Section 1.4 Lines - Montgomery Collegefaculty.montgomerycollege.edu/maronne/Ma180/power... · Find an equation for the line that contains the point ( 1,3) and is perpendicular to

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

−


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.








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)





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


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




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− = − +







−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= −




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




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


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





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 = −


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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Section 1.4 Lines - Montgomery Collegefaculty.montgomerycollege.edu/maronne/Ma180/power... · Find an equation for the line that contains the point ( 1,3) and is perpendicular to