2020 AMC 12A Problems/Problem 12
Problem
Line in the coordinate plane has equation
. This line is rotated $45\degree$ (Error compiling LaTeX. Unknown error_msg) 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 $45\degree$ (Error compiling LaTeX. Unknown error_msg). $45\degree$ (Error compiling LaTeX. Unknown error_msg) creates an isosceles right triangle since the sum of the angles of the triangle must be $180\degree$ (Error compiling LaTeX. Unknown error_msg) and one angle is $90\degree$ (Error compiling LaTeX. Unknown error_msg) which means the last leg angle must also be $45\degree$ (Error compiling LaTeX. Unknown error_msg). 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 $\opair{5,3}$ (Error compiling LaTeX. Unknown error_msg). Construct another line from
to
, the vector $\opair{3,-5}$ (Error compiling LaTeX. Unknown error_msg). This is
and equal to the original line segment. The difference between the two vectors is $\opair{2,8}$ (Error compiling LaTeX. Unknown error_msg), 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