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