Conditions for an Affine Transformation to Be an Isometry
An affine transformation is an isometry if and only if the following conditions hold: $$ \begin{cases} a^2 + a'^2 = 1 \\ b^2 + b'^2 = 1 \\ ab + a'b' = 0 \end{cases} $$
Proof
Consider the general form of an affine transformation:
\[
\begin{cases}
x' = a x + b y + c \\ \\
y' = a' x + b' y + c'
\end{cases}
\]
For the transformation to be well-defined, the determinant of the linear part must be nonzero:
\[ \begin{pmatrix} a & b \\ a' & b' \end{pmatrix} \ne 0 \]
Let us take two arbitrary points in the plane, \( A(x_A, y_A) \) and \( B(x_B, y_B) \).
Their Euclidean distance is:
$$ \overline{AB} = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2} $$
$$ \overline{AB} = \sqrt{ (\Delta x)^2 + (\Delta y)^2 } $$
Under the affine transformation, the corresponding images are \( A'(x'_A, y'_A) \) and \( B'(x'_B, y'_B) \):
\[
\begin{cases}
x_A' = a x_A + b y_A + c \\ \\
y_A' = a' x_A + b' y_A + c'
\end{cases}
\]
\[
\begin{cases}
x_B' = a x_B + b y_B + c \\ \\
y_B' = a' x_B + b' y_B + c'
\end{cases}
\]
The Euclidean distance between their images is then:
$$ \overline{A'B'} = \sqrt{(x'_B - x'_A)^2 + (y'_B - y'_A)^2} $$
Substituting the expressions for \( x'_A, x'_B, y'_A, \) and \( y'_B \), we obtain:
$$ \overline{A'B'} = \sqrt{[a(x_B - x_A) + b(y_B - y_A)]^2 + [a'(x_B - x_A) + b'(y_B - y_A)]^2} $$
Letting \( \Delta x = x_B - x_A \) and \( \Delta y = y_B - y_A \) for simplicity, we can rewrite this as:
$$ \overline{A'B'} = \sqrt{[a \Delta x + b \Delta y]^2 + [a' \Delta x + b' \Delta y]^2} $$
Expanding the squared terms gives:
$$ \overline{A'B'} = \sqrt{a^2(\Delta x)^2 + 2ab(\Delta x \Delta y) + b^2(\Delta y)^2 + a'^2(\Delta x)^2 + 2a'b'(\Delta x \Delta y) + b'^2(\Delta y)^2} $$
Grouping similar terms, we get:
$$ \overline{A'B'} = \sqrt{(a^2 + a'^2)(\Delta x)^2 + (b^2 + b'^2)(\Delta y)^2 + 2(ab + a'b')(\Delta x \Delta y)} $$
For the transformation to be an isometry, the Euclidean distance between any two points must remain unchanged:
$$ \overline{AB} = \overline{A'B'} $$
That is,
$$ \sqrt{ (\Delta x)^2 + (\Delta y)^2 } = \sqrt{ (a^2 + a'^2)(\Delta x)^2 + (b^2 + b'^2)(\Delta y)^2 + 2(ab + a'b')(\Delta x \Delta y) } $$
This equality holds precisely when:
$$ \sqrt{ (\Delta x)^2 + (\Delta y)^2 } = \sqrt{ \underbrace{(a^2 + a'^2)}_{=1}(\Delta x)^2 + \underbrace{(b^2 + b'^2)}_{=1}(\Delta y)^2 + 2\underbrace{(ab + a'b')}_{=0}(\Delta x \Delta y) } $$
By comparing coefficients, we conclude that an affine transformation preserves Euclidean distance if and only if the following conditions are satisfied:
$$ \begin{cases} a^2 + a'^2 = 1 \\ b^2 + b'^2 = 1 \\ ab + a'b' = 0 \end{cases} $$
This completes the proof.
And so forth.
