2020 AMC 12A Problems/Problem 12
Contents
[hide]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
Solution 2
Since the slope of the line is , and the angle we are rotating around is x, then
Hence, the slope of the rotated line is . Since we know the line intersects the point , then we know the line is . Set to find the x-intercept, and so
~Solution by IronicNinja
See Also
2020 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
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All AMC 12 Problems and Solutions |
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