Question

Find the reflection of the point (x,y) in the line y=mx+c​

Answer 1

`\displaystyle \left((-(m^(2)-1)\, x + 2\, m\, y - 2\, m \, c)/(m^(2) + 1),\, ((m^(2) - 1)\, y + 2\, m \, x + 2\, c)/(m^(2) + 1)\right)`.

Explanation:

Consider the line that is perpendicular to
`y = m\, x + c` and goes through
`(x,\, y)`.

Both
`(x,\, y)` 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
`y = m\, x + c`.

Hence, if the vector between
`(x,\, y)` and that intersection could be found, adding twice that vector to
`(x,\, y)\!` would yield the coordinates of the reflection.

Since this new line is perpendicular to line
`y = m\, x + c`, the slope of this new line would be
`(-1/m)`.

Hence,
`\langle 1,\, -1/m\rangle` would be a direction vector of this new line.

`\langle m,\, -1\rangle` (a constant multiple of
`\langle 1,\, -1/m\rangle` would also be a direction vector of this new line.)

Both
`(x,\, y)` 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
`\lambda`,
`\langle x,\, y \rangle + \lambda\, \langle m,\, -1 \rangle = \langle x + \lambda \, m,\, y - \lambda \rangle` would be the position vector of the reflection of
`(x,\, y)` (the position vector of
`(x,\, y)\!` is
`\langle x,\, y \rangle`.)

`( x + \lambda \, m,\, y - \lambda )` would be the coordinates of the intersection between the new line and
`y = m\, x + c`.
`\lambda\, \langle m,\, -1 \rangle` would be the vector between
`(x,\, y)` and that intersection.

Since that intersection is on the line
`y = m\, x + c`, its coordinates should satisfy:

`y - \lambda = m\, (x + \lambda \, m) + c`.

Solve for
`\lambda`:

`y - \lambda = m\, x + m^(2)\, \lambda + c`.

`\displaystyle \lambda = (y - m\, x - c)/(m^(2) + 1)`.

Hence, the vector between the position of
`(x,\, y)` and that of the intersection would be:

`\begin{aligned} & \lambda\, \langle m,\, -1 \rangle \\= \; & \left\langle (m\, (y - m\, x - c))/(m^(2) + 1),\, ((-1)\, (y - m\, x - c))/(m^(2) + 1)\right\rangle \\ =\; &\left\langle (-m^(2)\, x + m\, y - m\, c )/(m^(2) + 1),\, (-y + m\, x + c)/(m^(2) + 1)\right\rangle \end{aligned}`.

Add twice the amount of this vector to position of
`(x,\, y)` to find the position of the reflection,
`\langle x,\, y \rangle + 2\, \lambda \,\langle m,\, -1 \rangle`.

`x`-coordinate of the reflection:

`\begin{aligned} & x + 2\, \lambda\, m \\ = \; & x + (-2\, m^(2)\, x + 2\, m \, y - 2\, m \, c)/(m^(2) + 1) \\ =\; & (-(m^(2) - 1) \, x + 2\, m \, y - 2\, m \, c)/(m^(2) + 1)\end{aligned}`.

`y`-coordinate of the reflection:

`\begin{aligned} & y + (-2\, \lambda)\\ = \; & y + (- 2\, y + 2\, m\, x + 2\, c)/(m^(2) + 1) \\ =\; & ((m^(2) - 1) \, y + 2\, m \, x + 2\, m \, c)/(m^(2) + 1)\end{aligned}`.