So as not to bury the lead, the answer is (A), namely that the range of $(x-1)/(y-2)$ will be the interval $(-1/3, 1/3)$. The solution essentially consists of two parts.
First part: finding the domain.
We want to find the points $(x, y)$ for which $ |x-1|+|y+1| < 1 $ If your geometric intuition is better than mine, you might observe that in three dimensions the region in question is the projection on the $x$-$y$ plane of intersection of a horizontal trough (the $|y+1|$ part) and a vertical trough (the $|x-1|$ part). A less brilliant way is to consider the four possible cases for the absolute values:
- $x\ge 1, y\ge -1\text{, so }|x-1|+|y+1| = x+y$
- $x\ge 1, y< -1\text{, so }|x-1|+|y+1| = x-y-2$
- $x< 1, y\ge -1\text{, so }|x-1|+|y+1| = -x+y+2$
- $x< 1, y< -1\text{, so }|x-1|+|y+1| = -x-y$
In the first case, the domain will be bounded by $x\ge1,y\ge1,x+y<1$. This is the interior (and the left and bottom edges) of a triangle with vertices $(1,0),(1,-1),(2,-1)$. Each of the other cases also triangular and the union of all four regions is the interior of a square with vertices $(1,0),(2, -1),(1, -2),(0,-1).$
Second part: bounding the function.
We now need to find the upper and lower bounds on the values of $ f(x, y)=\frac{x-1}{y-2} $ on the square we found in the first part. For any constant $c\ne0$, the level curves, $f(x,y)=c$, will be the lines $ y=\frac{1}{c}\;x+\left(\frac{2c-1}{c}\right) $ and if $c=0$ the level curve will be the vertical line $x=1$. Graph a few of these lines and you'll notice that all the level lines will intersect at the point $(1, 2)$ (which isn't a possible point on $f(x,y)$, but we don't care about that, since it's outside of our square).
Now all you have to do is look at the $c$ values for which these lines intersect the square. It's very easy to see that we're looking at $-1/3 so we have our answer. Here's a picture (with apologies in advance to the ten percent or so of the color-blind viewers out there).
