This question is regarding a detail in the calculation for an answer in the question Find a formula in terms of $a$, $b$ and $c$, for reflection in an arbitrary line $ax+by+c=0 \in R^2$.
The posted answer gives us a system of two linear equations, and while it seems straightforward I believe I'm missing a detail in the calculation and can't figure it out.
The system is:
$$ \begin{cases} y_1 - y_0 = \dfrac{b}{a}(x_1 - x_0) \\ a\left(\dfrac{x_0 + x_1}{2}\right) + b\left( \dfrac{y_0 + y_1}{2} \right) + c = 0 \end{cases} $$
The first equation is motivated by the fact that if we have some point $P(x_0, y_0) \in \Bbb{R}^2$ then we want the line going through $P$ and its reflection $P'$ to be orthogonal to the line $ax + by + c = 0$.
The second equation is motivated by the fact that we want the midpoint between $P$ and $P'$ to lie on the original line.
My approach is to first change each equation so that we may add or subtract the two equations to solve for $x_1$ or $y_1$ first.
Equation 1:
$$ \begin{align} y_1 - y_0 & = \dfrac{b}{a}(x_1 - x_0) \\ ay_1 - ay_0 & = b(x_1 - x_0) \\ bx_1 - ay_1 & = bx_0 - ay_0 \\ abx_1 - a^2 y_1 & = abx_0 - a^2 y_0 \end{align} $$
Equation 2:
$$ \begin{align} a\left( \dfrac{x_0 + x_1}{2} \right) + b\left( \dfrac{y_0 + y_1}{2} \right) + c & = 0 \\ a(x_0 + x_1) + b(y_0 + y_1) + 2c & = 0 \\ ax_1 + by_1 & = -ax_0 -by_0 -2c \\ abx_1 + b^2y_1 & = -abx_0 - b^2y_0 -2bc \end{align} $$
So now we have:
$$ \begin{cases} abx_1 - a^2y_1 = \phantom{-}abx_0 - a^2y_0 \\ abx_1 + b^2y_1 = -abx_0 -b^2y_0 -2bc \end{cases} $$
subtracting the two results in:
$$ (a^2 + b^2)y_1 = -2abx_0 + (a^2 - b^2)y_0 -2bc $$
My confusion stems from the fact that the correct answer for $y_1$ is:
$$ y_1 = -y_0 - \dfrac{2bc}{a^2 + b^2} $$
I've tried backtracking and putting that:
$$ -2abx_0 + (a^2 - b^2)y_0 = -(a^2 + b^2)y_0 $$
and get the result:
$$ y_0 = \dfrac{b}{a}x_0 $$
but I don't know where this came from.
Could anybody point out where I went wrong? I feel like I simply missed something but can't seem to find it. Thanks in advance.
Your derivation is correct and the solution for $y_1$ is
$$ y_1 =\frac{ -2b(ax_0 +c)+ (a^2 - b^2)y_0}{a^2 + b^2} $$
Note that, if $P(x_0,y_0)$ is on the given line, i.e. $ax_0+c=-by_0$, substitute into above solution to get $y_1=y_0$, i.e. the reflection is itself as expected. (The answer given is incorrect.)