How to find the coordinates of the reflection of the point $(0,1)$ in the line $y=mx$?

Question

So I understand the perpendicular line is $y=-\frac{1}mx+c$ and I think the point of intersection is $(\frac{-1+m}{m^2}),(\frac{-1+m}{m^3}+c)$ based on my calculation but I think I have made an error. Any help is greatly appreciated

Answer 1

Let $d_1$ be a line defined by the equation $y = mx$ and $d_2$ the line perpendicular to $d_1$ and passing through $(0,1)$. Consider the case $m \neq 0.$

Since $d_2 \perp d_1$ the angular coefficient of $d_2$ is $-\frac 1 m $ so $d_2$ has the following equation $$ y = - \frac 1 m x + c$$ where $c = 1$ since $(0,1)$ is a point of $d_2$. The intersection of the two lines is the solution of the system of equations

$$ \begin{cases} y = - \frac 1 m x +1\\ y = mx.\end{cases} $$

By substitution $mx = - \frac 1m x + 1 \iff (m+\frac 1 m)x = 1 \iff \frac{m^2 + 1}{m}x = 1 \iff x = \frac{m}{m^2 + 1}.$ So the intersection of the two lines is the point
$$ P = (\frac{m}{m^2+1}, \frac{m^2}{m^2 + 1}) $$

Let $A = (0,1)$ and $A'$ the reflection of $(0,1)$ along the line $y = mx$ then we must have $ \vec{PA} = - \vec{PA'}$. This means that $$ \vec{OA'} = \vec{OP} + \vec{PA'} = \vec{OP} - \vec{PA}$$ but $ \vec {PA} = \vec {OA} - \vec {OP}$ and we have

$$ \vec{OA'} = 2\vec{OP} - \vec{OA}.$$

Therefore $$A' = (2 \frac{m}{m^2 + 1},2 \frac{m^2}{m^2+1} -1).$$

Answer 2

Here I propose an alternative solution based on linear algebra.

Let us consider the linear transformation $T:\textbf{R}^{2}\to\textbf{R}^{2}$ be a the linear transformation given by the reflection along the line $y = mx$. Then one hast that $T(1,m) = (1,m)$ and $T(-m,1) = (m,-1)$.

Thus if we consider the basis $\mathcal{B}' = \{(1,m),(-m,1)\}$, it results that \begin{align*} [T]_{\mathcal{B}'} = \begin{bmatrix} 1 & 0\\ 0 & -1\\ \end{bmatrix} \end{align*}

Now let us consider the standard basis of $\textbf{R}^{2}$: $\mathcal{B} = \{(1,0),(0,1)\}$. Then one has that \begin{align*} [T]_{\mathcal{B}} = [T]_{\mathcal{B}'}^{\mathcal{B}}[T]_{\mathcal{B}'}[T]_{\mathcal{B}}^{\mathcal{B}'} \end{align*} Then it remains to determine the matrices $[T]_{\mathcal{B}'}^{\mathcal{B}}$ and $[T]_{\mathcal{B}}^{\mathcal{B}'}$. To begin with, notice that \begin{align*} [T]_{\mathcal{B}'}^{\mathcal{B}} = \begin{bmatrix} 1 & -m\\ m & 1 \end{bmatrix} \Rightarrow [T]_{\mathcal{B}}^{\mathcal{B}'} = ([T]_{\mathcal{B}'}^{\mathcal{B}})^{-1} = \frac{1}{1+m^{2}} \begin{bmatrix} 1 & m\\ -m & 1 \end{bmatrix} \end{align*}

Finally, if we denote $(0,1)$ by $v$, one has that \begin{align*} [T]_{\mathcal{B}} = \frac{1}{1+m^{2}} \begin{bmatrix} 1-m^{2} & 2m\\ 2m & m^{2}-1 \end{bmatrix} \Rightarrow [Tv]_{\mathcal{B}} = [T]_{\mathcal{B}}[v]_{\mathcal{B}} = \left(\frac{2m}{m^{2}+1},\frac{m^{2}-1}{m^{2}+1}\right) \end{align*}

Hopefully this helps.