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

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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?

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

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