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

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

Consider the real plane, ℝ². Define the relation S on ℝ² by:

(x,y)S(u,v) when x² + y² = u² + v²

(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!

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

Originally Posted by yoman360
Consider the real plane, ℝ². Define the relation S on ℝ² by: (x,y)S(u,v) when x² + y² = u² + v²
(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?

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

Originally Posted by Plato
Question: Is it true that $(1,0)\mathcal{S}(0,1)$ true?
yes, because 1² + 0² = 0² + 1² ==> 1 = 1

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

Originally Posted by Plato
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 x²+y² = 1