Coordinate plane.

[goyalsau]

If equation ||x/2|| + ||y/2|| = 5 encloses a certain region on the coordinate plane, what is the area of this region?

50

20

400

100

200

OA - 200
I Don't know How to approach this question.
If you know then please share it

[limestone]

First, simplify the equation a little bit

||x/2|| + ||y/2|| = 5 or |x/2| + |y/2| =5
or |x| + |y| =10

Now, list all the cases:

x,y>0 : x+y = 10 or y = -x +10 (I)
x<0;y>0: -x+y = 10 or y = x+10 (II)
x<0;y<0: -x -y = 10 or y = -x -10 (III)
x>0; y<0: x-y=10 or y = x-10 (IV)

As these four line are derived from the same absolute equation, then they must form a square.
To see it clearer, plug in some values and draw out all those lines in the coordinate plane.
And doing so you will find that each side of this square is 10* sqrt(2), then the area must be 200.

[neerajkumar1_1]

according to me
the equation |x/2| + |y/2| = 5 represents a equation of a circle
(original equation of a circle is x^2 + y^2 = r^2)

so here
at y=0
|x/2|=5
hence x/2=+5 or -5
therefore x= 10 or -10
u will get the same values for y
hence it will make a circle with a radius 10

area would be pi * r^2 = 3 *100 = 300
which is not there in the answer ..

well thats a start... i guess somebody can figure out what i am missing...

[neerajkumar1_1]

First, simplify the equation a little bit

||x/2|| + ||y/2|| = 5 or |x/2| + |y/2| =5
or |x| + |y| =10

Now, list all the cases:

x,y>0 : x+y = 10 or y = -x +10 (I)
x<0;y>0: -x+y = 10 or y = x+10 (II)
x<0;y<0: -x -y = 10 or y = -x -10 (III)
x>0; y<0: x-y=10 or y = x-10 (IV)

As these four line are derived from the same absolute equation, then they must form a square.
To see it clearer, plug in some values and draw out all those lines in the coordinate plane.
And doing so you will find that each side of this square is 10* sqrt(2), then the area must be 200.

i guess limestone got it right.. u can ignore my reply...

[limestone]

I have just uploaded this image, and sorry for my bad drawing.

And if you want to know why (I): y=-x+10 is perpendicular to (II) y=x+10 use the coefficients of x in both lines.
The product of their x's coeffecients is -1, which said that they are perpendicular to each other.

Side's length = Sqrt ( 10^2 + 10^2) = 10*sqrt(2)[/img][/url]

[manish.jaipatna]

Since the absolute value of x+y is given as 10 why can't be the length of any side of square is 10 and area is 100. Lets take following values

x=2 y =8 (|x|+|y|=10) - I
x=-2 y =8 (|x|+|y|=10) - II
x=-2 y =-8 (|x|+|y|=10) - III
x=2 y =-8 (|x|+|y|=10) - IV

then each side has length 10 & square is 100.[/quote]