2006 AMC 10A Problem 11

Simplify the equation first, then describe every point that satisfies it.

1. Expand: $(x + y)^2 = x^2 + 2xy + y^2$.
2. The equation becomes $x^2 + 2xy + y^2 = x^2 + y^2$. Take $x^2 + y^2$ away from both sides:
$$2xy = 0, \qquad xy = 0.$$
3. A product is zero exactly when one of the factors is zero. So a point works if $x = 0$ or if $y = 0$.
4. The points with $x = 0$ form the $y$-axis, like $(0, 5)$. The points with $y = 0$ form the $x$-axis, like $(3, 0)$. Check $(3, 0)$ in the original: $(3 + 0)^2 = 9$ and $3^2 + 0^2 = 9$. It works.

So the graph is two lines that cross at the origin.

The answer is $\boxed{\textbf{(C)}\ \text{two lines}}$.

The graph of an equation is the set of all points that make it true. Simplify as much as you can, then ask "which points work?" When the simplified equation says a product equals zero, the graph is everything that makes the first factor zero together with everything that makes the second factor zero.

Expand the left side: $(x + y)^2 = x^2 + 2xy + y^2$. Cancelling $x^2 + y^2$ from both sides leaves
$$2xy = 0 \quad\Longrightarrow\quad xy = 0.$$
A product is zero exactly when a factor is zero, so the solutions are all points with $x = 0$ (the $y$-axis) together with all points with $y = 0$ (the $x$-axis). That is two lines crossing at the origin.

The answer is $\boxed{\textbf{(C)}\ \text{two lines}}$.