Summer is Coming! Join the Game of Timers Competition to Win Epic Prizes. Registration is Open. Game starts Mon July 1st.

 It is currently 15 Jul 2019, 23:22 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # In the x-y plane, find the area of the region enclosed by the graph of

Author Message
TAGS:

### Hide Tags V
Status: Preparing for GMAT
Joined: 25 Nov 2015
Posts: 1027
Location: India
GPA: 3.64
In the x-y plane, find the area of the region enclosed by the graph of  [#permalink]

### Show Tags

21 00:00

Difficulty:   75% (hard)

Question Stats: 52% (02:38) correct 48% (02:40) wrong based on 101 sessions

### HideShow timer Statistics 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

_________________
Please give kudos, if you like my post

When the going gets tough, the tough gets going...
Math Expert V
Joined: 02 Aug 2009
Posts: 7764
In the x-y plane, find the area of the region enclosed by the graph of  [#permalink]

### Show Tags

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
_________________
Director  P
Joined: 31 Jul 2017
Posts: 515
Location: Malaysia
GMAT 1: 700 Q50 V33 GPA: 3.95
WE: Consulting (Energy and Utilities)
In the x-y plane, find the area of the region enclosed by the graph of  [#permalink]

### Show Tags

1
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 $$(1,-2)$$
_________________
If my Post helps you in Gaining Knowledge, Help me with KUDOS.. !!
Director  P
Joined: 31 Jul 2017
Posts: 515
Location: Malaysia
GMAT 1: 700 Q50 V33 GPA: 3.95
WE: Consulting (Energy and Utilities)
In the x-y plane, find the area of the region enclosed by the graph of  [#permalink]

### Show Tags

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 = $$4*\frac{1}{2}*3*3 = 18$$
_________________
If my Post helps you in Gaining Knowledge, Help me with KUDOS.. !!
Math Expert V
Joined: 02 Aug 2009
Posts: 7764
Re: In the x-y plane, find the area of the region enclosed by the graph of  [#permalink]

### Show Tags

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..
_________________
Veritas Prep GMAT Instructor D
Joined: 16 Oct 2010
Posts: 9428
Location: Pune, India
Re: In the x-y plane, find the area of the region enclosed by the graph of  [#permalink]

### Show Tags

2
1
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 1934 times ]

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

_________________
Karishma
Veritas Prep GMAT Instructor

Manager  B
Joined: 04 Oct 2017
Posts: 81
Re: In the x-y plane, find the area of the region enclosed by the graph of  [#permalink]

### Show Tags

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 D
Joined: 16 Oct 2010
Posts: 9428
Location: Pune, India
Re: In the x-y plane, find the area of the region enclosed by the graph of  [#permalink]

### Show Tags

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 $$3\sqrt{2}$$.
_________________
Karishma
Veritas Prep GMAT Instructor

Intern  B
Joined: 21 Aug 2016
Posts: 3
Re: In the x-y plane, find the area of the region enclosed by the graph of  [#permalink]

### Show Tags

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
Non-Human User Joined: 09 Sep 2013
Posts: 11649
Re: In the x-y plane, find the area of the region enclosed by the graph of  [#permalink]

### Show Tags

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________ Re: In the x-y plane, find the area of the region enclosed by the graph of   [#permalink] 11 Jun 2019, 20:00
Display posts from previous: Sort by

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