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

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<math>\textbf{(E)}</math> reflection about the <math>y</math>-axis.
 
<math>\textbf{(E)}</math> reflection about the <math>y</math>-axis.
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==Solution==
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Consider a point <math>(x, y)</math>. Reflecting it about the <math>x</math>-axis will map it to <math>(x, -y)</math>, and rotating it counterclockwise about the origin b <math>90^{\circ}</math> will map it to <math>(y, x)</math>. The operation that undoes this is reflection about the line <math>y = x</math>, so the answer is <math>\boxed{\textbf{(D)}}</math>.
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==See Also==
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{{AMC10 box|year=2016|ab=A|num-b=15|num-a=17}}
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{{MAA Notice}}

Revision as of 22:49, 3 February 2016

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 (Problems • Answer Key • Resources)
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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