1. ## Absolute Value Question

I have a question here and I don't really understand what it means:

Find the area enclosed by the polygon whose equation is $||x|-2|+||y|-2|=4$

Would anybody be able to explain first, what it is asking, and second some pointers/hints? I haven't done absolute value before - I have looked it up although I haven't understood too much.

2. ## Re: Absolute Value Question

Originally Posted by BobtheBob
Find the area enclosed by the polygon whose equation is $||x|-2|+||y|-2|=4$

Would anybody be able to explain first, what it is asking, and second some pointers/hints?
Asking for hints is natural, but the fact that you don't understand what the question is asking is surprising. The set of points (x, y) satisfying the equation ||x| - 2| + ||y| - 2| = 4 is a polygon (plus one point (0, 0)). You need to find its area.

The idea of solving problems with absolute values is to break them into several cases where you can get rid of absolute values. Here there are four cases: (1) x >=0, y >=0; (2) x < 0, y >= 0; (3) x < 0, y < 0; (4) x >= 0, y < 0. By considering each of those cases, you rewrite |x| as either x or -x, and similarly for y.

I suggest you start with case (1), which has four subcases: (a) x - 2 >= 0, y - 2 >= 0; (b) x - 2 < 0, y - 2 >= 0; (c) x - 2 < 0, y - 2 < 0 and (d) x - 2 >= 0, y - 2 < 0. Find the set of points satisfying each of (a)—(d). In each of those subcases you can expand |x - 2| as either x - 2 or -(x - 2), and similarly for |y - 2|. After that, you don't have absolute values anymore, just equations of lines. Draw those lines, but only inside the area satisfying the corresponding subcase.

After you finish with case (1), note that if (x, y) satisfies the original equation, then so do (-x, y), (-x, -y) and (x, -y). Therefore, the polygon is symmetric with respect to the x- and y-axes. Thus, you can draw the parts corresponding to (2)—(4) just by reflecting the figure for (1) with respect to the axes.

3. ## Re: Absolute Value Question

Ok thanks very much for the help! I understand it far more now.
Thanks again.

4. ## Re: Absolute Value Question

The expression y - 2 changes sign when y = 2. So, above the horizontal line y = 2 there are points (x, y) with y > 2, i.e., y - 2 > 0, and below there are points (x, y) with y - 2 < 0. A similar thing happens for sloped lines, but this is not needed in this problem. In general, if you need need to find the area where a linear expression ax + by > 0, draw the line ax + by = 0. On one side of this line, ax + by > 0, on the other, ax + by < 0. You can take a concrete point on either side and calculate the value of ax + by to know which side is which.

In the picture below

the lines x = 2 and y = 2 break the plane into four areas. Areas are hatched so that the color of horizontal lines corresponds to the values of x - 2 and the color of vertical lines correspond to the values of y - 2. If the lines are blue, the corresponding expression is negative, and if they are read, the expression is positive.