Difference between revisions of "2020 AMC 12A Problems/Problem 12"
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== Solution == | == Solution == | ||
− | The slope of the line is <math>\frac{3}{5}</math>. We must transform it by <math>45^{\circ}</math>. <math>45^{\circ}</math> creates an isosceles right triangle since the sum of the angles of the triangle must be <math>180^{\circ}</math> and one angle is <math>90^{\circ}</math> which means the last leg angle must also be <math>45^{\circ}</math>. In the isosceles right triangle, the two legs are congruent. We can, therefore, construct an isosceles right triangle with a line of <math>\frac{3}{5}</math> slope on graph paper. That line with <math>\frac{3}{5}</math> slope starts at <math>(0,0)</math> and will go to <math>(5,3)</math>, the vector <math><5,3></math>. Construct another line from <math>(0,0)</math> to <math>(3,-5)</math>, the vector <math><3,-5></math>. This is <math>\perp</math> and equal to the original line segment. The difference between the two vectors is <math><2,8></math>, which is the slope <math>4</math>, and that is the slope of line <math>k</math>. Furthermore, the equation <math>3x-5y+40=0</math> passes straight through <math>(20,20)</math> since <math>3(20)-5(20)+40=60-100+40=0</math>, which means that any rotations about <math>(20,20)</math> would contain <math>(20,20)</math>. We can create a line of slope <math>4</math> through <math>(20,20)</math>. The <math>x</math>-intercept is therefore <math>20-\frac{20}{4} = \boxed{\textbf{(B) } 15.}</math>~lopkiloinm | + | The slope of the line is <math>\frac{3}{5}</math>. We must transform it by <math>45^{\circ}</math>. |
+ | <math>45^{\circ}</math> creates an isosceles right triangle since the sum of the angles of the triangle must be <math>180^{\circ}</math> and one angle is <math>90^{\circ}</math> which means the last leg angle must also be <math>45^{\circ}</math>. | ||
+ | In the isosceles right triangle, the two legs are congruent. We can, therefore, construct an isosceles right triangle with a line of <math>\frac{3}{5}</math> slope on graph paper. That line with <math>\frac{3}{5}</math> slope starts at <math>(0,0)</math> and will go to <math>(5,3)</math>, the vector <math><5,3></math>. Construct another line from <math>(0,0)</math> to <math>(3,-5)</math>, the vector <math><3,-5></math>. This is <math>\perp</math> and equal to the original line segment. The difference between the two vectors is <math><2,8></math>, which is the slope <math>4</math>, and that is the slope of line <math>k</math>. | ||
+ | Furthermore, the equation <math>3x-5y+40=0</math> passes straight through <math>(20,20)</math> since <math>3(20)-5(20)+40=60-100+40=0</math>, which means that any rotations about <math>(20,20)</math> would contain <math>(20,20)</math>. We can create a line of slope <math>4</math> through <math>(20,20)</math>. The <math>x</math>-intercept is therefore <math>20-\frac{20}{4} = \boxed{\textbf{(B) } 15.}</math>~lopkiloinm |
Revision as of 14:12, 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
~lopkiloinm