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

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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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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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New post 05 Feb 2018, 19:36
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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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Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html


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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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New post 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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New post 05 Feb 2018, 21:35
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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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New post 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..
Kudos for your observation.
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.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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New post 05 Feb 2018, 23:54
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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
IMG_8406.jpg [ 1.38 MiB | Viewed 1172 times ]


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

Answer (C)
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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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New post 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 \((3\sqrt{2})^2 = 18\)

Answer (C)


Hi Karishma,

Thank you for the explanation. How did you determine the side of the square???
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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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New post 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 \((3\sqrt{2})^2 = 18\)

Answer (C)


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

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New post 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   [#permalink] 26 Feb 2018, 23:30
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