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# 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 +  [#permalink]

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Updated on: 26 Oct 2017, 06:46
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62% (01:57) correct 38% (02:05) wrong based on 883 sessions

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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

Originally posted by papillon86 on 08 Nov 2009, 14:11.
Last edited by Bunuel on 26 Oct 2017, 06:46, edited 3 times in total.
Edited the question and added the OA
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Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

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08 Nov 2009, 14:34
17
20
papillon86 wrote:
In 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

Need help in solving equations involving Mod......
help?

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.
Attachment:

MSP17971c13h40gd024h6g10000466ge1e9df941i96.gif [ 1.86 KiB | Viewed 25094 times ]

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Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

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08 Nov 2009, 17:41
8
3
srini123 wrote:
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

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).

Hope it's clear.
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Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

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13 Dec 2012, 04:06
2
eaakbari wrote:
Quote:

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

Any absolute values such as |x| = 5 could mean that x = 5 or x = -5.

Derive both (-) and (+) possibilities.

For the problem: |x+y| + |x-y| = 4

We could derive two possibilities for |x+y| could be -(x+y) and (x+y)
We could derive two possibilities for |x-y| could be -(x-y) and (x-y)

This is the reason why we have 4 derived equations.

(x+y) + (x-y) = 4
(x+y) - (x-y) = 4
-(x+y) + (x-y) = 4
-(x+y) - (x-y) = 4

Just simplify those...

If you want more practice on this question type: http://burnoutorbreathe.blogspot.com/2012/12/absolute-values-solving-for-area-of.html
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Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

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21 May 2011, 05:46
1
|x-y| = x-y if x-y > 0

|x-y| = -(x-y) if x-y < 0

x+y > 0 => x > -y then x !> y

x+y + x - y = 4

x = 2

-x - y + x - y = 4 (if x < -y, then x !< y)

y = -2

x + y -x + y = 4

=> y = 2

-x-y + x - y = 4

=> y = -2

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Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

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27 Feb 2012, 23:42
1
devinawilliam83 wrote:
Hi,
Can this be solved by graphing. If yes .. how do we graph the equation with 2 mod parts

VeritasPrepKarishma wrote:
VinuPriyaN wrote:
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

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.

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:
Attachment:

Square.gif [ 1.86 KiB | Viewed 29933 times ]
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.
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Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

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15 Apr 2014, 05:01
1
dheeraj24 wrote:
Yeah karishma,

I totally agree with your explanation, but the point is, why couldn't we draw the slant lines for the points (2,0), (-2,0), (0,2) and (0,-2) instead of horizontal lines and consider the length of diagonal rather than length of side for the original question (|x+y| + |x-y| = 4).

Because you are asked the area of the region bounded by |x+y| + |x-y| = 4.
This equation gives you ONLY horizontal/vertical lines passing through points (2,0), (-2,0), (0,2) and (0,-2) such as x = 2, y = 2, x = -2, y = -2.

Note that x = 2 is the equation of a line (it is not a coordinate) which passes through point (2, 0) and is parallel to the y axis. Similarly, y = 2 is the equation of a line which is parallel to x axis and passes through the point (0, 2) and so on. I think you are taking x = 2 as a coordinate but that is not the case. A coordinate has a value for y too. x =2 is the equation of a line. It implies that x coordinate is always 2 and y can be anything. So all points lying on a line passing through x = 2 and parallel to y axis satisfy this criteria.
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Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

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15 Apr 2014, 09:16
1
PathFinder007 wrote:
Hi Karishma,

I am still not clear

in question |x/2| + |y/2| = 5 we are getting following cordinates.

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

and in question |x+y| + |x-y| = 4. we are having following cordinates
x=2
x=-2
y=2
y=-2

why we are drawing graph differently?

Thanks

For $$|x+y| + |x-y| = 4$$ the equations are:

$$x = 2$$;
$$y = 2$$;
$$y = -2$$;
$$x = -2$$.

For $$|\frac{x}{2}| + |\frac{y}{2}| = 5$$ the equations are:

$$y=-10-x$$;
$$y=10+x$$;
$$y=10-x$$;
$$y=x-10$$.

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Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

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20 Jun 2015, 03:53
1
jayanthjanardhan wrote:
Bunuel wrote:
srini123 wrote:
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

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).

Hope it's clear.

Hi bunnel,

How did u rhombus for this one and a square for the other one?...I got the limits for both the questions, but could not figure out they turn out to be a square and rhombus!...

Even that is a square but never forget that A Square is a specific type of Rhombus only

I hope, You can understand that the Product of the slopes of the adjacent sides is -1 in that figure which proves the angle between the adjacent sides as 90 degree

a Square is a "Rhombus with all angles 90 degrees". So calling it a Rhombus won't be wrong either but you are right about the figure being a Square.
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Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

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08 Nov 2009, 15:27
Bunuel wrote:
papillon86 wrote:
In 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

Need help in solving equations involving Mod......
help?

I've never seen such kind of question in GMAT before.

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.

Why cant we consider (4,0) and (0,4) as points on graph ? then area would be different... , right?
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Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

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08 Nov 2009, 15:39
srini123 wrote:
Why cant we consider (4,0) and (0,4) as points on graph ? then area would be different... , right?

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.
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Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

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08 Nov 2009, 16:58
Bunuel wrote:
srini123 wrote:
Why cant we consider (4,0) and (0,4) as points on graph ? then area would be different... , right?

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.

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
_________________
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Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

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08 Nov 2009, 20:23
Thanks Bunuel , once again wonderful explanation +1 Kudos..

have a good day...
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Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

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22 May 2011, 07:21
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
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Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

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22 May 2011, 08:38
VinuPriyaN wrote:
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

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.
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Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

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15 Feb 2012, 00:25
prashantbacchewar wrote:
In the X-Y plane, the area of the region bounded by the graph of |x + y| + |x – y| = 4 is
(1) 8
(2) 12
(3) 16
(4) 20
(5) 24

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.
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Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

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27 Feb 2012, 23:33
Hi,
Can this be solved by graphing. If yes .. how do we graph the equation with 2 mod parts

VeritasPrepKarishma wrote:
VinuPriyaN wrote:
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

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.
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Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

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05 Dec 2012, 23:17
(1) derive all equations from |x+y| + |x-y| = 4

x+y+x-y =4 ==> x=2
x+y-x+y =4 ==> y=2
-x-y+x-y =4 ==> y=-2
-x-y-x+y =4 ==> x=-2

(3) Notice you have formed a square region bounded by x=2, y=2, y=-2 and x=-2 lines
(4) Area = 4*4 = 16

For more detailed solutions for similar question types:
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Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

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13 Dec 2012, 04:26
Bunuel wrote:
srini123 wrote:
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

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).

Hope it's clear.

Hii Bunuel.
What is the best approach of finding the points of intersection in order to make the square.
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Re: In the x-y plane, the area of the region bounded by the graph of |x +  [#permalink]

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13 Dec 2012, 04:30
Marcab wrote:
Bunuel wrote:
srini123 wrote:
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

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).

Hope it's clear.

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

I'd say substituting x=0 and y=0 in the equations of lines and making a drawing.
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Re: In the x-y plane, the area of the region bounded by the graph of |x +   [#permalink] 13 Dec 2012, 04:30

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