Difference between revisions of "2021 AMC 12B Problems/Problem 5"

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<math>\textbf{(A)} ~1 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ~5 \qquad\textbf{(D)} ~7 \qquad\textbf{(E)} ~9</math>
 
<math>\textbf{(A)} ~1 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ~5 \qquad\textbf{(D)} ~7 \qquad\textbf{(E)} ~9</math>
 
==Solution==
 
==Solution==
<math>\boxed{\textbf{(D)} ~7}</math>
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The final image of <math>P</math> is <math>(-6,3)</math>. We know the reflection rule for reflecting over <math>y=-x</math> is <math>(x,y) --> (-y, -x)</math>. So before the reflection and after rotation the point is <math>(-3,6)</math>.
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By definition of rotation, the slope between <math>(-3,6)</math> and <math>(1,5)</math> must be perpendicular to the slope between <math>(a,b)</math> and <math>(1,5)</math>. The first slope is <math>\frac{5-6}{1-(-3)} = \frac{-1}{4}</math>. This means the slope of <math>P</math> and <math>(1,5)</math> is <math>4</math>.
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Rotations also preserve distance to the center of rotation, and since we only "travelled" up and down by the slope once to get from <math>(3,-6)</math> to <math>(1,5)</math> it follows we shall only use the slope once to travel from <math>(1,5)</math> to <math>P</math>.
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Therefore point <math>P</math> is located at <math>(1+1, 5+4) = (2,9)</math>. The answer is <math>9-2 = 7 = \boxed{\textbf{(D)}}</math>.
  
 
== Video Solution by OmegaLearn (Rotation & Reflection tricks) ==
 
== Video Solution by OmegaLearn (Rotation & Reflection tricks) ==

Revision as of 21:44, 11 February 2021

Problem

The point $P(a,b)$ in the $xy$-plane is first rotated counterclockwise by $90\deg$ around the point $(1,5)$ and then reflected about the line $y = -x$. The image of $P$ after these two transformations is at $(-6,3)$. What is $b - a ?$

$\textbf{(A)} ~1 \qquad\textbf{(B)} ~3 \qquad\textbf{(C)} ~5 \qquad\textbf{(D)} ~7 \qquad\textbf{(E)} ~9$

Solution

The final image of $P$ is $(-6,3)$. We know the reflection rule for reflecting over $y=-x$ is $(x,y) --> (-y, -x)$. So before the reflection and after rotation the point is $(-3,6)$.

By definition of rotation, the slope between $(-3,6)$ and $(1,5)$ must be perpendicular to the slope between $(a,b)$ and $(1,5)$. The first slope is $\frac{5-6}{1-(-3)} = \frac{-1}{4}$. This means the slope of $P$ and $(1,5)$ is $4$.

Rotations also preserve distance to the center of rotation, and since we only "travelled" up and down by the slope once to get from $(3,-6)$ to $(1,5)$ it follows we shall only use the slope once to travel from $(1,5)$ to $P$.

Therefore point $P$ is located at $(1+1, 5+4) = (2,9)$. The answer is $9-2 = 7 = \boxed{\textbf{(D)}}$.

Video Solution by OmegaLearn (Rotation & Reflection tricks)

https://youtu.be/VyRWjgGIsRQ

~ pi_is_3.14

See Also

2021 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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 12 Problems and Solutions

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