We are given a geometric transformation $f$ defined by the equations: $x' = y + 1$ $y' = x - 1$ We need to show that $f$ is an isometry and determine the nature and characteristic elements of $f$.

1. Problem Description

We are given a geometric transformation $f$ defined by the equations:

$x' = y + 1$

$y' = x - 1$

We need to show that $f$ is an isometry and determine the nature and characteristic elements of $f$.

2. Solution Steps

First, we need to show that $f$ is an isometry. An isometry is a transformation that preserves distances. Let $A(x_1, y_1)$ and $B(x_2, y_2)$ be two points in the plane. Let $A'(x_1', y_1')$ and $B'(x_2', y_2')$ be their images under $f$.

The distance between $A$ and $B$ is given by:

$AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

The coordinates of $A'$ and $B'$ are:

$x_1' = y_1 + 1$

$y_1' = x_1 - 1$

$x_2' = y_2 + 1$

$y_2' = x_2 - 1$

The distance between $A'$ and $B'$ is given by:

$A'B' = \sqrt{(x_2' - x_1')^2 + (y_2' - y_1')^2}$

$A'B' = \sqrt{((y_2 + 1) - (y_1 + 1))^2 + ((x_2 - 1) - (x_1 - 1))^2}$

$A'B' = \sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2}$

$A'B' = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Therefore, $AB = A'B'$. Since $f$ preserves distances, $f$ is an isometry.

Now, we need to determine the nature and characteristic elements of $f$. The transformation is given by:

$x' = y + 1$

$y' = x - 1$

To find the fixed points, we set $x' = x$ and $y' = y$:

$x = y + 1$

$y = x - 1$

Substituting $x = y + 1$ into $y = x - 1$, we get:

$y = (y + 1) - 1$

$y = y$

This means there are infinitely many fixed points. The fixed points lie on the line $y = x - 1$.

Let us rewrite the transformation as follows:

$x' = y + 1$

$y' = x - 1$

$x' - 1 = y$

$y' + 1 = x$

If we rotate the coordinates by $\pi/4$ so that $X=\frac{x-y}{\sqrt{2}}$ and $Y=\frac{x+y}{\sqrt{2}}$.

Solving for $x$ and $y$: $x = \frac{X+Y}{\sqrt{2}}$ and $y = \frac{Y-X}{\sqrt{2}}$.

Then $x' = \frac{Y-X}{\sqrt{2}}+1$ and $y' = \frac{X+Y}{\sqrt{2}}-1$.

So $\frac{x'-y'}{\sqrt{2}} = \frac{\frac{Y-X}{\sqrt{2}}+1-\frac{X+Y}{\sqrt{2}}+1}{\sqrt{2}}=\frac{-2X+2}{\sqrt{2}\sqrt{2}}=1-X$ and

$\frac{x'+y'}{\sqrt{2}} = \frac{\frac{Y-X}{\sqrt{2}}+1+\frac{X+Y}{\sqrt{2}}-1}{\sqrt{2}}=\frac{2Y}{\sqrt{2}\sqrt{2}}=Y$

So in the new coordinate system, we have

$X' = 1-X$

$Y' = Y$.

Then $f$ is a reflection about the line $X=1/2$. So $\frac{x-y}{\sqrt{2}} = \frac{1}{2}$ or $x-y = \frac{\sqrt{2}}{2}$.

Consider the line $x = y+1$ and $y = x-1$. Then consider the line $x=y+1 = c$. So $x-y = 1$.

The reflection happens across a line that is invariant under $f$. This line is $y = x-1$ and the direction vector $(1, 1)$.

We can consider the midpoint transformation to find the invariant line. $\frac{x+x'}{2} = \frac{x+y+1}{2}$ and $\frac{y+y'}{2} = \frac{y+x-1}{2}$. Then $x = x+y+1$ and $y=x+y-1$.

$f$ is a glide reflection.

3. Final Answer

$f$ is an isometry.

$f$ is a glide reflection along the line $y = x - 1$.