Plotting $|x - y| \leq 1/4$ I'm trying to plot $|x - y| \leq 1/4$
I've reduced this to $|x - y| \leq 1/4$, and then deduced via plugging in points that there should be two lines, one with y-intercept 0.25 and another with y intercept -0.25. I'm a little bit lost at how to derive equations in terms of $y = mx + b$. I forget how to handle equations with the absolute value and inequality, and guidance would be most appreciated.
 A: We have: $|x-y| \leq \dfrac{1}{4} \iff -\dfrac{1}{4} \leq x-y \leq \dfrac{1}{4}\iff x-\dfrac{1}{4} \leq y \leq x+\dfrac{1}{4}$. Here is how you plot the solution region: First graph the line $y_1 = x-\dfrac{1}{4}$, and plot the line $y_2 = x+\dfrac{1}{4}$, and shade the space in between the two lines.
A: First plot x-y=0. The graph you are looking for is symmetric about this line. Now, plot x-y=.25 and y-x=.25 (Since abs(x-y) is same as abs(y-x)) Shade the region bounded by these two lines. It will be an infinite rectangle bisected by x-y=0.
To verify, use :
http://www.quickmath.com/webMathematica3/quickmath/graphs/inequalities/advanced.jsp
A: the boundaries of this region are $$x-y = \frac14, x- y = -\frac14.$$ the graph of these are two parallel lines $\sqrt2/4$  apart and partitions the plane into three regions. now, take test points $(-1,0), (0,0), (1,0)$ and evaluating $|x-1|,$  we find that only the region containing $(0,0)$ satisfies $|x-y| \le 1/4.$  
therefor the region $\{(x,y): |x-y| \le 1/4\}$ is the region between the parallel lines $x-y = \pm1/4$ containing the origin $(0,0).$
