1. Find the equations of the vertical and horizontal asymptotes.

The vertical asymptote occurs where the denominator is zero:
$x - 1 = 0 \implies x = 1$

The horizontal asymptote is found by examining the behaviour as $x \to \pm\infty$, or by polynomial division:
$y = \frac{x-2}{x-1} = \frac{x-1-1}{x-1} = 1 - \frac{1}{x-1}$
As $x \to \pm\infty$, $\frac{1}{x-1} \to 0$, so the horizontal asymptote is:
$y = 1$

2. Determine the exact coordinates of the intersections with the axes.

For the $x$-axis intersection, substitute $y = 0$:
$0 = \frac{x-2}{x-1} \implies x - 2 = 0 \implies x = 2$
Coordinates are $(2, 0)$.

For the $y$-axis intersection, substitute $x = 0$:
$y = \frac{0-2}{0-1} = 2$
Coordinates are $(0, 2)$.

3. Sketch the curve, clearly labelling the asymptotes and intercepts.

[Sketch of a hyperbola with dashed lines for asymptotes $x = 1$ and $y = 1$.
The top-left branch passes through the $y$-intercept $(0, 2)$ and approaches both asymptotes.
The bottom-right branch passes through the $x$-intercept $(2, 0)$ and approaches both asymptotes.]

1. Identify the equation of the line to be drawn.

To solve $\frac{x-2}{x-1} \le 2-x$ graphically, we use the sketched curve $C$ with equation $y = \frac{x-2}{x-1}$.
We must draw the additional line $y = 2-x$ on the same axes.

2. Determine the points of intersection between the curve and the line.

The line $y = 2-x$ has an $x$-intercept at $(2, 0)$ and a $y$-intercept at $(0, 2)$.
These coordinates exactly match the axes intercepts of the curve $C$ found in part (i), meaning the line intersects the curve exactly at the points $(0, 2)$ and $(2, 0)$.

3. Use the graph to deduce the solution to the inequality.

The inequality $\frac{x-2}{x-1} \le 2-x$ is satisfied where the curve $C$ lies below or on the line $y = 2-x$.
Observing the relative positions of the graphs and noting the vertical asymptote at $x = 1$:

• For $x \le 0$, the curve is below or on the line.
• For $0 < x < 1$, the curve is above the line.
• For $1 < x \le 2$, the curve is below or on the line (excluding the asymptote $x = 1$).
• For $x > 2$, the curve is above the line.

Thus, the required range of values is $x \le 0 \text{ or } 1 < x \le 2$.

1. Concisely find the solution to the inequality from part (ii).

From part (ii), we solve $\frac{x-2}{x-1} \le 2-x$:
$\frac{x-2}{x-1} - (2-x) \le 0$
$\frac{x-2 - (2-x)(x-1)}{x-1} \le 0$
$\frac{x(x-2)}{x-1} \le 0$
The critical values are $x=0$, $x=1$ (asymptote), and $x=2$. Evaluating the signs in the intervals between these values yields the solution for part (ii):
$x \le 0 \quad \text{or} \quad 1 < x \le 2$

2. Apply the substitution $x \to |x|$ into the solution from part (ii).

The inequality in (iii) is identical to (ii) except $x$ is replaced by $|x|$. Thus, we substitute $|x|$ into our solution:
$|x| \le 0 \quad \text{or} \quad 1 < |x| \le 2$

3. Solve the resulting modulus inequalities.

For the first condition, since $|x| \ge 0$ for all real $x$:
$|x| \le 0 \implies |x| = 0 \implies x = 0$

For the second condition, solving the compound modulus inequality $1 < |x| \le 2$ yields:
$-2 \le x < -1 \quad \text{or} \quad 1 < x \le 2$

Combining these conditions gives the final solution set:
$-2 \le x < -1 \quad \text{or} \quad x = 0 \quad \text{or} \quad 1 < x \le 2$