AMC 10 · 2019 · #5

Grade 6 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Line Intersection Vertices then Area

coordinate-geometryslope-interceptlinear-equations-two-vararea-trianglesisosceles-triangle coordinate-geometryidentify-subproblems ↑ Prerequisites: coordinate-geometryslope-interceptarea-triangles

📏 Medium solution 💡 3 insights

Problem

Two lines pass through the point (2,2) — one with slope one half and one with slope 2. A third line is x plus y equals 10. These three lines bound a triangle; find its area.

Pick an answer.

(A)

(B)

$4\sqrt{2}$

(C)

(D)

(E)

$6\sqrt{2}$

How to solve

Strategy Draw a Diagram

Tool #1: sketch the three lines and the triangle on a coordinate grid — this makes the three vertices visible. Tool #7 splits the job into three subproblems: (a) write the two line equations through (2,2), (b) find where each meets x+y=10, (c) compute the area from the three vertices. Tool #3 confirms the answer matches a single choice.

1STEP 1

Write the two line equations

A point and a slope fix each equation.

Line 1: y = 1/2x + 1, Line 2: y = 2x - 2

2STEP 2

Find the first crossing

Solve against the third line.

x + 1/2x + 1 = 10 → 3/2x = 9 → x = 6, y = 4

3STEP 3

Find the second crossing

The other line works the same way.

x + 2x - 2 = 10 → 3x = 12 → x = 4, y = 6

4STEP 4

Collect the three vertices

The given point is the third vertex.

A = (2, 2), B = (6, 4), C = (4, 6)

5STEP 5

Box it in

Put the triangle inside a square.

Bounding square area = 4 · 4 = 16

6STEP 6

Subtract the corner triangles

Removing three corners leaves the area.

Area = 16 - 1/2(4 · 2) - 1/2(2 · 2) - 1/2(2 · 4) = 16 - 4 - 2 - 4 = 6

7STEP 7

Match the choice

The area is 6.

6 → (C)

Answer

Cross-check using base-and-height. The triangle is isosceles: AB runs from (2,2) to (6,4), length √(16 + 4) = √(20) = 2√(5); AC runs from (2,2) to (4,6), length √(4 + 16) = 2√(5). Base BC from (6,4) to (4,6), length √(4 + 4) = 2√(2). The base BC lies on x + y = 10. Distance from A = (2,2) to the line x + y = 10 is (|2 + 2 - 10|)/√(2) = 6/√(2) = 3√(2). Area = 1/2 · 2√(2) · 3√(2) = 1/2 · 12 = 6. ✓

💡Key takeaway

This AMC 12 problem only needs Grade 6 area-by-pieces you already know: find the three triangle corners (2,2), (6,4), (4,6), surround them with a 4 × 4 square (area 16), then subtract the three right-triangle corners (4 + 2 + 4 = 10). The triangle's area is 16 - 10 = 6.