Answer:
[tex]\displaystyle \left(\frac{-(m^{2}-1)\, x + 2\, m\, y - 2\, m \, c}{m^{2} + 1},\, \frac{(m^{2} - 1)\, y + 2\, m \, x + 2\, c}{m^{2} + 1}\right)[/tex].
Step-by-step explanation:
Consider the line that is perpendicular to [tex]y = m\, x + c[/tex] and goes through [tex](x,\, y)[/tex].
Both [tex](x,\, y)[/tex] and the reflection would be on this new line. Besides, the two points would be equidistant from the intersection of this new line and line [tex]y = m\, x + c[/tex].
Hence, if the vector between [tex](x,\, y)[/tex] and that intersection could be found, adding twice that vector to [tex](x,\, y)\![/tex] would yield the coordinates of the reflection.
Since this new line is perpendicular to line [tex]y = m\, x + c[/tex], the slope of this new line would be [tex](-1/m)[/tex].
Hence, [tex]\langle 1,\, -1/m\rangle[/tex] would be a direction vector of this new line.
[tex]\langle m,\, -1\rangle[/tex] (a constant multiple of [tex]\langle 1,\, -1/m\rangle[/tex] would also be a direction vector of this new line.)
Both [tex](x,\, y)[/tex] and the aforementioned intersection are on this new line. Hence, their position vectors would differ only by a constant multiple of a direction vector of this new line.
In other words, for some constant [tex]\lambda[/tex], [tex]\langle x,\, y \rangle + \lambda\, \langle m,\, -1 \rangle = \langle x + \lambda \, m,\, y - \lambda \rangle[/tex] would be the position vector of the reflection of [tex](x,\, y)[/tex] (the position vector of [tex](x,\, y)\![/tex] is [tex]\langle x,\, y \rangle[/tex].)
[tex]( x + \lambda \, m,\, y - \lambda )[/tex] would be the coordinates of the intersection between the new line and [tex]y = m\, x + c[/tex]. [tex]\lambda\, \langle m,\, -1 \rangle[/tex] would be the vector between [tex](x,\, y)[/tex] and that intersection.
Since that intersection is on the line [tex]y = m\, x + c[/tex], its coordinates should satisfy:
[tex]y - \lambda = m\, (x + \lambda \, m) + c[/tex].
Solve for [tex]\lambda[/tex]:
[tex]y - \lambda = m\, x + m^{2}\, \lambda + c[/tex].
[tex]\displaystyle \lambda = \frac{y - m\, x - c}{m^{2} + 1}[/tex].
Hence, the vector between the position of [tex](x,\, y)[/tex] and that of the intersection would be:
[tex]\begin{aligned} & \lambda\, \langle m,\, -1 \rangle \\= \; & \left\langle \frac{m\, (y - m\, x - c)}{m^{2} + 1},\, \frac{(-1)\, (y - m\, x - c)}{m^{2} + 1}\right\rangle \\ =\; &\left\langle \frac{-m^{2}\, x + m\, y - m\, c }{m^{2} + 1},\, \frac{-y + m\, x + c}{m^{2} + 1}\right\rangle \end{aligned}[/tex].
Add twice the amount of this vector to position of [tex](x,\, y)[/tex] to find the position of the reflection, [tex]\langle x,\, y \rangle + 2\, \lambda \,\langle m,\, -1 \rangle[/tex].
[tex]x[/tex]-coordinate of the reflection:
[tex]\begin{aligned} & x + 2\, \lambda\, m \\ = \; & x + \frac{-2\, m^{2}\, x + 2\, m \, y - 2\, m \, c}{m^{2} + 1} \\ =\; & \frac{-(m^{2} - 1) \, x + 2\, m \, y - 2\, m \, c}{m^{2} + 1}\end{aligned}[/tex].
[tex]y[/tex]-coordinate of the reflection:
[tex]\begin{aligned} & y + (-2\, \lambda)\\ = \; & y + \frac{- 2\, y + 2\, m\, x + 2\, c}{m^{2} + 1} \\ =\; & \frac{(m^{2} - 1) \, y + 2\, m \, x + 2\, m \, c}{m^{2} + 1}\end{aligned}[/tex].
