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Sketching of graph

  1. Feb 12, 2010 #1
    1. The problem statement, all variables and given/known data
    Sketch (x+y+3)^2 + (x-y-3)^2 = 0


    2. Relevant equations



    3. The attempt at a solution
    How?? I have totally no idea.
     
  2. jcsd
  3. Feb 12, 2010 #2

    Mentallic

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    Expand, simplify and complete the square. Now think about what the solutions are for [itex]a^2+b^2=0[/itex] for all real numbers a and b, and see if you can apply this idea to your problem.
     
  4. Feb 12, 2010 #3

    LCKurtz

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    For that matter, think about [itex]a^2+b^2=0[/itex] before you expand the terms.
     
  5. Feb 13, 2010 #4

    Mentallic

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    Haha that's a much more effective method :smile:
     
  6. Feb 14, 2010 #5
    I tried using this method.
    Let a be x, b be y+3
    (a+b)^2 + (a-b)^2 = 0
    2a^2 + 2b^2 = 0
    (x)^2 + (y+3)^2 = 0
    But if I bring x^2 over,
    (y+3)^2 = -x^2
    I doubt there is such curve and if there is, it is undefined isn't it?

    thanks for the reply.
     
  7. Feb 14, 2010 #6

    LCKurtz

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    You are right, it isn't much of a graph. The left side of your last equation is non-negative and the right side is non-positive. The only way they can be equal is if both sides are zero. Can you find any (x,y) that does that? If so, whatever you find constitutes your graph.
     
  8. Feb 15, 2010 #7
    Ok. So, (0,-3) is the only value? Is the graph just a dot?

    Thanks.
     
  9. Feb 15, 2010 #8

    LCKurtz

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    Yes. Also note that in your original equation:

    (x+y+3)2 + (x-y-3)2 = 0

    both terms must be zero, so the intersection of the two lines:

    x+y+3=0 and x - y - 3 = 0

    is the only point which satisfies the equation. This agrees with your answer.
     
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