2021 AMC Fall 10B Problem 17

Reflecting in $\ell$ and then in $m$, both through the origin, is the same as rotating about the origin by twice the angle from $\ell$ to $m$.

Compare $P = (-1, 4)$ with $P'' = (4, 1)$. Both are at distance $\sqrt{17}$ from the origin, and $(-1, 4) \mapsto (4, 1)$ is the map $(x, y) \mapsto (y, -x)$, a $90^\circ$ clockwise rotation. So the angle from $\ell$ to $m$ is $45^\circ$ clockwise.

Line $\ell$ is $y = 5x$, direction vector $(1, 5)$. Rotating a vector $(a, b)$ clockwise by $45^\circ$ gives $\frac{1}{\sqrt2}(a + b,\ b - a)$, so $(1, 5)$ becomes $\frac{1}{\sqrt2}(6, 4)$, which points along $(3, 2)$.

Thus $m$ passes through the origin with slope $\frac23$: $y = \frac23 x$, or $2x - 3y = 0$.

The answer is $\boxed{\textbf{(D)}\ 2x-3y=0}$.

A reflection reverses orientation, so two reflections preserve it and fix the common point: the result is a rotation, by twice the angle between the mirrors. Recognizing $P \to P''$ as a rotation by a nice angle avoids computing $P'$ entirely. Checking $|P| = |P''|$ is a quick confirmation that the data are consistent.

Compute $P'$ directly: the projection of $(-1, 4)$ onto direction $(1, 5)$ is $\frac{19}{26}(1, 5)$, so $P' = 2 \cdot \frac{19}{26}(1, 5) - (-1, 4) = \left(\frac{32}{13}, \frac{43}{13}\right)$. Line $m$ is the perpendicular bisector of $P'P''$ and passes through the origin, so it passes through the midpoint $\left(\frac{42}{13}, \frac{28}{13}\right)$, giving slope $\frac{28}{42} = \frac23$.

• Rotating line l by the full 90 degrees instead of half of it, or turning it counterclockwise, which produces the wrong slope.
• Reflecting $P''$ back over $\ell$ (wrong order of reflections) or making an arithmetic slip in the reflection formula.