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Math Help - Sketch a graph for all (x,y) that are related to the point (1,0)

  1. #1
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    Sketch a graph for all (x,y) that are related to the point (1,0)

    Consider the real plane, ℝ. Define the relation S on ℝ by:

    (x,y)S(u,v) when x + y = u + v


    (a) Sketch a graph for all (x,y) that are related to the point (1,0)

    My attempt:

    I think the answer is x=1 since the graph of x=1 intersects the point (1,0) and all circles with a radius of 1 or higher.

    Or do I need to sketch a graph that intersects (1,0) and ALL circles of any given radius.

    Or it something more simple that i'm not thinking about

    Thanks!
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  2. #2
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    Re: Sketch a graph for all (x,y) that are related to the point (1,0)

    Quote Originally Posted by yoman360 View Post
    Consider the real plane, ℝ. Define the relation S on ℝ by: (x,y)S(u,v) when x + y = u + v
    (a) Sketch a graph for all (x,y) that are related to the point (1,0)
    Question: Is it true that (1,0)\mathcal{S}(0,1) true?

    What about (1,0)\mathcal{S}\left( {\tfrac{{\sqrt 2 }}{2}, - \tfrac{{\sqrt 2 }}{2}} \right)~?.

    And if a^2+b^2=1, is it true that (1,0)\mathcal{S}(a,b) true?
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  3. #3
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    Re: Sketch a graph for all (x,y) that are related to the point (1,0)

    Quote Originally Posted by Plato View Post
    Question: Is it true that (1,0)\mathcal{S}(0,1) true?
    yes, because 1 + 0 = 0 + 1 ==> 1 = 1


    Quote Originally Posted by Plato View Post
    What about (1,0)\mathcal{S}\left( {\tfrac{{\sqrt 2 }}{2}, - \tfrac{{\sqrt 2 }}{2}} \right)~?.
    No because 1 = 1

    Quote Originally Posted by Plato View Post
    And if a^2+b^2=1, is it true that (1,0)\mathcal{S}(a,b) true?
    oh i see so the answer is the sketch of the graph x+y = 1
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