2012 AMC 10B Problem 3

P1-10~1 minPrint

Coordinate Geometry Transformations & Symmetry Set up the equation/formula and compute; no special trick needed

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?

Show solution without answering

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.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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