直线 $y = x - 4$ 与曲线 $y = \frac{2}{x - 3}$ 相交于点 $A$ 和点 $B$。 求 $A B$ 的长度。

1. Set up an equation to find the $x$-coordinates of the intersection points

$x - 4 = \frac{2}{x - 3}$

2. Rearrange into a standard quadratic equation

$(x - 4)(x - 3) = 2$
$x^2 - 7x + 12 = 2$
$x^2 - 7x + 10 = 0$

3. Solve the quadratic equation for $x$

$(x - 2)(x - 5) = 0$
$x = 2 \text{ or } x = 5$

4. Find the corresponding $y$-coordinates

When $x = 2$, $y = 2 - 4 = -2$.
When $x = 5$, $y = 5 - 4 = 1$.
The points of intersection are $(2, -2)$ and $(5, 1)$.

5. Calculate the length of the line segment $AB$

$AB = \sqrt{(5 - 2)^2 + (1 - (-2))^2}$
$AB = \sqrt{3^2 + 3^2}$
$AB = \sqrt{18} = 3\sqrt{2} \approx 4.24$