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M06#5 Absolute value

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Senior Manager
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M06#5 Absolute value [#permalink]  23 Jan 2011, 19:37
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If equation |X| + |Y|= 5 encloses a certain region on the coordinate plane, what is the area of that region?

5
10
25
50
100

[Reveal] Spoiler:
The end points of the square are (0, 5) (5, 0) (0, -5) (-5, 0); each side of the square equals and the area of the figure is that expression squared or 50. You can also find the area of a square by multiplying its diagonals by each other and dividing by 2. For example: .

How do we know that the area is square and not something else ?
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Re: M06#5 Absolute value [#permalink]  24 Jan 2011, 02:27
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Expert's post
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shrive555 wrote:
If equation |X| + |Y|= 5 encloses a certain region on the coordinate plane, what is the area of that region?

5
10
25
50
100

[Reveal] Spoiler:
The end points of the square are (0, 5) (5, 0) (0, -5) (-5, 0); each side of the square equals and the area of the figure is that expression squared or 50. You can also find the area of a square by multiplying its diagonals by each other and dividing by 2. For example: .

How do we know that the area is square and not something else ?

Below is the the region we get by joining X and Y intercepts which are (0, 5), (0, -5), (5, 0), (-5, 0):
Attachment:

MSP254019e3df2hii74417g00000g7c2817hhbe0a3b.gif [ 2.86 KiB | Viewed 2204 times ]
Now, diagonals of the rectangle are equal (10 and 10), and also are perpendicular bisectors of each other (as the are on X and Y axises), so it must be a square. Area of a square equals to diagonal^2/2=10^2/2=50.

Similar question:
cmat-club-test-question-m25-101963.html
graphs-modulus-help-86549.html

Hope it helps.
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Re: M06#5 Absolute value [#permalink]  25 Jan 2011, 08:51
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Thanks B

whats the difference between |x+y| + |x-y| = 4 -----(A) and |x|+|y| =5 ------- (B)

For B we find the area through diagonal and for A we simply take one side and square it to get the area.
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Re: M06#5 Absolute value [#permalink]  25 Jan 2011, 09:01
1
KUDOS
Expert's post
shrive555 wrote:
Thanks B

whats the difference between |x+y| + |x-y| = 4 -----(A) and |x|+|y| =5 ------- (B)

For B we find the area through diagonal and for A we simply take one side and square it to get the area.

You can find the area of a square in either way: $$area_{square}=side^2=\frac{diagonal^2}{2}$$.

The difference between the square given by |x+y|+|x-y|=4 and the square given by |x|+|y| =5 is that the first one is "horizontal" square and the second one is the square turned by 45 degrees.
|x+y|+|x-y|=4:
Attachment:

MSP86219e41c890f2h195700001928g4634f4e6h4h.gif [ 1.79 KiB | Viewed 2165 times ]

|x|+|y| =5:
Attachment:

MSP232619e41b4bhdg1dh6e00003fi9762d3fb98db7.gif [ 2.74 KiB | Viewed 2164 times ]

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Re: M06#5 Absolute value [#permalink]  26 Feb 2011, 15:40
Bunuel, why cant we assume other values for |x| + |y| = 5

it could also be |-3| + |8|...please advise. i may have a concept wrong
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Ajit

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Re: M06#5 Absolute value [#permalink]  26 Feb 2011, 16:42
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ajit257 wrote:
Bunuel, why cant we assume other values for |x| + |y| = 5

it could also be |-3| + |8|...please advise. i may have a concept wrong

We are not assuming the values for x and y. We are expanding |x| + |y| = 5 to get the figure.

If x>0 and y>0 we'll have x+y=5;
If x>0 and y<0 we'll have x-y=5;
If x<0 and y>0 we'll have -x+y=5;
If x<0 and y<0 we'll have -x-y=5;

So we have 4 lines (4 linear equations which will give 4 segments) when drawn these lines give the square shown in my original post. But there is an easier way to get this figure: get X and Y intercepts (which are (0, 5), (0, -5), (5, 0), (-5, 0)) and join them.

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Re: M06#5 Absolute value [#permalink]  26 Feb 2011, 16:45
Bunuel...you're the best....kudos to you
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m06#5 [#permalink]  22 Apr 2012, 05:26
If equation |x|+|y|=5 encloses a certain region on the coordinate plane, what is the area of that region?

A) 5
B) 10
C) 25
D) 50
E) 100

I thought the equation somehow resembled that of a circle centered around the origin. After plugging in some value:

e.g x=2 then y=3 or y=-3
x=1 y=4 or -4

I figured that the region might be a circle with a radius of 5, but the explanation says that the figure is a square. Why can't this be a circle?
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Re: m06#5 [#permalink]  22 Apr 2012, 05:31
Expert's post
BN1989 wrote:
If equation |x|+|y|=5 encloses a certain region on the coordinate plane, what is the area of that region?

A) 5
B) 10
C) 25
D) 50
E) 100

I thought the equation somehow resembled that of a circle centered around the origin. After plugging in some value:

e.g x=2 then y=3 or y=-3
x=1 y=4 or -4

I figured that the region might be a circle with a radius of 5, but the explanation says that the figure is a square. Why can't this be a circle?

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Re: m06#5 [#permalink]  22 Apr 2012, 05:44
Bunuel wrote:
BN1989 wrote:
If equation |x|+|y|=5 encloses a certain region on the coordinate plane, what is the area of that region?

A) 5
B) 10
C) 25
D) 50
E) 100

I thought the equation somehow resembled that of a circle centered around the origin. After plugging in some value:

e.g x=2 then y=3 or y=-3
x=1 y=4 or -4

I figured that the region might be a circle with a radius of 5, but the explanation says that the figure is a square. Why can't this be a circle?

Yes, why can't it be a circle?
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Re: m06#5 [#permalink]  22 Apr 2012, 05:52
Expert's post
BN1989 wrote:
Bunuel wrote:
BN1989 wrote:
If equation |x|+|y|=5 encloses a certain region on the coordinate plane, what is the area of that region?

A) 5
B) 10
C) 25
D) 50
E) 100

I thought the equation somehow resembled that of a circle centered around the origin. After plugging in some value:

e.g x=2 then y=3 or y=-3
x=1 y=4 or -4

I figured that the region might be a circle with a radius of 5, but the explanation says that the figure is a square. Why can't this be a circle?

Yes, why can't it be a circle?

Well it cannot be a circle because it's a square. The reason why it's a square is given in previous posts (you can also see a diagram there).

Next, in an xy plane, the circle with center (a, b) and radius r is the set of all points (x, y) such that:
$$(x-a)^2+(y-b)^2=r^2$$. There is no resemblance whatsoever between this equation and the equation given in the question.

Hope it's clear.
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Re: M06#5 Absolute value [#permalink]  14 Apr 2014, 21:04
Hi Bunuel,

I understood your concept of equations and plotting points. But, i have a small doubt.

Why can't the graph in the second question (|x|+|y| =5 )be a horizontal square than a square turned by 45 degrees in the first question (|x+y|+|x-y|=4)

My question is there any reason why in the second question we have plotted a graph which is turned by 45 degrees rather than a horizontal square.

Help is appreciated.

Bunuel wrote:
shrive555 wrote:
Thanks B

whats the difference between |x+y| + |x-y| = 4 -----(A) and |x|+|y| =5 ------- (B)

For B we find the area through diagonal and for A we simply take one side and square it to get the area.

You can find the area of a square in either way: $$area_{square}=side^2=\frac{diagonal^2}{2}$$.

The difference between the square given by |x+y|+|x-y|=4 and the square given by |x|+|y| =5 is that the first one is "horizontal" square and the second one is the square turned by 45 degrees.
|x+y|+|x-y|=4:
Attachment:
MSP86219e41c890f2h195700001928g4634f4e6h4h.gif

|x|+|y| =5:
Attachment:
MSP232619e41b4bhdg1dh6e00003fi9762d3fb98db7.gif
Re: M06#5 Absolute value   [#permalink] 14 Apr 2014, 21:04
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