Section 1.3 lines

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Section 1.3
Lines

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Find the slope of the line containing the points (-1, 4) and (2, -3).
3 4 7
2 1 3
m
− −
= = −
+
4 3 7
1 2 3
m
+
= = −
− −
The average rate of change of y with respect to x is
7
3
−

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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 1
m
− −
=
− −
5
4
= −
( )( )5
4 1
4
y x− = − − −
( )
5
4 1
4
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
3
2
y x= −
3
2
3x – 2y = 6
– 2y = −3x+6
3
3
2
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
6
2
y x= +
3
Slope ; -intercept 6
2
y= =
4 6y x− = −
3
2
y x=
3
Slope ; -intercept 0
2
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
3
4
y x= −
3
So a line parallel to this one would have a slope of .
4
( )1 1y y m x x− = −
( )
3
3 ( 1)
4
y x− = − −
3 15
4 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 .
3
4
perpendicular
4
3
m = −
−5 −4 −3 −2 −1 1 2 3 4 5
−5
−4
−3
−2
−1
1
2
3
4
5
1
3
4
m =
2
4
3
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
3
4
y x= −
4
So a line perpendicular to this one would have a slope of .
3
−
( )1 1y y m x x− = −
( )
4
3 ( 1)
3
y x− = − − −
4 5
3 3
y x= − +−8 −6 −4 −2 2 4 6
−6
−4
−2
2
4
6
8
x
y