Fall 2021 AMC 10 B

The reflection through $\ell$ followed by the reflection through $m$ is equivalent to a rotation about the intersection of the two lines by twice the angle formed by the two lines. As the result of the two reflections in the present case, $P$ was rotated by $90^{\circ }$ clockwise around the origin to $P^{\prime \prime }(4,1)$. Therefore the line $\ell$ must be rotated by $45^{\circ }$ clockwise about the origin to obtain line $m$.

The slope of line $\ell$ is $5$, and a quick sketch shows that the slope of line $m$ is a positive number less than $5$. The equation of $m$ can be calculated if given one point on $m$ other than the origin $O$. Point $A(1,5)$ is on line $\ell$. Let $B(x,y)$ be the intersection of line $m$ with the line through $A$ perpendicular to line $\ell$. Then $△BAO$ is an isosceles right triangle with right angle at $A$. Then the slope of line $AB$ is $-\frac{1}{5}$, so $y-5=-\frac{1}{5}(x-1)$. Point $B$ is on the circle with radius $OA$ centered at $A$. Therefore $(x-1{)}^2+(y-5{)}^2=26$. The two solutions of this pair of equations are $(6,4)$ and $(-4,6)$. Because $B$ is in the first quadrant, it must be $(6,4)$, so the slope of $m$ is $\frac{2}{3}$, and its equation can be written $2x-3y=0$.