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Thread: MX1 Linear

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    Junior Member supR's Avatar
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    MX1 Linear

    Hey guys, really need some help with this question!

    5F Q9 from Cambridge MX1

    Ty!

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    Re: MX1 Linear

    Our specific example, given the lines 2x - 3y + 13 = 0 and x + y - 1 = 0, we know that the line that passes through their point of intersection must be given by:

    a) (2x - 3y + 13) + k(x + y - 1) = 0

    b) (2x - 3y + 13) + k(x + y - 1) = (2 + k)x - (3 - k)y + (13 - k) = 0 from which we can easily form an expression for the gradient as m = (2 + k)/(3 - k)

    Since it must be parallel to: 4x+3y-1=0
    then m = -4/3


    So,
    m = (2+k)/(3 - k)
    -4/3 = (2 + k)/(3 - k)
    k = 18

    Sub k value into (2x - 3y + 13) + k(x + y - 1)=0

    20x + 15y - 5 = 0

    Simplify 20x + 15y - 5 = 0

    20x + 15y - 5 = 0
    4x + 3y - 1 = 0

    Hence, l3
    Last edited by OkDen; 26 Aug 2017 at 9:01 PM.
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    Re: MX1 Linear

    Thank you so much!! That makes so much sense!
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    Re: MX1 Linear

    Quote Originally Posted by supR View Post
    Thank you so much!! That makes so much sense!
    No problem. You're welcome!
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