Find the slope of the tangent to the curve

The given curve is $y=\frac{x-1}{x-2}$.

$\therefore \frac{d y}{d x}=\frac{(x-2)(1)-(x-1)(1)}{(x-2)^{2}}$

$=\frac{x-2-x+1}{(x-2)^{2}}=\frac{-1}{(x-2)^{2}}$

Thus, the slope of the tangent at x = 10 is given by,

$\left.\left.\frac{d y}{d x}\right]_{x=10}=\frac{-1}{(x-2)^{2}}\right]_{x=10}=\frac{-1}{(10-2)^{2}}=\frac{-1}{64}$

Hence, the slope of the tangent at $x=10$ is $\frac{-1}{64}$.