 6
x = 2
3(0) + 2y = 6
2y = 6
y = 3
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1 : 3 2 12L x y− + = 2 : 6 4 0L x y− =
2 3 12y x= +3 62
y x= +
3Slope ; -intercept 62
y= =
4 6y x− = −32
y x=
3Slope ; -intercept 02
y= =
−6 −5 −4 −3 −2 −1 1 2 3
−2
−1
1
2
3
4
5
6
x
y
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Find an equation for the line that contains the point ( 1,3) and is parallel to the lin 3 4 1e .2x y =
−−
4 3 12.y x− = − +
3 34
y x= −
3So a line parallel to this one would have a slope of .4
( )1 1y y m x x− = −
( )33 ( 1)4
y x− = − −
3 154 4
y x= +−8 −6 −4 −2 2 4 6
−6
−4
−2
2
4
6
8
x
y
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Find the slope of a line perpendicular to a line with slope . 34
perpendicular43
m = −
−5 −4 −3 −2 −1 1 2 3 4 5
−5
−4
−3
−2
−1
1
2
3
4
5
134
m =
243
m = −
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Find an equation for the line that contains the point ( 1,3) and is perpendicular to the li 2ne 3 1 .4x y− =
−
4 3 12.y x− = − +
3 34
y x= −
4So a line perpendicular to this one would have a slope of .3
−
( )1 1y y m x x− = −
( )43 ( 1)3
y x− = − − −
4 53 3
y x= − +−8 −6 −4 −2 2 4 6
−6
−4
−2
2
4
6
8
x
y
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