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2012 AMC 10B Problem 3

2012 AMC 10B Problem 3AMC 10 Step by Step

Name: Date: October 28, 2026

Coordinate Geometry · difficulty 1 of 5 · about 1 min · position P1-10

The point in the xy-plane with coordinates $(1000, 2012)$ is reflected across the line $y = 2000$ . What are the coordinates of the reflected point?

A) $(998,2012)$
B) $(1000,1988)$
C) $(1000,2024)$
D) $(1000,4012)$
E) $(1012,2012)$

Answer

Key insight

The mirror line y = 2000 is flat, so the x-coordinate stays 1000. The point is 12 above the line, so its reflection is 12 below it.

Solution

The line $y = 2000$ is a horizontal line. Reflecting across a horizontal line moves a point straight up or straight down, never sideways. So the $x$ -coordinate stays $1000$ .
The point has $y = 2012$ . That is $2012 - 2000 = 12$ units above the line.
A mirror image is the same distance from the mirror but on the other side. So the image is $12$ units below the line, at $y = 2000 - 12 = 1988$ .
The reflected point is $(1000, 1988)$ .

The answer is $\boxed{\textbf{(B)}\ (1000, 1988)}$ .

Why this works

The mirror line sits exactly halfway between a point and its reflection. So measure how far the point is from the line, then go that same distance past the line. For a horizontal line only $y$ changes; for a vertical line only $x$ changes.

The trap

Changing the x-coordinate instead of the y-coordinate, or adding 12 again and going to 2024 instead of going to the other side of the line.

Concise solution

The line $y = 2000$ is horizontal, so a reflection across it moves a point straight up or down: the $x$ -coordinate stays $1000$ .

The point sits $2012 - 2000 = 12$ units above the line, so its image sits $12$ units below it, at $y = 2000 - 12 = 1988$ .

The answer is $\boxed{\textbf{(B)}\ (1000, 1988)}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 17: Analytic Geometry
Introduction to Geometry, Chapter 16: The More Things Change. . . (§16.1, §16.2, §16.3, §16.4)
Introduction to Algebra, Chapter 8: Graphing Lines

Techniques

Set up the equation/formula and compute; no special trick needed