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Math Help - Standard Form of an Equation

  1. #1
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    Standard Form of an Equation

    Write in standard form, the equation with integral coefficients for the set of coplanar points described. For each point, its distance from the fixed point (-1,-2) is \frac{1}{4} its distance from the fixed point (6,-4).

    Is this answer right - 15x^2 + 15y^2 + 44x + 56y + 28 = 0 ?
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  2. #2
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    Let the point be (x,y).

    We can use the distance formula.

    \sqrt{(x+1)^{2}+(y+2)^{2}}=\frac{1}{4}\sqrt{(x-6)^{2}+(y+4)^{2}}
    Last edited by galactus; March 17th 2008 at 02:41 PM.
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  3. #3
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    Quote Originally Posted by galactus View Post
    Let the point be (x,y).

    We can use the distance formula.

    (x+1)^{2}+(y+2)^{2}=\frac{1}{4}\left[(x-6)^{2}+(y+4)^{2}\right]
    Should't the equation be (x+1)^{2}+(y+2)^{2}=\frac{1}{16}\left[(x-6)^{2}+(y+4)^{2}\right] instead. After your square both sides to get rid of the radical you also square \frac {1}{4} to get \frac {1}{16} and from there on solve. So is my answer right in my 1st post?
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  4. #4
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    Yes, you're correct. I forgot to include my radical signs.
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  5. #5
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    Quote Originally Posted by galactus View Post
    Yes, you're correct. I forgot to include my radical signs.
    Alright thnx. I just wanted to make sure I answered the equation right?
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