Intuitional explanation needed - Determining if a Point is in a turned square

Question

This should be viewed as a turned square.

Now, for the Point P(X,Y) why is it that if $|X|+|Y|\leq1$ then the Point is inside the square?

Answer 1

In comes down to the constraints of the lines. To be inside the square, a point $(x,y)$ must be on the proper side of each line that binds the square. For example, the line in the first quadrant is $x+y=1$, and all valid points are below this line. This means that smaller $y$ values are acceptable, and the inequality reads $y\leq-x+1$. The list of the four constraints for the square are

$$y\leq-x+1\;\;\;\;\mbox{(Quadrant 1)}$$

$$y\leq x+1\;\;\;\;\mbox{(Quadrant 2)}$$

$$y\geq -x-1\;\;\;\;\mbox{(Quadrant 3)}$$

$$y\geq x-1\;\;\;\;\mbox{(Quadrant 4)}$$

In each quadrant, $x$ and $y$ retain their signs; that is, once we specify the quadrant, we know what the signs of $x$ and $y$ are. For example, in quadrant 1, $x$ and $y$ are both positive, so $|x|=x$ and $|y|=y$. Another example is in quadrant 2, $|y|=y$ and $|x|=-x$. This allows us to change all of our equations to

$$|y|\leq-|x|+1\;\;\;\;\mbox{(Quadrant 1)}$$

$$|y|\leq -|x|+1\;\;\;\;\mbox{(Quadrant 2)}$$

$$-|y|\geq |x|-1\;\;\;\;\mbox{(Quadrant 3)}$$

$$-|y|\geq |x|-1\;\;\;\;\mbox{(Quadrant 4)}$$

All of these equations are the same. They all read $|x|+|y|\leq1$. I hope this helps!