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)} | + | |
+ | 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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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
Contents
Problem
The point in the -plane is first rotated counterclockwise by around the point and then reflected about the line . The image of after these two transformations is at . What is
Solution
The final image of is . We know the reflection rule for reflecting over is . So before the reflection and after rotation the point is .
By definition of rotation, the slope between and must be perpendicular to the slope between and . The first slope is . This means the slope of and is .
Rotations also preserve distance to the center of rotation, and since we only "travelled" up and down by the slope once to get from to it follows we shall only use the slope once to travel from to .
Therefore point is located at . The answer is .
Video Solution by OmegaLearn (Rotation & Reflection tricks)
~ pi_is_3.14
See Also
2021 AMC 12B (Problems • Answer Key • Resources) | |
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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