AMC 10 · 2014 · #14

Grade 8 geometry-2d
Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity
coordinate-geometryslope-interceptarea-triangles convert-to-algebraidentify-subproblems ↑ Prerequisites: coordinate-geometryarea-triangles
📏 Medium solution 💡 2 insights
Problem
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 △ APQ.

Pick an answer.

(A)
45
(B)
48
(C)
54
(D)
60
(E)
72

AMC 10 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Introduce a Variable

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.

1STEP 1

Place the points on a grid

A=(6,8). Both intercepts lie on the y-axis and add to 0, so they are opposites: P=(0,b), Q=(0,-b).

A=(6,8), P=(0,b), Q=(0,-b)
2STEP 2

Write the three side lengths

Distance formula from A: AP²=36+(8-b)², AQ²=36+(8+b)², and PQ runs down the y-axis so PQ²=(2b)².

AP²=36+(8-b)², AQ²=36+(8+b)², PQ²=4b²
3STEP 3

The right angle gives Pythagoras

Perpendicular at A means the angle there is 90°, so AP²+AQ²=PQ². Expanding, -16b and +16b cancel: 200+2b²=4b².

AP²+AQ²=PQ² → 200+2b²=4b²
4STEP 4

Solve for the intercept

Subtract 2b² from both sides: 200=2b², so b²=100 and b=10. The intercepts are P=(0,10), Q=(0,-10).

200=2b² → b²=100 → b=10
5STEP 5

Compute the area

Take PQ=20 as the base; the height is A's x-coordinate 6. Half of 20 times 6 gives area 60.

Area=1/2 · PQ · 6=1/2 · 20 · 6=60 → (D)
Answer
60
Check the right angle directly. With P=(0,10) and Q=(0,-10), line AP has slope 8−106=−13\frac{8-10}{6}=-\frac{1}{3} and line AQ has slope 8+106=3\frac{8+10}{6}=3. Their product is −13⋅3=−1-\frac{1}{3}\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.
💡Key takeaway

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.

  • Place the points on a grid
  • Write the three side lengths
  • The right angle gives Pythagoras
  • Solve for the intercept
  • Compute the area

More like this

Same archetype — closest grade level first.