Coordinate geometry

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Coordinate geometry

by nidhis.1408 » Thu Oct 18, 2012 3:42 pm
In which quadrant of the coordinate plane does the point (x, y) lie?

(1) |xy| + x|y| + |x|y + xy > 0
(2) -x < -y < |y|

This a manhattan problem and I am not able to understand the solution given.

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by Anurag@Gurome » Thu Oct 18, 2012 5:48 pm
nidhis.1408 wrote:In which quadrant of the coordinate plane does the point (x, y) lie?

(1) |xy| + x|y| + |x|y + xy > 0
(2) -x < -y < |y|

This a manhattan problem and I am not able to understand the solution given.
Statement 1: |xy| + x|y| + |x|y + xy > 0
As |xy| + x|y| + |x|y + xy > 0
-> |x||y| + x|y| + |x|y + xy > 0
-> |y|*(|x| +x) + y*(|x| + x) > 0
-> (|x| + x)*(|y| + y) > 0

This means (|x| + x) and (|y| + y) are of same sign and none of them is equal to zero.
Now, the values of (|x| + x) can never be negative because,
(1) |x| + x = x + x = 2x , for x positive
(2) |x| + x = -x + x = 0 , for x negative

Same for (|y| + y).
Therefore, (|x| + x) and (|y| + y) both are positive as none of them can be zero. This also means that x and y are positive too. So, the point (x, y) lies in the first quadrant.

Sufficient.

Statement 2: -x < -y < |y|
Observe that,
(1) For y positive, |y| = y. Thus, |y| > -y
(2) For y negative, |y| = -y. Thus, |y| is not greater than -y.

Therefore, for |y| to greater than -y, y must be positive.

Now, -x < -y. This implies x > y. As y is positive, x must be positive. The point (x, y) lies in the first quadrant.

Sufficient.

The correct answer is D.
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