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Math Help - Scalar Equations

  1. #1
    Senior Member
    Joined
    Nov 2008
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    425

    Scalar Equations

    Find the scalar equation of l1 in each case:

    a) with point (2,-3) and parallel to l2: (x,y)=(1,3)+t(5,2)
    b) with y intercept -2 and perpendicular to l2: (x,y)=(4,-1)+t(6,-5)

    can someone check my work? let me know if i have my answers correct or wrong.

    a) 5x+2y+c=0
    5(2)+2(-3)+c=0
    c=-4

    therefore s eqtn is 5x+2y-4=0

    b) 5x+6y+c=0
    5(0)+6(-2)+c=0
    c=12

    therefore s eqtn is 5x+6y+12=0

    thanks in advance
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  2. #2
    MHF Contributor
    Joined
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    Hi

    You did the wrong way

    a) with point (2,-3) and parallel to l2: (x,y)=(1,3)+t(5,2)
    (5,2) is a direction vector of l2 and of l1 since l1 is parallel to l2
    A Cartesian equation of l1 is therefore 2x-5y+c=0
    Substituting (2,-3) gives 2x-5y-19=0

    You can check very easily
    l1 equation is (x,y)=(2,-3)+t(5,2)
    Take t=-1 => (x,y)=(-3,-5) and 2(-3)-5(-5)-19=0

    Remember that a line whose equation is ax+by+c=0 is parallel to the vector (-b,a) and perpendicular to the vector (a,b)

    b) I let you correct your answer
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