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Equation for set of points

  1. Sep 5, 2007 #1
    1. The problem statement, all variables and given/known data

    Determine the eqution for the set of all points (x,y) so that the distnce of (x,y) from (4,0) is twce the distance of (x,y) from (1,0). Describe the set geometrically.

    2. Relevant equations



    3. The attempt at a solution

    After looking at this for a little while I figured it would be good to set the distance from the points from x,y to 4,0 equal to twice the distance from x,y to 1,0, I plugged these into the distance formula and got:

    rad((4-x)^2 + (-y)^2) = 2*rad((1-x)^2+(-y)^2)

    I'm kind of stuck from there and I don't know if I'm on the right track or not could someone please help. Thanks.
     
    Last edited: Sep 6, 2007
  2. jcsd
  3. Sep 5, 2007 #2

    rock.freak667

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    Well that seems to be the way to do it..but maybe you'll need another equation with x and y so you can solve simultaneously and get values for x and y

    here is some help: the points (1,0) and (4,0) are on the x-axis right? maybe you could just sketch the x-axis...put a point (x,y) and the points (1,0) and (4,0).. then draw the triangle formed such that the dist. of (x,y) from (1,0) is [tex]l[/tex] and the other distance is [tex]2l]/tex] then see if you can get another equation to help you from that
     
  4. Sep 5, 2007 #3

    Dick

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    You are off to a good start. Now square both sides of your distance equation and see if you can simplify the algebra.
     
  5. Sep 5, 2007 #4

    EnumaElish

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    That would be the case if there is only one such point. Multiple points will be described as {y = f(x) such that the distance of (x,y) from (4,0) is twice the distance of (x,y) from (1,0)}. Wouldn't it?
     
  6. Sep 5, 2007 #5

    rock.freak667

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    well [tex]\sqrt{(4-x)^2 + (-y)^2} = 2\sqrt{(1-x)^2+(-y)^2} [/tex]

    squaring both sides and simplifying would give [tex](4-x)^2+y^2=4(1-x)^2+4y^2[/tex]
    giving

    [tex] (4-x)^2-4(1-x)^2-3y^2=0[/tex] which has two unknowns...either that or I keep missing something
     
  7. Sep 5, 2007 #6

    Dick

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    If you keep expanding you can see the x^2 terms will cancel too. So you can put y to be ANYTHING and get a corresponding x. There are an infinite number of solutions. The equation defines a CURVE containing an infinite number of points.
     
  8. Sep 5, 2007 #7
    I ended up getting 12 = 3x^2 +3y^2 for my equation which is half a circle.
     
  9. Sep 5, 2007 #8

    rock.freak667

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    oh..i didnt read through the question, i found the point (x,y) not describe the set geometrically
     
  10. Sep 5, 2007 #9
    I think the question says find the locus and not the equation.
     
  11. Sep 5, 2007 #10
    locus? whats that..the question says to give the equation...
     
  12. Sep 5, 2007 #11

    rock.freak667

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    Doesn't the locus just describe the equation of the circle?
     
  13. Sep 5, 2007 #12

    Dick

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    That looks good. But why do you say it's only 'half' of a circle? If you can describe the curve, that is the locus.
     
  14. Sep 5, 2007 #13
    i solved it to get y to one side and graphed it on my calculator and it was a half a circle
     
  15. Sep 5, 2007 #14

    Dick

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    In solving it you took only a plus square root. The negative is also a solution.
     
  16. Sep 5, 2007 #15
    ohhh thanks a lot that is right i totally forgot about that

    I am kind of confused about how I would go about getting it though when I had a square root on both sides of the equation I then squared the entire equation to simplify, would it be at this point that it would become + or - ?
     
    Last edited: Sep 5, 2007
  17. Sep 6, 2007 #16

    Dick

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    You are right to be cautious, but what is inside of your radicals is always non-negative. So squaring can't introduce any false roots. It's just when you took the square root of y^2.
     
  18. Sep 6, 2007 #17

    HallsofIvy

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    No, that's the wrong direction. squaring both sides of x= 2 gives x2= 4. If you start with x2= 4 and take the square root of both sides, then you get [itex]x= \pm 2[/itex]
     
  19. Sep 6, 2007 #18
    So it ends up being a full circle when you have the negative root for an answer too right?
    thanks a lot for the help :)
     
    Last edited: Sep 6, 2007
  20. Sep 10, 2007 #19
    hey i'm just making sure that it is correct that it ends up being a circle because I heard people talking in class today and they were saying they thought it was a triangle..but I'm pretty sure that it is a circle and this is correct
     
  21. Sep 10, 2007 #20
    triangle

    is this correct because I have to hand it in and people in my class were saying it should be a triangle, but I really think it should be a circle
     
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