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Math Help - Loci

  1. #1
    Member purplec16's Avatar
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    Loci

    Determine the locus of the point P(x,y) which moves so that its distance from (-3,-2) is twice its perpendicular distance from 2x+3=0
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  2. #2
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    Re: Loci

    Are you sure it's 2x + 3 = 0 and not 2x + 3 = y?
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Re: Loci

    \sqrt{(x+3)^2+(y+2)^2}=2\left|\dfrac{2x+3}{\sqrt{2  ^2+0^2}}\right| now, take squares.


    Edited: Sorry, I didn't see Prove It's post.
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  4. #4
    Member purplec16's Avatar
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    Re: Loci

    You have to use the same formula on each side this is how i did mines sorry i was unable to load it because it was thundering outside

    I first solve for x so x=-3/2 and y is just y so we have the point (-3/2,y)

    Using the distance formula

    \sqrt{(x+3)^2+(y+2)^2}=2*(\sqrt{(x+\frac{3}{2})^2+  (y-y)^2})

    x^2+6x+9+y^2+4y+4=2x^2+6x+9/4+0

    -x^2+y^2+0x+4y+17/2=0
    Last edited by Ackbeet; June 25th 2011 at 11:20 AM.
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  5. #5
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    Re: Loci

    If it is, in fact, the vertical line 3x+2= 0 or x= -2/3, then we have
    \sqrt{(x+3)^2+ (y+2)^2}= 2|x+ 3/2|

    Square both sides, combine like terms, and complete the square. I believe this is a hyperbola.
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