Reflection of a point about the line $y = 2 x$

The reflection point of $P(a,b)$ , call it $P'(a',b')$ , must lie on a line that is normal to $y=2x$ . Let this line equal to:

$\large y-b = -\frac{1}{2}(x-a) \Rightarrow y = -\frac{1}{2}x + \frac{a+2b}{2}$

which intersects $y=2x$ in the point $Q(x,y)$ :

$\large 2x =-\frac{1}{2}x + \frac{a+2b}{2} \Rightarrow x = \frac{a+2b}{5}, y = \frac{2a+4b}{5}$ . If $Q$ is the midpoint of $P$ and $P'$ , then we have:

$\large \frac{a+a'}{2} = \frac{a+2b}{5} \Rightarrow a' = \frac{-3a+4b}{5}$ ;

$\large \frac{b+b'}{2} = \frac{2a+4b}{5} \Rightarrow b' = \frac{4a+3b}{5}$

Hence, the reflection point $P'$ has coordinates $\boxed{x = \frac{-3a+4b}{5}, y = \frac{4a+3b}{5}},$ or Choice A.

The reflection has eigenvectors (1, 2) with eigenvalue sqrt(5) and (-2, 1) with -sqrt(5).

Matrix A=P. D. P^(-1) and A.(a, b) yields to the solution.

The standard way to solve this type of problem is through linear algebra.

The image $\mathbf{p'}$ of a point $\mathbf{p}$ under a reflection across the line $\mathbf{n}^T (\mathbf{r} - \mathbf{r}_0) = 0$ is given by

$\mathbf{p'} = \mathbf{r}_0 + \mathbf{M} (\mathbf{p} - \mathbf{r}_0)$

where the reflection matrix $\mathbf{M} = \mathbf{I} - 2 \hspace{4pt} \dfrac{\mathbf{n} \mathbf{n}^T }{ \mathbf{n}^T \mathbf{n} }$

In this problem, the reflection line is $y = 2 x$ , so in standard form, it is given by $2 x - y = 0$ , thus $\mathbf{r}_0 = (0, 0)$ and $\mathbf{n} = (2, -1)$

The reflection matrix is, therefore,

$\mathbf{M} = \begin{bmatrix} 1 && 0 \\ 0 && 1 \end{bmatrix} - \dfrac{2}{2^2 + (-1)^2 } \begin{bmatrix} 2 \\ -1 \end{bmatrix} \begin{bmatrix} 2 && -1 \end{bmatrix}$

Simplifying,

$\mathbf{M} = \begin{bmatrix} 1 && 0 \\ 0 && 1 \end{bmatrix} - \dfrac{2}{5 } \begin{bmatrix} 4 && -2 \\ -2 && 1 \end{bmatrix}$

Which becomes,

$\mathbf{M} = \displaystyle \dfrac{1}{5} \begin{bmatrix} -3 && 4 \\ 4 && 3 \end{bmatrix}$

With $\mathbf{r}_0 = \mathbf{0}$ , the reflection relation becomes

$\mathbf{p'} = \mathbf{M} \mathbf{p}$

Since $\mathbf{p} = (a, b)$ , then $\mathbf{p'} = \left( \dfrac{-3 a + 4 b}{5} , \dfrac{4 a + 3 b }{5} \right)$