# Sketching of graph

## Homework Statement

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

## The Attempt at a Solution

How?? I have totally no idea.

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Mentallic
Homework Helper
Expand, simplify and complete the square. Now think about what the solutions are for $a^2+b^2=0$ for all real numbers a and b, and see if you can apply this idea to your problem.

LCKurtz
Homework Helper
Gold Member
Expand, simplify and complete the square. Now think about what the solutions are for $a^2+b^2=0$ for all real numbers a and b, and see if you can apply this idea to your problem.
For that matter, think about $a^2+b^2=0$ before you expand the terms.

Mentallic
Homework Helper
For that matter, think about $a^2+b^2=0$ before you expand the terms.
Haha that's a much more effective method

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?

LCKurtz
Homework Helper
Gold Member
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.

Ok. So, (0,-3) is the only value? Is the graph just a dot?

Thanks.

LCKurtz
Homework Helper
Gold Member
Ok. So, (0,-3) is the only value? Is the graph just a dot?

Thanks.
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.