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# In the x-y plane, find the area of the region enclosed by the graph of

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Status: Preparing for GMAT
Joined: 25 Nov 2015
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In the x-y plane, find the area of the region enclosed by the graph of  [#permalink]

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05 Feb 2018, 11:09
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Question Stats:

55% (02:43) correct 45% (02:37) wrong based on 120 sessions

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In the x-y plane, find the area (in sq. units) of the region enclosed by the graph of |x-1| + |y+2|=3.
A) 4.5
B) 9
C) 18
D) 36
E) 42

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In the x-y plane, find the area of the region enclosed by the graph of  [#permalink]

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05 Feb 2018, 19:36
souvonik2k wrote:
In the x-y plane, find the area (in sq. units) of the region enclosed by the graph of |x-1| + |y+2|=3.
A) 4.5
B) 9
C) 18
D) 36
E) 42

See attached figure..the point of intersection is (1,-2)
When x-1 is 0,
$$|y+2|=3.....y=3-2=1$$ and
$$y=-2-3=-5$$
So this becomes base $$1-(-5)=6$$

Now let y+2=0,
$$|x-1|=3.....x=4$$ and
$$x=1-3=-2$$

Two triangles with base 6..
Area of the one with altitude 4 = $$\frac{1}{2} *4*6=12$$
Area of other =$$\frac{1}{2} *6*2=6$$..

Total =$$12+6=18$$

C
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In the x-y plane, find the area of the region enclosed by the graph of  [#permalink]

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05 Feb 2018, 21:26
chetan2u wrote:
souvonik2k wrote:
In the x-y plane, find the area (in sq. units) of the region enclosed by the graph of |x-1| + |y+2|=3.
A) 4.5
B) 9
C) 18
D) 36
E) 42

See attached figure
When x is 0,
$$|y+2|=3.....y=3-2=1$$ and
$$y=-2-3=-5$$
So this becomes base $$1-(-5)=6$$

Now let y=0,
$$|x-1|=3.....x=4$$ and
$$x=1-3=-2$$

Two triangles with base 6..
Area of the one with altitude 4 = $$\frac{1}{2} *4*6=12$$
Area of other =$$\frac{1}{2} *6*2=6$$..

Total =$$12+6=18$$

C

Hi chetan2u,

Please correct me if I am wrong in my understanding.

Here, we have to take $$x - 1 = 0$$ (Instead of $$x = 0$$) & $$y + 2 = 0$$ (instead of $$y = 0$$).
When, $$x - 1 = 0, |y+2| = 3, y = 1,-5$$
$$y + 2 = 0, |x-1|= 3, x = 4,-2.$$
However, in this case the origin should be transformed to $$(1,-2)$$
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In the x-y plane, find the area of the region enclosed by the graph of  [#permalink]

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05 Feb 2018, 21:35
4
souvonik2k wrote:
In the x-y plane, find the area (in sq. units) of the region enclosed by the graph of |x-1| + |y+2|=3.
A) 4.5
B) 9
C) 18
D) 36
E) 42

Good Question souvonik2k.. +1 to you..!!

Let $$X = x - 1, Y = y + 2$$
We have, $$|X| + |Y| = 3.$$ Now, take $$X = 0, Y = 3$$. Take $$Y = 0, X = 3$$. Now, Draw a line from $$(0,3) & (3,0)$$. (Alternatively, You can also think of X intercept & Y intercept in Ist Quadrant).

So, area of triangle in first Quadrant will be $$A = \frac{1}{2}*3*3.$$
As this is mod, total area will be A = $$4*\frac{1}{2}*3*3 = 18$$
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Re: In the x-y plane, find the area of the region enclosed by the graph of  [#permalink]

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05 Feb 2018, 21:43
rahul16singh28 wrote:
chetan2u wrote:
souvonik2k wrote:
In the x-y plane, find the area (in sq. units) of the region enclosed by the graph of |x-1| + |y+2|=3.
A) 4.5
B) 9
C) 18
D) 36
E) 42

See attached figure
When x is 0,
$$|y+2|=3.....y=3-2=1$$ and
$$y=-2-3=-5$$
So this becomes base $$1-(-5)=6$$

Now let y=0,
$$|x-1|=3.....x=4$$ and
$$x=1-3=-2$$

Two triangles with base 6..
Area of the one with altitude 4 = $$\frac{1}{2} *4*6=12$$
Area of other =$$\frac{1}{2} *6*2=6$$..

Total =$$12+6=18$$

C

Hi chetan2u,

Please correct me if I am wrong in my understanding.

Here, we have to take $$x - 1 = 0$$ (Instead of $$x = 0$$) & $$y + 2 = 0$$ (instead of $$y = 0$$).
When, $$x - 1 = 0, |y+2| = 3, y = 1,-5$$
$$y + 2 = 0, |x-1|= 3, x = 4,-2.$$
However, in this case the origin should be transformed to $$(1,-2)$$

Hi..
Rahul, you are absolutely correct, the origin will shift to (1,-2)..
Although answer will remain the same as there are two perpendicular diagonals of length 6, so area = 1/2*6*6=18..
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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Re: In the x-y plane, find the area of the region enclosed by the graph of  [#permalink]

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05 Feb 2018, 23:54
2
souvonik2k wrote:
In the x-y plane, find the area (in sq. units) of the region enclosed by the graph of |x-1| + |y+2|=3.
A) 4.5
B) 9
C) 18
D) 36
E) 42

|x-1| + |y+2|=3

gives us 4 lines.

x > 1 and y > -2
(x - 1) + (y + 2) = 3
y = -x + 2

x < 1 and y > -2
-(x - 1) + (y + 2) = 3
y = x

x > 1 and y < -2
(x - 1) - (y + 2) = 3
y = x -6

x < 1 and y < -2
-(x - 1) - (y + 2) = 3
y = -x - 4

This is the figure the 4 lines enclose:
Attachment:

IMG_8406.jpg [ 1.38 MiB | Viewed 1392 times ]

They form a square with each side as $$3\sqrt{2}$$ so area of the figure will be $$(3\sqrt{2})^2 = 18$$

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Posts: 59
Re: In the x-y plane, find the area of the region enclosed by the graph of  [#permalink]

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25 Feb 2018, 03:47
VeritasPrepKarishma wrote:
souvonik2k wrote:
In the x-y plane, find the area (in sq. units) of the region enclosed by the graph of |x-1| + |y+2|=3.
A) 4.5
B) 9
C) 18
D) 36
E) 42

|x-1| + |y+2|=3

gives us 4 lines.

x > 1 and y > -2
(x - 1) + (y + 2) = 3
y = -x + 2

x < 1 and y > -2
-(x - 1) + (y + 2) = 3
y = x

x > 1 and y < -2
(x - 1) - (y + 2) = 3
y = x -6

x < 1 and y < -2
-(x - 1) - (y + 2) = 3
y = -x - 4

This is the figure the 4 lines enclose:
Attachment:
IMG_8406.jpg

They form a square with each side as $$3\sqrt{2}$$ so area of the figure will be $$(3\sqrt{2})^2 = 18$$

Hi Karishma,

Thank you for the explanation. How did you determine the side of the square???
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8406
Location: Pune, India
Re: In the x-y plane, find the area of the region enclosed by the graph of  [#permalink]

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26 Feb 2018, 04:43
Kezia9 wrote:
VeritasPrepKarishma wrote:
souvonik2k wrote:
In the x-y plane, find the area (in sq. units) of the region enclosed by the graph of |x-1| + |y+2|=3.
A) 4.5
B) 9
C) 18
D) 36
E) 42

|x-1| + |y+2|=3

gives us 4 lines.

x > 1 and y > -2
(x - 1) + (y + 2) = 3
y = -x + 2

x < 1 and y > -2
-(x - 1) + (y + 2) = 3
y = x

x > 1 and y < -2
(x - 1) - (y + 2) = 3
y = x -6

x < 1 and y < -2
-(x - 1) - (y + 2) = 3
y = -x - 4

This is the figure the 4 lines enclose:
Attachment:
IMG_8406.jpg

They form a square with each side as $$3\sqrt{2}$$ so area of the figure will be $$(3\sqrt{2})^2 = 18$$

Hi Karishma,

Thank you for the explanation. How did you determine the side of the square???

The length of the diagonals is 6 (from -2,-2 to 4, -2) which means half the diagonal is 3. Considering each side of the square to be the hypotenuse of a right triangle, since the two legs are 3 each, the hypotenuse will be $$3\sqrt{2}$$.
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Intern
Joined: 21 Aug 2016
Posts: 3
Re: In the x-y plane, find the area of the region enclosed by the graph of  [#permalink]

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26 Feb 2018, 23:30
Hi Chetan, I am confused as to how you did this problem. How did you manage to figure out the altitude is 4? Can you provide more color on how you did this? I tried to draw the figure but it takes a long time.

Best,

Jimmy
Re: In the x-y plane, find the area of the region enclosed by the graph of &nbs [#permalink] 26 Feb 2018, 23:30
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# In the x-y plane, find the area of the region enclosed by the graph of

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