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If equation |x/2| + |y/2| = 5 enclose a certain region

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If equation |x/2| + |y/2| = 5 enclose a certain region [#permalink]

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New post 20 Jun 2015, 03:10
jayanthjanardhan wrote:
Hi GMATInight,

thanks a lot for the expalnantion. I get the logic now.

However, kindly refer to this link below.

in-the-x-y-plane-the-area-of-the-region-bounded-by-the-86549-40.html#p1540033

Its a similar problem, but the diagram we end getting is a square and not a rhombus...what am i missing here?


Even that is a square but never forget that a Square is a specific type of Rhombus only

I hope, You can understand that the Product of the slopes of the adjacent sides is -1 in that fugure which proves the angle between the adjacent sides as 90 degree

a Square is a "Rhombus with all angles 90 degrees". So calling it a Rhombus won;t be wrong either but you are right about the figure being a Square.
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Re: If equation |x/2| + |y/2| = 5 enclose a certain region [#permalink]

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New post 23 Sep 2015, 00:35
I think there is a pattern for this sort of problem.

1. x<0 and y<0 --> −x−y=10 --> y=−10−x;

2. x<0 and y≥0 --> −x+y=10 --> y=10+x;

3. x≥0 and y<0 --> x−y=10 --> y=x−10;

4. x≥0 and y≥0 --> x+y=10 --> y=10−x;

From above, we see that 1 & 4 are parallel, since they have the same slope, as this would be the case for 2 & 4.
From above, we see that intersection points on x and y-axis is -10 and 10.

If you get to this far, you know it is a sqaure.

Afterwards, as written by many ppl, it is just finding an area of a square, 200.
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Re: If equation |x/2| + |y/2| = 5 enclose a certain region [#permalink]

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New post 09 Jun 2018, 20:38
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Re: If equation |x/2| + |y/2| = 5 enclose a certain region   [#permalink] 09 Jun 2018, 20:38

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