Distinct lines $\ell$ and $m$ lie in the $xy$-plane. They intersect at the origin. Point $P(-1, 4)$ is reflected about line $\ell$ to point $P'$, and then $P'$ is reflected about line $m$ to point $P''$. The equation of line $\ell$ is $5x - y = 0$, and the coordinates of $P''$ are $(4,1)$. What is the equation of line $m?$
$(\textbf{A})\: 5x+2y=0\qquad(\textbf{B}) \: 3x+2y=0\qquad(\textbf{C}) \: x-3y=0\qquad(\textbf{D}) \: 2x-3y=0\qquad(\textbf{E}) \: 5x-3y=0$
Denote $O$ as the origin.
Even though the problem is phrased as a coordinate bash, that looks disgusting. Instead, let's try to phrase this problem in terms of Euclidean geometry, using the observation that $\angle POP'' = 90^{\circ}$, and that both $\ell$ and $m$ must pass through $O$ in order to preserve the distance from $P$ to the origin.
[asy] unitsize(1.4cm); draw((0,3)--(0,0)--(3,0), dashed); dot((0,3)); dot((3,0)); label("$P$", (0,3), W); label("$P''$", (3,0), S);  draw((0,0)--(1.5,4.5)); label("$\ell$", (1.5,4.5), N); draw((0,0)--(4,2)); label("$m$", (4,2), E);  dot((1.8,2.4)); label("$P'$", (1.8,2.4), N); label("$O$",(0,0));  dot((1,3)); dot((2.5,1.25)); label("$A$", (1,3), E); label("$B$", (2.5,1.25), N); [/asy]
($A$ and $B$ are just defined as points on lines $\ell$ and $m$.)
Because of how reflections work, we have that $\angle AOP' = \angle POA$ and $\angle P'OB = \angle BOP''$; adding these two equations together and using angle addition, we have that $\angle AOB = \angle POA + \angle BOP''$. Since the sum of both sides combined must be $90^{\circ}$ by angle addition,
\[\angle AOB = 45^{\circ}.\]
This is helpful! We can now return to using coordinates, with this piece of information in mind:
[asy] unitsize(0.2cm); markscalefactor = 0.08; import graph; Label f;  f.p=fontsize(9);  xaxis(-2,6,Ticks(f, 2.0));  yaxis(-1,6,Ticks(f, 2.0)); dot((-1,4)); label("$P$", (-1,4), W); dot((4,1)); label("$P''$", (4,1), W);  draw((0,0)--(1.2,6)); label("$\ell$", (1.2,6), N); dot((0.5,2.5));label("$(0.5,2.5)$", (0.5,2.5), E);label("$A$", (0.5,2.5), W); dot((3,2));label("$B$", (3,2), E); draw((0.5,2.5)--(3,2), dashed);  draw((0,0)--(6,4)); label("$m$", (6,4), E);  draw(anglemark((6, 4), (0, 0), (1, 5))); label("$45^{\circ}$", (0.54,0.75)); [/asy]
The $45^{\circ}$ angle is a little bit unwieldy in the coordinate plane, so we should try to make a $45-45-90$ triangle. Let $A$ be a point on $\ell$; to make $A$ fit nicely in the diagram, let it be $(0.5,2.5)$. Now, let's draw a perpendicular to $\ell$ through point $A$, intersecting $m$ at point $B$. $OAB$ is a $45-45-90$ triangle, so $B$ is a $90$ degree counterclockwise rotation from $O$ about $A$. Therefore, the coordinates of $B$ are
\[(0.5+2.5,2.5-0.5) = (3,2).\]
So, $(3,2)$ is a point on line $m$, which we already know passes through the origin; therefore, $m$'s equation is $y=\frac{2x}{3} \implies \boxed{\textbf{(D) } 2x-3y = 0}.$
~ihatemath123
(We never actually had to use the information of the exact coordinates of $P$; as long as $\angle POP'' = 90^{\circ}$, when we move $P$ around, this will not affect $m$'s equation.)