How to find the reflection of a function about a straight line of the form $y = mx + c$? I know the reflection of $f(x)$ about the axis $y = c$ is $2c - f(x)$. I also know the reflection of $f(x)$ about the axis $x = c$ is $f(2c - x)$. But I don't know what's the reflection of $f(x)$ about the line $y = mx + c$. An example is illustrated below.

 A: Given: A function $y=f(x)$ from $\mathbb{R}$ into $\mathbb{R}$.
Find: The set $f^*$ of points consisting of the reflections of points of $f$ in the line $L$ defined by the equation $y=mx+c$.
Let $(x(t),y(t))\in f$ and let $(x^*(t),y^*(t)$ denote its reflection in the line $L$.
Then the line containing $(x(t),y(t))$ and $(x^*(t),y^*(t))$ has slope $-\dfrac{1}{m}$ and the midpoint of the line segment connecting the two points lies on $L$.
These two facts can be stated as equations.
\begin{eqnarray}
 \dfrac{y^*(t)-y(t)}{x^*(t)-x(t)}&=&-\dfrac{1}{m}\\
    \dfrac{y(t)+y^*(t)}{2}&=&m\left(\dfrac{x(t)+x^*(t)}{2}\right)+c
\end{eqnarray}
With a bit of algebra, these can be re-written as a system of two linear equations in $x^*(t)$ and $y^*(t)$.
\begin{eqnarray}
 x^*(t)+my^*(t)&=&x(t)+my(t)\\
 -mx^*(t)+y^*(t)&=&mx(t)-y(t)+2c
\end{eqnarray}
Using the elimination method and replacing $y(t)$ with $f(a)$ we find that
\begin{eqnarray}
 x^*(t)&=&-\dfrac{(m^2-1)x(t)-2m(y(t)-c)}{m^2+1} \\
 y^*(t)&=&\dfrac{(m^2-1)y(t)+2(mx(t)+c)}{m^2+1} 
\end{eqnarray}
Example: Let $y=x$ so that $m=1$ and $c=0$. Substituting into these last two equations gives the result
\begin{eqnarray}
x^*(t)&=&y(t)\\
y^*(t)&=&x(t)
\end{eqnarray}
which we know is correct since reflecting about $y=x$ inverts the function.
For a more complex example, find parametric equations for $y=x^2$ reflected about the line $y=-2x+1$.
Substituting into the equations and simplifying gives
\begin{eqnarray}
x^*(t)&=&-\frac{1}{5}\left(4t^2+3t-4\right)\\
y^*(t)&=&\frac{1}{5}\left(3t^2-4t+2\right)
\end{eqnarray}
Desmos graph at this link: 
https://www.desmos.com/calculator/hou5bqgopb

A: I will use the form $Ax+By=C$ for now. Anyway, to flip a point $(x_o,y_o)$ around the line $Ax+By=C$, first find $Ax_o+By_o$. Subtract that from $C$ to get a new constant. Add that new constant to $C$ to get $C_i$. Your new point will lie on the line $Ax+By=C_i$. Now find $Bx_o-Ay_o$, which will get you a constant $C_P$ which you can use to find the line perpendicular to the line you want to flip the point over that goes through the point you want to flip over. Solve the system of equations $Ax+By=C_i$ and $Bx-Ay=C_P$. The result is the point flipped over the line. Now, assuming that you do have a function of one variable in terms of another, you can substitute $y=f(x)$ and get the whole thing flipped over the line.
