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If equation |x|+|y|= 5 encloses a certain region on the

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Intern
Joined: 20 Jun 2007
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If equation |x|+|y|= 5 encloses a certain region on the [#permalink]

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05 Jun 2008, 17:45
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If equation |x|+|y|= 5 encloses a certain region on the graph, what is the area of that region?

a) 5
b) 10
c) 25
d) 50
e) 100

Can someone help explain this little bit more into detail? Would greatly be appreciated

The end point of the square are (0, 5) (5, 0) (0, -5) (-5, 0); each side of the square equals to $$5*\sqrt{2}$$ and the area of the figure is that expression squared or 50.

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Director
Joined: 10 Sep 2007
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Re: graph equation question [#permalink]

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05 Jun 2008, 17:52
|x| and |y| are always >= 0, so their minimum value can be 0.
As We have to satisfy the inequality |X| + |Y| = 5, so if one of them is 0 then other one has to be 5.
Taking Y as 0 and X as 5 or -5 give 2 coordinates (5,0), (-5,0)
Similarly if X is 0 then Y can be 5 or -5 give 2 coordinates (0,5), (0,-5)

You have 4 coordinates of this graph. Due to symmetry you can say that it is square and not rohombus.
Length of any side = Sqrt(25 + 25) = 5Sqrt(2)

So Area of Square = Side * Side = 5Sqrt(2) * 5Sqrt(2) = 50

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Intern
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Re: graph equation question [#permalink]

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05 Jun 2008, 18:18
can't it x & y also be 3 & 2 or 4 & 1. it would still satisfy the equation no?

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Manager
Joined: 21 Mar 2008
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Re: graph equation question [#permalink]

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05 Jun 2008, 22:04
dzelkas wrote:
can't it x & y also be 3 & 2 or 4 & 1. it would still satisfy the equation no?

3 and 2 and 4 and 1 are points on this line. But we are trying to see what kind of a figure it is and hence looking for intersection of the line with x (y=0) and y(x=) axis to give us reference coordinates.Thats how we come up with the coordinates as (0, 5) and (5, 0).
Similarly , for the other 3 equations( one each for a variation of the sign of x and y), we will have the other 2 corodinates(we will have common corodinates as well) and finally get the figure we intend to find the area of.
You actually dont need to go all the way. By the symmetry of the equation, you can figure out that it will be a square.

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Director
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Re: graph equation question [#permalink]

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05 Jun 2008, 22:36
vertex at (-5,0) (0,5) (5,0) (0,-5)
length = sqrt(50)
area = 50
D

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Re: graph equation question   [#permalink] 05 Jun 2008, 22:36
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