AMC 10 · 2014 · #14
Grade 8 geometry-2dThe y-intercepts, P and Q, of two perpendicular lines intersecting at the point A(6,8) have a sum of zero. What is the area of △APQ?
Pick an answer.
AMC 10 2014 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: Two perpendicular lines cross at $A(6,8)$. One hits the $y$-axis at $P$, the other at $Q$, and the two $y$-intercepts add up to $0$. Find the area of $\triangle APQ$.
Givens: Both lines pass through $A(6,8)$; The two lines are perpendicular to each other; $P$ and $Q$ are their $y$-intercepts, so both sit on the $y$-axis; The $y$-coordinates of $P$ and $Q$ sum to $0$; Answer choices: (A) $45$, (B) $48$, (C) $54$, (D) $60$, (E) $72$
Unknowns: The area of triangle $APQ$
Understand
Restated: Two perpendicular lines cross at $A(6,8)$. One hits the $y$-axis at $P$, the other at $Q$, and the two $y$-intercepts add up to $0$. Find the area of $\triangle APQ$.
Givens: Both lines pass through $A(6,8)$; The two lines are perpendicular to each other; $P$ and $Q$ are their $y$-intercepts, so both sit on the $y$-axis; The $y$-coordinates of $P$ and $Q$ sum to $0$; Answer choices: (A) $45$, (B) $48$, (C) $54$, (D) $60$, (E) $72$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #1 Draw a Diagram, #7 Identify Subproblems
The whole puzzle hinges on the unknown height of the intercepts, so Tool #4 (Introduce a Variable) lets me call the top intercept $b$ and the bottom one $-b$ — the "sum is zero" condition handed me that pairing for free. Tool #1 (Draw a Diagram) turns the words into points on a grid so I can see the right angle at $A$. Tool #7 (Identify Subproblems) splits the work into two clean pieces: first pin down $b$ using the right angle, then compute the area from the base and height.
Execute — Answer: D
6.NS.C.8 Step 1 Place the points on a grid
- Put $A$ at $(6,8)$.
- Both intercepts lie on the $y$-axis, so they look like $(0,\text{height})$.
- Because the two heights add to $0$, they are opposites: name the top one $P=(0,b)$ and the bottom one $Q=(0,-b)$.
- So $P$ and $Q$ are mirror images across the origin.
💡 Two numbers that add to zero are the same distance from $0$, one up and one down.
8.G.B.8 Step 2 Write the three side lengths
- Use the distance between two points for each side of $\triangle APQ$.
- From $A=(6,8)$ to $P=(0,b)$ the legs are $6$ across and $8-b$ up, so $AP^2=6^2+(8-b)^2$.
- To $Q=(0,-b)$ they are $6$ across and $8+b$ up, so $AQ^2=6^2+(8+b)^2$.
- The side $PQ$ runs straight down the $y$-axis from $(0,b)$ to $(0,-b)$, a length of $2b$, so $PQ^2=(2b)^2$.
💡 The gap between two points is the hypotenuse of a little right triangle made of the across-gap and up-gap.
8.G.B.7 Step 3 The right angle gives Pythagoras
- The two lines are perpendicular and meet at $A$, so the angle at $A$ is $90^\circ$ and $PQ$ is the hypotenuse of a right triangle.
- That means $AP^2+AQ^2=PQ^2$.
- Substitute and expand: $36+(8-b)^2+36+(8+b)^2=4b^2$ becomes $72+(64-16b+b^2)+(64+16b+b^2)=4b^2$, and the $-16b$ and $+16b$ cancel.
💡 A right angle is exactly the condition that turns the three sides into a Pythagorean equation.
8.EE.C.7 Step 4 Solve for the intercept
- From $200+2b^2=4b^2$, subtract $2b^2$ from both sides to get $200=2b^2$, so $b^2=100$ and $b=10$.
- So the intercepts are $P=(0,10)$ and $Q=(0,-10)$.
- (Taking $b=-10$ just swaps the names of $P$ and $Q$, so it gives the same triangle.)
💡 Collecting the $b^2$ terms leaves a plain equation you can undo with a square root.
6.G.A.1 Step 5 Compute the area
- Take $PQ$ as the base.
- It runs from $(0,10)$ to $(0,-10)$, so its length is $20$.
- The matching height is how far $A$ sits from that base line — the $y$-axis — which is just the $x$-coordinate of $A$, namely $6$.
- Area of a triangle is half base times height: $\tfrac12\cdot 20\cdot 6=60$.
- That is choice (D).
💡 With the base flat on the $y$-axis, the height is simply how far sideways the tip $A$ reaches.
6.NS.C.8 Put $A$ at $(6,8)$. Both intercepts lie on the $y$-axis, so they look like $(0,\ 8.G.B.8 Use the distance between two points for each side of $\triangle APQ$. From $A=(6 8.G.B.7 The two lines are perpendicular and meet at $A$, so the angle at $A$ is $90^\cir 8.EE.C.7 From $200+2b^2=4b^2$, subtract $2b^2$ from both sides to get $200=2b^2$, so $b^2 6.G.A.1 Take $PQ$ as the base. It runs from $(0,10)$ to $(0,-10)$, so its length is $20$ Review
Reasonableness: Check the right angle directly. With $P=(0,10)$ and $Q=(0,-10)$, line $AP$ has slope $\frac{8-10}{6}=-\frac13$ and line $AQ$ has slope $\frac{8+10}{6}=3$. Their product is $-\frac13\cdot 3=-1$, confirming the lines really are perpendicular. The base $20$ and height $6$ give area $60$, matching (D), and it lands sensibly among the choices.
Alternative: Shortcut using the midpoint of the hypotenuse. Since $P$ and $Q$ are opposites, the midpoint of $PQ$ is the origin $O$. In a right triangle the distance from the right-angle vertex to the midpoint of the hypotenuse equals half the hypotenuse, so $OA=OP=OQ$. But $OA=\sqrt{6^2+8^2}=10$, so the hypotenuse $PQ=2\cdot 10=20$. With height $6$ (the $x$-coordinate of $A$), the area is $\tfrac12\cdot 20\cdot 6=60$ again.
CCSS standards used (min grade 8)
6.NS.C.8Solve real-world problems by graphing points in all four quadrants (Placing $A=(6,8)$ and the paired intercepts $P=(0,b)$, $Q=(0,-b)$ on the coordinate plane.)8.G.B.8Apply the Pythagorean theorem to find distance between two points in a coordinate system (Writing $AP^2$, $AQ^2$, and $PQ^2$ from the coordinates of $A$, $P$, and $Q$.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Using the right angle at $A$ to write $AP^2+AQ^2=PQ^2$.)8.EE.C.7Solve linear equations in one variable (Solving $200+2b^2=4b^2$ to get $b=10$.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Computing the area as $\tfrac12\cdot 20\cdot 6=60$ with base $PQ$ and height $6$.)
⭐ Two intercepts that add to zero sit symmetric on the $y$-axis, and the right angle at $A$ turns the three sides into a Pythagorean equation — then base times height, halved, gives the area.
⭐ Two intercepts that add to zero sit symmetric on the $y$-axis, and the right angle at $A$ turns the three sides into a Pythagorean equation — then base times height, halved, gives the area.
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