# Question regarding calculation of a system of two linear equations for finding the equation of a reflection in a line

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.

• It’s no wonder you’re confused since the final solution in that answer you linked is incorrect. It’s easy to verify that the point doesn’t lie on the perpendicular through $(x_0,y_0)$ as it should.
– amd
Apr 12, 2020 at 5:00
• Incidentally, there’s a more illuminating way to write the resulting formula for the reflection: $(x_0,y_0)-2{ax_0+by_0+c\over a^2+b^2}(a,b)$.
– amd
Apr 12, 2020 at 5:02

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.)
• Thanks for the answer! Just to be sure, the answer that I wrote for $y_1$ is incorrect right? I had a hunch because I simply couldn't make out the answer no matter what I did.
• @Seankala - you are correct up to the point $(a^2 + b^2)y_1 = -2abx_0 + (a^2 - b^2)y_0 -2bc$ Apr 12, 2020 at 3:40