Line $l$ in the coordinate plane has equation $3x-5y+40=0$. This line is rotated $45^{\circ}$ counterclockwise about the point $(20,20)$ to obtain line $k$. What is the $x$-coordinate of the $x$-intercept of line $k?$
$\textbf{(A) } 10 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 25 \qquad \textbf{(E) } 30$
[AMC 12A 2020 Question-12 Solution-1.png](https://artofproblemsolving.com/wiki/index.php/File:AMC_12A_2020_Question-12_Solution-1.png)
The slope of the line is $\frac{3}{5}$. We must transform it by $45^{\circ}$. 
$45^{\circ}$ creates an isosceles right triangle, since the sum of the angles of the triangle must be $180^{\circ}$ and one angle is $90^{\circ}$. This means the last leg angle must also be $45^{\circ}$. 
In the isosceles right triangle, the two legs are congruent. We can therefore construct an isosceles right triangle with a line of $\frac{3}{5}$ slope on graph paper. That line with $\frac{3}{5}$ slope starts at $(0,0)$ and will go to $(5,3)$, the vector $$.
Construct another line from $(0,0)$ to $(3,-5)$, the vector $$. This is $\perp$ and equal to the original line segment. The difference between the two vectors is $$, which is the slope $4$, and that is the slope of line $k$. 
Furthermore, the equation $3x-5y+40=0$ passes straight through $(20,20)$ since $3(20)-5(20)+40=60-100+40=0$, which means that any rotations about $(20,20)$ would contain $(20,20)$. We can create a line of slope $4$ through $(20,20)$. The $x$-intercept is therefore $20-\frac{20}{4} = \boxed{\textbf{(B) } 15.}$ ~lopkiloinm ~ShawnX (diagram)