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Math Help - Conics, Lines questions

  1. #1
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    Conics, Lines questions

    1)Find the equation(s) of the locus of point P that is a constant distance of 4 units from the locus defined by y=2x+1.

    No idea how to go about doing this

    2) Find the equation of the locus of a point P that is equidistant from A(-4,4 and B(4,2).

    same for this question
    plz help me.
    THX in advance
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  2. #2
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    Quote Originally Posted by Solid8Snake View Post
    1)Find the equation(s) of the locus of point P that is a constant distance of 4 units from the locus defined by y=2x+1.

    No idea how to go about doing this

    2) Find the equation of the locus of a point P that is equidistant from A(-4,4 and B(4,2).

    same for this question
    plz help me.
    THX in advance
    to #1:

    The locus consists of 2 parallels to the given line with a distance of 4.

    The equation of the given line is: -2x+y-1=0

    Now use the formula to calculate the distance of a point P(s, t) to a straight line:

    d = \dfrac{As + Bt + C}{\sqrt{A^2+B^2}}

    The distance of point P(x, y) to the given line is therefore:

    d = 4 = \dfrac{-2x + y - 1}{\sqrt{2^2+1^2}}~\implies~y = -2x+1 \pm 4\sqrt{5}


    to #2

    The locus is the perpendicular bisector of the line segment AB.

    Let P have the coordinates (x, y). Calculate the distance AP and the distance BP. Both must be equal. Solve for y.
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  3. #3
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    I don't quite understand how you +-4root5 on the either side, care to explain?
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  4. #4
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    Quote Originally Posted by Solid8Snake View Post
    I don't quite understand how you +-4root5 on the either side, care to explain?
    The distance of point P(x, y) to the given line is therefore:

    4 = \dfrac{-2x + y - 1}{\sqrt{2^2+1^2}}~\implies~4 = \dfrac{-2x + y - 1}{\sqrt{5}}

    Now multiply both sides of the equation by \sqrt{5} and solve for y.
    You'll get the result I've given in my previous post.
    Attached Thumbnails Attached Thumbnails Conics, Lines questions-gerade_parallglabstd.png  
    Last edited by earboth; May 18th 2009 at 11:44 AM.
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  5. #5
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    Thank you very much!
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