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Locus of a point

  1. Jul 20, 2009 #1
    1. Find the equation of the locus of a point P which is equidistant from the y-axis and the point (3,-1)


    2. I think that I need to use

    SqRoot (x-x)sq + (y-y)sq = x
    SqRoot (x-3)sq + (y+1)sq = x

    Expanding is where I get stuck



    Cheers
     
  2. jcsd
  3. Jul 20, 2009 #2

    statdad

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    Homework Helper

    If [tex] (a,b) [/tex] is the point, you need work with

    Distance from [tex] (a,b) [/tex] to [tex] (3,-1) [/tex] = distance from [tex] (a,b) [/tex] to the [tex]y-[/tex] axis. Part of this equation is

    [tex]
    \sqrt{(a-0)^2 + (b-b)^2} = \sqrt{a^2} = |a|
    [/tex]

    What is the other part? remember your solution will be an equation, not a single number.
     
  4. Jul 20, 2009 #3
    Hi there
    There is no other part to this question. I think the equation should be y=x Sq + 3

    Cheers (very confused).
     
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