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Math Help - Equation of a line

  1. #1
    Ash
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    Equation of a line

    I need assistance in learning the steps and answering these problems.


    #1 The equation of L1 and L2 are given. I need to write an equations of L3 and an equation of L4 according to the given conditions.


    L1: x - 2y = 6 L2: 4 - y = x

    1. L3 is parallel to L1 and has the same y-intercept as L2.




    2. L4 is perpendicular to L1 and has the same x-intercept as L2.





    #2 How do I write the equation of L1.

    The equation of L2 is x = -1. L1 is parallel to L2 and contains the point (3,5).

    Do I write x = 3, y =5 since those are the points L2 contains?
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  2. #2
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by Ash View Post
    I need assistance in learning the steps and answering these problems.


    #1 The equation of L1 and L2 are given. I need to write an equations of L3 and an equation of L4 according to the given conditions.


    L1: x - 2y = 6 L2: 4 - y = x

    1. L3 is parallel to L1 and has the same y-intercept as L2.




    2. L4 is perpendicular to L1 and has the same x-intercept as L2.





    #2 How do I write the equation of L1.

    The equation of L2 is x = -1. L1 is parallel to L2 and contains the point (3,5).

    Do I write x = 3, y =5 since those are the points L2 contains?
    L1: x - 2y = 6 L2: 4 - y = x

    First I will write L1 and L2 in slope-intercept form:
    L1: y = 1/2*x - 3
    L2: y = -x + 4

    1. L3 is parallel to L1 and has the same y-intercept as L2.
    The slope of L1 is m1 = 1/2
    The y-intercept of L2 is b = 4
    From these, L3 becomes
    y = 1/2*x + 4


    2. L4 is perpendicular to L1 and has the same x-intercept as L2.
    The slope perpendicular to L1 is m2 = -1/m1 = -2
    The x-intercept of L2 is found by setting y = 0: 0 = -x + 4 --> x = 4, which gives us the point (4,0)
    From these, L4 becomes
    y = -2x + b ... plug in (4,0)
    0 = -2(4) + b --> b = 8
    y = -2x + 8


    # The equation of L2 is x = -1. L1 is parallel to L2 and contains the point (3,5).

    L2: x = -1
    A like parallel to this has the same form as this equation: x = c
    Using this form, we can solve for c using the point given: (3,5)
    From these, L1 becomes
    x = c ... plug in (3,5)
    3 = c
    x = 3

    That is pretty straight forward. It's just a vertical line in the form x = c passing through the point (3,5). The only vertical line that passes through this point is the line x = 3.
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