2020 AMC 12A Problems/Problem 12
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
~lopkiloinm
See Also
2020 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 1 |
Followed by Problem 3 |
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All AMC 12 Problems and Solutions |
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