Difference between revisions of "2020 AMC 12A Problems/Problem 12"
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== Problem == | == Problem == | ||
| − | Line <math>l</math> in the coordinate plane has equation <math>3x-5y+40=0</math>. This line is rotated <math>45\ | + | Line <math>l</math> in the coordinate plane has equation <math>3x-5y+40=0</math>. This line is rotated <math>45^{\circ}</math> counterclockwise about the point <math>(20,20)</math> to obtain line <math>k</math>. What is the <math>x</math>-coordinate of the <math>x</math>-intercept of line <math>k?</math> |
<math>\textbf{(A) } 10 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 25 \qquad \textbf{(E) } 30</math> | <math>\textbf{(A) } 10 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 25 \qquad \textbf{(E) } 30</math> | ||
Revision as of 14:08, 1 February 2020
Problem
Line 
in the coordinate plane has equation 
. This line is rotated 
counterclockwise about the point 
to obtain line 
. What is the 
-coordinate of the -intercept of line 
Solution
The slope of the line is 
. We must transform it by . creates an isosceles right triangle since the sum of the angles of the triangle must be 
and one angle is 
which means the last leg angle must also be . In the isosceles right triangle, the two legs are congruent. We can, therefore, construct an isosceles right triangle with a line of slope on graph paper. That line with slope starts at 
and will go to 
, the vector 
. Construct another line from to 
, the vector 
. This is 
and equal to the original line segment. The difference between the two vectors is 
, which is the slope 
, and that is the slope of line . Furthermore, the equation passes straight through since 
, which means that any rotations about would contain . We can create a line of slope through . The -intercept is therefore 
