In the x-y plane, the area of the region bounded by the graph of |x +

Question

In the x-y plane, the area of the region bounded by the graph of |x + y| + |x - y| = 4 is

A. 8
B. 12
C. 16
D. 20
E. 24

Bunuel · Accepted Answer

OK, there can be 4 cases: |x+y| + |x-y| = 4 A. x+y+x-y = 4 --> x=2 B. x+y-x+y = 4 --> y=2 C. -x-y +x-y= 4 --> y=-2 D. -x-y-x+y=4 --> x=-2 The area bounded by 4 graphs x=2, x=-2, y=2, y=-2 will be square with the side of 4 so the area will be 4*4=16. MSP17971c13h40gd024h6g10000466ge1e9df941i96.gif Answer: C

Bunuel · Answer

I think this one is different.

|\frac{x}{2}| + |\frac{y}{2}| = 5

After solving you'll get equation of four lines:

y=-10-x 
 y=10+x 
 y=10-x 
 y=x-10

These four lines will also make a square, BUT in this case the  diagonal   will be 20 so the  Area=\frac{20*20}{2}=200 . Or the  Side= \sqrt{200} , area=200.

If you draw these four lines you'll see that the figure (square) which is bounded by them is turned by 90 degrees and has a center at the origin. So the side will not be 20.

Also you made a mistake in solving equation. The red part is not correct. You should have the equations written above.

In our original question when we were solving the equation |x+y| + |x-y| = 4 each time x or y were cancelling out so we get equations of a type  x=some value  twice and  y=some value  twice. And these equations give the lines which are parallel to the Y or X axis respectively so the figure bounded by them is a "horizontal" square (in your question it's "diagonal" square). 
 https://gmatclub.com/forum/download/file.php?id=16459

Hope it's clear.

srini123 · Answer

Why cant we consider (4,0) and (0,4) as points on graph ? then area would be different... , right?

Bunuel · Answer

First of all we are not considering points separately, as we have X-Y plane and roots of equation will represent lines, we'll get the figure bounded by this 4 lines. The equations for the lines are:

x=2
x=-2
y=2
y=-2

This lines will make a square with the side 4, hence area 4*4=16.

Second: points (4,0) or (0,4) doesn't work for |x+y| + |x-y| = 4.

srini123 · Answer

Thanks Bunuel, I used similar method for a similar question and I got wrong answer
the question was

what is the area bounded by graph |x/2| + |y/2| = 5 ?

I got hunderd since
x=10
x=-10
y=10
y=-10

isnt the area 400 ? the answer given was 200, please explain

srini123 · Answer

Thanks Bunuel , once again wonderful explanation +1 Kudos..

have a good day...

VinuPriyaN · Answer

Given |x-y| + |x+y| = 4

I don't understand why can't |x-y| and |x+y| be 1 and 3 instead of 2 and 2! (which again equals 4)

Can any one please explain this to me?

Thanks & Regards,
Vinu

KarishmaB · Answer

Look at the solution given by Bunuel above. When you solve it, you get four equations.
One of them is x = 2 which means that x = 2 and y can take any value. If y = 1, |x-y| = 1 and |x+y| = 3. 
For different values of y, |x-y| and |x+y| will get different values. We are not discounting any of them.

Bunuel · Answer

Merging similar topics. Please ask if anything remains unclear. Some questions on the same subject to practice: m06-5-absolute-value-108191.html graphs-modulus-help-86549.html m06-q5-72817.html if-equation-encloses-a-certain-region-110053.html Hope it helps.

devinawilliam83 · Answer

Hi,
Can this be solved by graphing. If yes .. how do we graph the equation with 2 mod parts

Bunuel · Answer

Yes, it can be done by graphing. |x+y| + |x-y| = 4 can expand in four different wasy: A. x+y+x-y = 4 --> x=2 B. x+y-x+y = 4 --> y=2 C. -x-y +x-y= 4 --> y=-2 D. -x-y-x+y=4 --> x=-2 So you can draw all these four lines x=2, x=-2, y=2, y=-2 to get a square with the side of 4: Square.gif See more examples here: m06-5-absolute-value-108191.html graphs-modulus-help-86549.html m06-q5-72817.html if-equation-encloses-a-certain-region-110053.html Hope it helps.

Marcab · Answer

Hii Bunuel.
What is the best approach of finding the points of intersection in order to make the square.

Bunuel · Answer

I'd say substituting x=0 and y=0 in the equations of lines and making a drawing.

honchos · Answer

The side of the square can't be 4, instead its sqrt(8)

Bunuel · Answer

The side of the square IS 4: MSP39361d6ehgde6ie87a8800003827f7f92a367c60.gif

PathFinder007 · Answer

HI Bunnel,

Have one doubt on this.

Similar kind of question is posted on following link

https://gmatclub.com/forum/m25-q19-76535.html

why here we are doing different then defined on above link.

on the above link we have square of side 20 then why we are not getting ans as 20*20 = 400

Bunuel · Answer

In that link the square does NOT have the side of 10, it has the side of 10\sqrt{2} and the diagonal of 20: https://gmatclub.com/forum/download/file.php?id=16459 The figure from original question is different: https://gmatclub.com/forum/download/file.php?id=22999 Check this: in-the-x-y-plane-the-area-of-the-region-bounded-by-the-86549.html#p649401 Hope it helps.

PathFinder007 · Answer

If we add all the lines in this way. then its diagonal is 4 so its side should be 2root2.

In this case the area would be 8

KarishmaB · Answer

The square is formed by lines x = 2, y = 2, x = -2 and y = -2 from the mod equation given in the question. You cannot turn it the way you did because you made the square smaller when you turned it. The side of the square needs to be 4 and diagonal needs to be 4*root2. Instead when you turned it, you made the diagonal 4 and side 4/root2 = 2*root2. That is incorrect. 
You can turn the square but it will not cut the axis at 2 or -2. It will cut the axis at 2*root2 or -2*root2. Then the square remains the same area-wise.

In the x-y plane, the area of the region bounded by the graph of |x +

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In the x-y plane, the area of the region bounded by the graph of |x +

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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My Rewards