Reflection (Euclidean Plane)
The set of reflections about a line in \(\mathbb{R}^{2}\) is a subgroup of the isometry group on \(\mathbb{R}^{2}\). Furthermore, it is a generator for that group.
Definition
More strictly, a reflection of a point \(P\) about a line \(l\) maps \(P\) to \(P'\) such that \(l\) is the perpendicular bisector of the line segment \(PP'\).
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We denote the reflection about \(l\) by \(\sigma_{l}\).
To find a formula for the reflection about a line \(l\), we can use two facts about the relationship between the starting point, its reflection, and the line.
Let \(P = \begin{bmatrix} x \\ y \end{bmatrix}\), the point we are trying to reflect about a line \(l = \{(x, y) \in \mathbb{R}^{2} : ax + by + c = 0\}\). Let \(P'\) be that reflection, given by \(P' = \begin{bmatrix} x' \\ y' \end{bmatrix}\).
Now, the first fact to note is that the vector from \(P\) to \(P'\) is perpendicular to the line \(l\), or zero. Since the line \(ax + by = 0\) is parallel to \(l\), and \((b, -a)\) lies on this line, we use that as the direction vector for our line, and hence:
\[\begin{bmatrix}
b \\
-a
\end{bmatrix}
\cdot
\begin{bmatrix}
x' - x \\
y' - y
\end{bmatrix}
=
b(x' - x) - a(y' - y)
= 0 \\\]
The other fact we can use is that the midpoint of \(PP'\) must lie on the line \(l\), and therefore:
\[ a\left(\frac{x + x'}{2}\right) + b \left(\frac{y + y'}{2}\right) + c = 0.\]
We then just need to solve these equations simultaneously for \(x'\) and \(y'\). Doing this in full generality, we start by writing the two equations as one vector equation, and then manipulate them to recover the points \(P\) and \(P'\) by expressing them in terms of matrices.
\[\begin{align*}
\begin{bmatrix}
b(x' - x) - a(y' - y) \\
a\left(\frac{x + x'}{2}\right) + b \left(\frac{y + y'}{2}\right) + c
\end{bmatrix}
&=
\begin{bmatrix}
0 \\
0
\end{bmatrix} \\
\begin{bmatrix}
bx' - bx - ay' + ay \\
ax + ax' + by + by' + 2c
\end{bmatrix}
&=
\begin{bmatrix}
0 \\
0
\end{bmatrix} \\
\begin{bmatrix}
bx' - ay' \\
ax' + by'
\end{bmatrix}
&=
\begin{bmatrix}
bx - ay \\
-ax - by - 2c
\end{bmatrix} \\
\begin{bmatrix}
b & -a \\
a & b
\end{bmatrix}
\begin{bmatrix}
x' \\ y'
\end{bmatrix}
&=
\begin{bmatrix}
b & -a \\
-a & -b
\end{bmatrix}
\begin{bmatrix}
x \\ y
\end{bmatrix}
-
\begin{bmatrix}
0 \\ 2c
\end{bmatrix} \\
\begin{bmatrix}
x' \\ y'
\end{bmatrix}
&=
\begin{bmatrix}
b & -a \\
a & b
\end{bmatrix}^{1}
\left(
\begin{bmatrix}
b & -a \\
-a & -b
\end{bmatrix}
\begin{bmatrix}
x \\ y
\end{bmatrix}
-
\begin{bmatrix}
0 \\ 2c
\end{bmatrix}
\right) \\
\begin{bmatrix}
x' \\ y'
\end{bmatrix}
&=
\frac{1}{a^{2} + b^{2}}
\begin{bmatrix}
b^{2} - a^{2} & -2ab \\
-2ab & a^{2} - b^{2}
\end{bmatrix}
\begin{bmatrix}
x \\ y
\end{bmatrix}
-
\frac{1}{a^{2} + b^{2}}
\begin{bmatrix}
2ac \\ 2bc
\end{bmatrix} \\
\end{align*}\]
Note that the inverse above exists, since if \(a = 0\) and \(b = 0\), the form \(ax + by + c = 0\) is not a line.
While calculating this formula may be useful in computing the reflection across many lines, often it is easiest to solve the system of equations without calculating the inverse.
Reflections are Isometries
Every reflection is an isometry.
Proof
The fact that a reflection is an isometry follows from the fact that the formula above is of the form \(A\vec{x} + \vec{b}\) where \(A\) is an orthogonal matrix. To verify this fact, we just take the inner product of the two columns, as well as the norm of each column, to show that the columns are ortholinear.
\[\begin{align*}
\begin{bmatrix}
b^{2} - a^{2} \\
-2ab
\end{bmatrix}
\cdot
\begin{bmatrix}
-2ab \\
a^{2} - b^{2}
\end{bmatrix}
&= -2ab(b^{2} - a^{2}) -2ab(a^{2} - b^{2}) \\
&= 2ab(a^{2} - b^{2}) -2ab(a^{2} - b^{2}) \\
&= 0
\end{align*}\]
\[\begin{align*}
\left\|
\frac{1}{a^{2} + b^{2}}
\begin{bmatrix}
b^{2} - a^{2} \\
-2ab
\end{bmatrix}
\right\|
&= \sqrt{
\left(
\frac{b^{2} - a^{2}}{a^{2} + b^{2}}
\right)^{2}
\left(
\frac{-2ab}{a^{2} + b^{2}}
\right)^{2}
} \\
&= \sqrt{
\frac{b^{4} - 2a^{2}b^{2} + a^{4} + 4a^{2}b^{2}}{a^{4} + b^{4} + 2a^{2}b^{2}}
} \\
&= \sqrt{
\frac{b^{4} + a^{4} + 2a^{2}b^{2}}{a^{4} + b^{4} + 2a^{2}b^{2}}
} \\
&= 1
\end{align*}\]
\[\begin{align*}
\left\|
\frac{1}{a^{2} + b^{2}}
\begin{bmatrix}
-2ab \\
a^{2} - b^{2} \\
\end{bmatrix}
\right\|
&= \sqrt{
\left(
\frac{-2ab}{a^{2} + b^{2}}
\right)^{2}
\left(
\frac{a^{2} - b^{2}}{a^{2} + b^{2}}
\right)^{2}
} \\
&= \sqrt{
\frac{4a^{2}b^{2} + b^{4} - 2a^{2}b^{2} + a^{4}}{a^{4} + b^{4} + 2a^{2}b^{2}}
} \\
&= \sqrt{
\frac{b^{4} + a^{4} + 2a^{2}b^{2}}{a^{4} + b^{4} + 2a^{2}b^{2}}
} \\
&= 1
\end{align*}\]
