Difference between revisions of "2016 AMC 10A Problems/Problem 16"

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== Problem ==
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A triangle with vertices <math>A(0, 2)</math>, <math>B(-3, 2)</math>, and <math>C(-3, 0)</math> is reflected about the <math>x</math>-axis, then the image <math>\triangle A'B'C'</math> is rotated counterclockwise about the origin by <math>90^{\circ}</math> to produce <math>\triangle A''B''C''</math>. Which of the following transformations will return <math>\triangle A''B''C''</math> to <math>\triangle ABC</math>?
 
A triangle with vertices <math>A(0, 2)</math>, <math>B(-3, 2)</math>, and <math>C(-3, 0)</math> is reflected about the <math>x</math>-axis, then the image <math>\triangle A'B'C'</math> is rotated counterclockwise about the origin by <math>90^{\circ}</math> to produce <math>\triangle A''B''C''</math>. Which of the following transformations will return <math>\triangle A''B''C''</math> to <math>\triangle ABC</math>?
  

Revision as of 12:06, 4 February 2016

Problem

A triangle with vertices $A(0, 2)$, $B(-3, 2)$, and $C(-3, 0)$ is reflected about the $x$-axis, then the image $\triangle A'B'C'$ is rotated counterclockwise about the origin by $90^{\circ}$ to produce $\triangle A''B''C''$. Which of the following transformations will return $\triangle A''B''C''$ to $\triangle ABC$?

$\textbf{(A)}$ counterclockwise rotation about the origin by $90^{\circ}$.

$\textbf{(B)}$ clockwise rotation about the origin by $90^{\circ}$.

$\textbf{(C)}$ reflection about the $x$-axis

$\textbf{(D)}$ reflection about the line $y = x$

$\textbf{(E)}$ reflection about the $y$-axis.

Solution

Consider a point $(x, y)$. Reflecting it about the $x$-axis will map it to $(x, -y)$, and rotating it counterclockwise about the origin b $90^{\circ}$ will map it to $(y, x)$. The operation that undoes this is reflection about the line $y = x$, so the answer is $\boxed{\textbf{(D)}}$.

See Also

2016 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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