AMC 10 · 2006 · #11
Grade 8 geometry-2dWhich of the following describes the graph of the equation (x+y)2=x2+y2?
Pick an answer.
AMC 10 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: We are given the equation $(x+y)^2 = x^2 + y^2$ and asked what its graph looks like in the coordinate plane. The graph is the set of all points $(x, y)$ that make the equation true. We must decide which description fits that set: the empty set, one point, two lines, a circle, or the entire plane.
Givens: The equation is $(x+y)^2 = x^2 + y^2$; $x$ and $y$ range over all real numbers; Answer choices describe the graph: (A) the empty set, (B) one point, (C) two lines, (D) a circle, (E) the entire plane
Unknowns: Which shape the set of points $(x, y)$ satisfying the equation forms
Understand
Restated: We are given the equation $(x+y)^2 = x^2 + y^2$ and asked what its graph looks like in the coordinate plane. The graph is the set of all points $(x, y)$ that make the equation true. We must decide which description fits that set: the empty set, one point, two lines, a circle, or the entire plane.
Givens: The equation is $(x+y)^2 = x^2 + y^2$; $x$ and $y$ range over all real numbers; Answer choices describe the graph: (A) the empty set, (B) one point, (C) two lines, (D) a circle, (E) the entire plane
Plan
Primary tool: #15 Organize Information in More Ways
Secondary: #3 Eliminate Possibilities, #1 Draw a Diagram
The equation looks tangled, but Tool #15 (Organize Information in More Ways) says to rewrite it into a form you already recognize. Expanding the squared side and cancelling the matching pieces turns the whole thing into a tiny statement, $xy = 0$, whose graph is easy to name. Tool #3 (Eliminate Possibilities) then uses the multiple-choice setup: once you know the graph is unbounded, a circle and a single point are out, and a quick test point rules out the empty set and the whole plane. Tool #1 (Draw a Diagram) confirms the survivors as two familiar lines. The whole difficulty melts once the equation is rewritten.
Execute — Answer: C
6.EE.A.3 Step 1 Expand the squared side
- The only messy part is $(x+y)^2$.
- Multiply it out using the rule $(x+y)^2 = x^2 + 2xy + y^2$.
- Now the equation reads $x^2 + 2xy + y^2 = x^2 + y^2$.
- Writing the equation this way lines up matching terms on the two sides.
💡 Opening the square exposes the extra middle term $2xy$ that the right side does not have.
6.EE.A.3 Step 2 Cancel and simplify
- Both sides carry an $x^2$ and a $y^2$.
- Subtract those from each side and they disappear, leaving only $2xy = 0$.
- Dividing by $2$ gives the clean condition $xy = 0$.
- This single equation says everything the original one did.
💡 The identical squares on both sides carry no information, so removing them leaves just the cross term.
6.EE.B.5 Step 3 Use the zero-product rule
- A product of two numbers equals zero only when at least one of the factors is zero.
- So $xy = 0$ happens exactly when $x = 0$ or $y = 0$.
- Every point on the graph satisfies one of these two conditions, and every point satisfying one of them is on the graph.
💡 You cannot multiply two nonzero numbers and land on zero, so one of them must be zero.
8.F.A.3 Step 4 Read off the graph
- The equation $x = 0$ is the set of all points whose first coordinate is zero — that is the vertical $y$-axis, a line.
- The equation $y = 0$ is the horizontal $x$-axis, another line.
- The graph is the two axes together: two lines.
- That matches choice $\textbf{(C)}$.
💡 Setting one coordinate to zero traces out a whole axis, and there are two of them.
6.EE.A.3 The only messy part is $(x+y)^2$. Multiply it out using the rule $(x+y)^2 = x^2 6.EE.A.3 Both sides carry an $x^2$ and a $y^2$. Subtract those from each side and they di 6.EE.B.5 A product of two numbers equals zero only when at least one of the factors is ze 8.F.A.3 The equation $x = 0$ is the set of all points whose first coordinate is zero — t Review
Reasonableness: Test some points against the original equation. At $(2, 0)$: left side $(2+0)^2 = 4$, right side $2^2 + 0^2 = 4$ — equal, and this point is on the $x$-axis. At $(0, 3)$: left side $9$, right side $9$ — equal, and it sits on the $y$-axis. At $(1, 1)$, an off-axis point: left side $(1+1)^2 = 4$, right side $1 + 1 = 2$ — not equal, so it is correctly excluded. Points work exactly when they lie on an axis, so the graph is the two axes: unbounded, so not a circle (choice D) and not a single point (choice B); a test point works, so not empty (choice A); and $(1,1)$ fails, so not the whole plane (choice E). Choice (C), two lines, is the only fit.
Alternative: Factor instead of describing points. From $2xy = 0$ you get $xy = 0$, which factors the graph directly into the union of the solution set of $x = 0$ and the solution set of $y = 0$. Each factor set is a straight line, and a union of two distinct lines is exactly a 'two lines' graph — no point-testing needed.
CCSS standards used (min grade 8)
6.EE.A.3Apply the properties of operations to generate equivalent expressions (Expanding $(x+y)^2$ into $x^2 + 2xy + y^2$ and cancelling the matching $x^2$ and $y^2$ terms to reach $xy = 0$.)6.EE.B.5Understand solving an equation or inequality as a process of finding values (Reading $xy = 0$ as a condition on points and using the zero-product rule to split it into $x = 0$ or $y = 0$.)8.F.A.3Interpret the equation y = mx + b as defining a linear function (Recognizing that each of $x = 0$ and $y = 0$ graphs as a straight line (the two coordinate axes), so the graph is two lines.)
⭐ Expand the square, cancel the matching pieces, and the whole equation shrinks to $xy = 0$ — a product is zero only when a factor is zero, so the graph is the two axis lines.
⭐ Expand the square, cancel the matching pieces, and the whole equation shrinks to $xy = 0$ — a product is zero only when a factor is zero, so the graph is the two axis lines.
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