The area bounded by the curves |x + y| = 1 and|x - y| = 1 is

Question

The area bounded by the curves |x + y| = 1 and |x - y| = 1 is

A. 3
B. 4 
C. 2
D. 1
E. None

Bunuel · Accepted Answer

The area bounded by the curves |x + y| = 1, |x - y| = 1 is A. 3 B. 4 C. 2 D. 1 E. None |x+y|=1 represents two lines: x+y=1 and x+y=-1 --> y=1-x and y=-1-x. Find the x and y intercept of these lines to plot; |x-y|=1 represents two lines: x-y=1 and x-y=-1 --> y=x-1 and y=x+1. Find the x and y intercept of these lines to plot; Notice that these lines are mirror images of each other. Here is a square you get when you plot them: Area.gif Notice that the diagonal of this square is equal to 2 (the difference between x intercepts). Area of a square is diagonal^2/2=2^2/2=2. Answer: C. Check similar questions 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 area-of-region-126117.html Hope it helps.

fivezero7 · Answer

hi anni, there is no need to plot it, once you have mastered the art of visualization. the way i did it is as under. take x+y=1 it intersects the axes at (1,0) and (0,1) and makes a right angled triangle with the axes with each side (other than the hypotenuse) as 1 hence the area of this triangle is 0.5*1*1 = 0.5 sq. units. all other lines are symmetrical and form 3 more congruent triangles with the axes at different points. so the total area will be 0.5+0.5+0.5+0.5 = 2 sq units. hope i am clear

fivezero7 · Answer

hi,
these equations represent four different equations
x+y=1
x+y=-1
x-y=1
x-y=-1
once you plot these on the graph, you will easily find the area under these equations.

just for the record, the answer shall be 2.

anni · Answer

thank you for the reply, is it required to plot these lines and then calculate?
can we get directly i mean is there any formula to solve this problem?
thanks

Cmplkj123 · Answer

One more method:

solve following equns:

X+Y=1
-X-y=1
X-y=1
-X+y=1

Out put X= +_ 1
Y=+_1

Plot the values of X and Y on graph, you will see the square

Now use pythagoras thm to find diagonal,which will be the side of that square.

= Sqaure root 2

square it and ans will be 2

mbaiseasy · Answer

(1) Derive all equations

|x + y| = 1
eq1: x + y = 1
eq2: x + y = -1

|x - y| = 1
eq3: x - y = 1
eq4: y - x = 1

(2) Plot your graph using x=0 and y=0.

eq1: 0,1 and 1,0
eq2: 0,-1 and -1,0
eq3: 0-1 and 1,0
eq4: 0,-1 and -1,0

(3) You will recognize a region that is a square with a diagonal of 2
(4) Calculate the area.

diagonal = side *  \sqrt{2} 
side =  \frac{2}{\sqrt{2}} 
side =  \sqrt{2}

Area =  side^2  =  \sqrt{2}^2  =  2

For detailed solutions for other similar problems.  https://burnoutorbreathe.blogspot.com/2012/12/absolute-values-solving-for-area-of.html

Bunuel · Answer

Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

Theory on Abolute Values: math-absolute-value-modulus-86462.html

DS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=37
PS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=58

Hard set on Abolute Values: inequality-and-absolute-value-questions-from-my-collection-86939.html

anik19890 · Answer

why only (0,1) (1,0) should only pick? for x+y=1, if i chose (10,-9) then it still become 1, can you please explain me what exactly the method of chosing number while finding area for this type of math

Icecream87 · Answer

Hi Bunuel, just a question, I am getting confused with this. Can't we just find all the solutions for x and y by using just one equation? 
If we take the first equation: 
|x + y| = 1
and say that y = 0 then x = 1 or -1
and if x = 0 then y =1 and -1

Would that be incorrect? And if that's the case, what is the difference between this equation |x + y| = 1 and this one |x|+|y|=1
Thanks a lot

mvictor · Answer

x+y=1 -> y=-x+1 -> slope -1.
x+y=-1 -> y=-x-1 -> slope -1. we have 2 parallel lines.

x-y=1 -> y=x-1 => slope 1.
x-y=-1 -> y=x+1 -> slope 1. -> we have another 2 parallel lines.

i simply drew the lines, and for the sake of getting the image, try x=1 then y=1 for all the equations.
i got a square like shape, with the points of intersection at (-1; 0); (0; 1); (1; 0); (0; -1)
since the diagonal of the square is the x axis, and it has a length of 2, we can apply the 45-45-90 triangle rule, and see that 2=x*sqrt(2), where x is the side of the square.
the side of the square is sqrt(2). now, we need to find the area -> sqrt(2) squared is equal to 2.

mvictor · Answer

you would find more points on THE LINE, but we are asked for the area of the figure when the 4 lines intersect.
the 4 points of intersection are as shown in the figure in bunuel's post.

Madhavi1990 · Answer

I have a query with the wording of the question 
it says : The area  bounded by the curves  |x + y| = 1, |x - y| = 1 is 
A. 3
B. 4 
C. 2
D. 1
E. None

Based on answers here it appears we consider them as four straight lines (not curves). Curves usually are written by x^2 + y^2 format; but since it said curves I thought it would be circular--> so in case its a non - square in the equation;I assume it to be lines not curves?? Will keep in mind in case I come across such language/ format later

sahilvijay · Answer

Answer is C:

An easy question instead <700 probably

|x+y|=1 , gives 2 equations x+y = + 1 and x+y =-1
|x-y| =1 gives x-y=1 and x-y = -1 
we know they will form an enclosed figure on x and y axis.
so directly find the points = put x=0 and y=0 
we get (0,1) (0,-1) (1,0) (-1,0)
when u actually plot the distance between opposite points are = 2 on each diogonal which cut at 90 degrees at O , origin
there fore

A = 0.5 D1D2 = 0.5x2x2 = 2 which is C

rahul16singh28 · Answer

The total Area =  4*\frac{1}{2}*1*1

baraa900 · Answer

the same question!! that's confusing!

Bunuel · Answer

Let me asks you: are we told that x and y are integers? Even if we were told that, is y=0 and x=+/-1 the only solution?

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

The area bounded by the curves |x + y| = 1 and|x - y| = 1 is

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

The area bounded by the curves |x + y| = 1 and|x - y| = 1 is

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards