Q: Find the area defined by $1 < |x-2|+|y+1| < 2$
After trying a lot, I asked my friend to solve this and she got the correct answer (which is 6) by shifting the origin to (2,-1) and then finding the area.
I don't really understand what she did. Shouldn't the limits 1 and 2 change when I shift the origin? I am really confused.
Is there any alternate way to solve this?
On
Each of the two inequalities defines the area outside or inside a square with sides titled by $\pi/4$ with respect to the $x$ and $y$ axes.
The intersection is the shaded region
The length of the sides of the big (small) square is $2\sqrt{2}$ ($\sqrt{2}$), so the total area is
$$ A = (2\sqrt{2})^2-\sqrt{2}^2=6. $$
On
Here is why your friend's method is a valid method.
You are looking for the area of a region, let's call it region $A$, defined by $1 < \lvert x-2 \rvert + \lvert y+1 \rvert < 2$.
Now define another region, call it $B$, by $1 < \lvert x \rvert + \lvert y \rvert < 2$.
Now consider the translation $T: (x,y) \mapsto (x-2,y+1)$. If $(x,y)$ is in $A$, then $1 < \lvert x-2 \rvert + \lvert y+1 \rvert < 2$, which implies that $(x-2,y+1)$ is in $B$. Moreover, if $(x-2,y+1)$ is in $B$ then $(x,y)$ is in $A$.
So $B$ is just the image of $A$ under the translation $T$. And since a translation preserves area, $B$ has the same area as $A$, and we can find the area of $A$ by measuring the area of $B$.
On
No, the limits do not need to change.
Suppose you take any "center" point $(a,b)$. You can measure the distance from any $(a,b)$ to any other point $(x,y)$ by $\text{distance}(x,y) = |x-a| +|y-b|$. This is not the usual way of measuring distance, of course, but it has many of the same properties. So the points where $\text{distance}(x,y) < 2$ form a region centered at $(a,b)$. Actually, this region is a diamond shape (a square rotated 45 degrees) as the other answer shows. But, the point is that this region has the same area no matter what center ($a,b)$ we use. Changing $(a,b)$ changes the location of our diamond, but the size of the region is defined by the "$2$", so it remains fixed.
In fact the diamond defined by $\text{distance}(x,y) < 2$ has area $8$.
The diamond defined by $\text{distance}(x,y) < 1$ has area $2$.
The area we want is just the larger area minus the smaller one, so it's $6$.
You can solve it directly if you wish, but it is not exactly easy. In fact, the problem could be solved using graphs.
The boundaries of the region in question are | x - 2 | + | y - 1 | = 1 and | x - 2 | + | y - 1 | = 2.
Let's take the latter one first. | y - 1 | = 2 - | x - 2 | when y - 1 = 2 - | x - 2 | OR when y - 1 = | x - 2 | - 2.
y - 1 = 2 - | x - 2 | simplifies to y = 3 - | x - 2 |.
y - 1 = | x - 2 | - 2 simplifies to y = | x - 2 | - 1.
By examining the slopes of the lines (1 and -1) that form the enclosed region indicated by the two graphs, you will observe that it is a square. The diagonals of the square measure 4 units each, so that its area is 4*4/2 = 8 square units.
By an identical process you will find that the other "side" of the three-sided inequality gives a square region whose area is 2*2/2 = 2 square units.
The area that you are looking for sits inside the two squares. Its area is 8 - 2 = 6 square units.
If you use a calculator utility such as desmos to look at the region, then the fact that the region lies between two squares will be clear.