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Re: In which quadrant of the coordinate plane does the point
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27 Mar 2013, 02:37
In which quadrant of the coordinate plane does the point (x, y) lie?
(1) |xy| + x|y| + |x|y + xy > 0
Case quadrant I (x,y)=(+,+)
\(|xy| + x|y| + |x|y + xy\)
\(xy +xy + xy + xy >0\)
The first term is positive, the second the third and the fourt also. The sum of 4 positive integers is >0. so quadrant I is possible
Case quadrant II (x,y)=(-,+)
The first term is positive(as always will be), the second is negative, the third is positive, the fourth is negative
\(|(-x)y| + (-x)|y| + |(-x)|y + (-x)y\)
\(xy -xy + xy - xy =0\) and not >0 so quadrant II is not possible
Case quadrant III (x,y)=(+,-)
The first term is positive(as always will be), the second is positive, the third is negative, the fourth is negative
\(|x(-y)| + x|(-y)| + |x|(-y) + x(-y)\)
\(xy + xy - xy - xy =0\) not >0 so quadrant III is not possible
Case quadrant IV (x,y)=(-,-)
The first term is positive(as always will be), the second is negative, the third is negative, the fourth is positive
\(|(-x)(-y)| + (-x)|(-y)| + |(-x)|(-y) + (-x)(-y)\)
\(xy -xy - xy + xy =0\) and not >0 so quadrant IV is not possible
SUFFICIENT
(2) -x < -y < |y|
\(|y|>-y\)
case y>0
\(y>-y\)
\(y>0\)
case y<0
\(-y>-y\)
So \(y>0\) must be the case here
We know that -y >-x and that y>0, we can sum these elements
\(-y+y>-x+0\)
\(0>-x\)
\(x>0\)
and given that x>0 and that y>0 the point is in the first quadrant
SUFFICIENT