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Math Help - How Would I Express This Equation Using the Distance Formula?

  1. #1
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    How Would I Express This Equation Using the Distance Formula?

    For each point, its distance from the fixed point (-3,0) is twice its distance from the fixed point (3,0).

    My book gives me the simplified answer, but I can't figure out how to express it in the distance formula to show my work.

    Also, if someone's kind enough, please express the following as well: "For each point, its distance from the fixed point (4,3) is 3 times its distance from the fixed point (-1,2)."

    FYI, this isn't my homework. I just need to know how to do this for an upcoming test.

    Oh, and an explanation of where to plug in numbers would be greatly appreciated.
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  2. #2
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    Quote Originally Posted by Noraowolf1 View Post
    For each point, its distance from the fixed point (-3,0) is twice its distance from the fixed point (3,0).

    My book gives me the simplified answer, but I can't figure out how to express it in the distance formula to show my work.

    Also, if someone's kind enough, please express the following as well: "For each point, its distance from the fixed point (4,3) is 3 times its distance from the fixed point (-1,2)."

    FYI, this isn't my homework. I just need to know how to do this for an upcoming test.

    Oh, and an explanation of where to plug in numbers would be greatly appreciated.
    I can see it now...

    Syllybus for Pre-Algebra:
    1) How to do stuff for tests.
    2) Where to plug stuff in.

    That is a fine course.

    Distance from (-3,0) is \sqrt{(x-(-3))^{2}+(y-0)^{2}}
    Simplify a bit \sqrt{(x+3)^{2}+y^{2}}

    Distance from (3,0) is \sqrt{(x-3)^{2}+(y-0)^{2}}
    Simplify a bit \sqrt{(x-3)^{2}+y^{2}}

    Twice the Distance
    \sqrt{(x+3)^{2}+y^{2}} = 2\cdot \sqrt{(x-3)^{2}+y^{2}}

    Everything is positive, so there is no harm in squaring both sides.
    (x+3)^{2}+y^{2} = 4\cdot ((x-3)^{2}+y^{2})

    Now what?
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  3. #3
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    Quote Originally Posted by TKHunny View Post
    I can see it now...

    Syllybus for Pre-Algebra:
    1) How to do stuff for tests.
    2) Where to plug stuff in.

    That is a fine course.

    Distance from (-3,0) is \sqrt{(x-(-3))^{2}+(y-0)^{2}}
    Simplify a bit \sqrt{(x+3)^{2}+y^{2}}

    Distance from (3,0) is \sqrt{(x-3)^{2}+(y-0)^{2}}
    Simplify a bit \sqrt{(x-3)^{2}+y^{2}}

    Twice the Distance
    \sqrt{(x+3)^{2}+y^{2}} = 2\cdot \sqrt{(x-3)^{2}+y^{2}}

    Everything is positive, so there is no harm in squaring both sides.
    (x+3)^{2}+y^{2} = 4\cdot ((x-3)^{2}+y^{2})

    Now what?
    Well, my book says the simplified answer is x^2+y^2-10x+9=0

    Oh, and it's actually Advanced Algebra II. Not Pre-Alg.
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  4. #4
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    Sweet! Take my final form and use your best algebra skills to get it to the book's form.
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