Difference between revisions of "2020 AMC 12A Problems/Problem 12"
Lopkiloinm (talk | contribs) (→Problem) |
Lopkiloinm (talk | contribs) (→Solution) |
||
| Line 7: | Line 7: | ||
== 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} | + | 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 14:09, 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
