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Math Help - Geometry/conics

  1. #1
    Member Awsom Guy's Avatar
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    Geometry/conics

    I hope this question goes here.
    Heres my question:
    The line y-2x+u=0 is 2sqrt5 units from the point (1,-3). Find the possible values of u.

    Here is my attempt:
    y-2x+u
    ((-3)-2(1)+u)/√(2^2+1^2 )=2√5
    (-5+u)/√5√5/1=(2√5)/1√5/1
    -5+u=10
    u=15 Or u=?

    I hope that makes sense, sorry about the mess I dont know how to use the program here.
    I can't find the other answer, but I know that the answer is u=15 or u=-5.
    Thanks.
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  2. #2
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    You forgot the absolute values in the perpendicular distance formula  d = \frac{|Am_1 + Bm_2 + C |}{\sqrt{A^2 + B^2}}

    So really, you need to solve  |u-5| = 10 , not just  u - 5 = 10
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  3. #3
    Member Awsom Guy's Avatar
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    Thanks

    does that mean it turns into u+5=10.
    Thanks for help.
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  4. #4
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    The absolute value means u - 5 can be a positive or negative number of the same magnitude.

    So aside from (u-5) = 10, you also have -(u-5) = 10
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  5. #5
    Member Awsom Guy's Avatar
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    thanks

    thanks, that really helped.
    I hope people continue to help.
    Thanks
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