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Same Quadrant

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by hai1 » Sun Dec 27, 2009 3:12 pm
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Ans: C

My analysis:
1) XY can be >0 if X>0 & Y>0 e.g: (2,3) or if X<0 & Y<0 e.g.: (-2.-3)
So we cannot say (-X,Y) are in the same quadrant as (-a,b)

2)ax can be >0; if a>0 & x>0 or if a<0 & x<0
we cannot say be 2 as it doesn't give the complete picture.

Combining 1 & 2:
ax can be >0; if x & a>0; if x>0 in xy> 0, y should be >0 e.g.: (x, y) = (2,3); e.g.: (-x, y) = (-2,3)
ax can be >0; if x & a<0; if x<0 in xy> 0, y should be <0 e.g.: (x, y) = (-2,-3); e.g.: (-x, y) = (2,-3)

So my answer is E instead of the answer pointed by the source.

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by kevincanspain » Sun Dec 27, 2009 6:01 pm
I wonder whether you have lost sight of the question. You seem to think that you need to determine in which quadrant the point (-x,y) is in.

Note that if the two points (-a,b) and (-b,a) are in the same quadrant, ab > 0 and the two points are either in quadrant ii or iv.

(1) xy > 0 implies that (-x,y) is in either ii or iv NOT SUFF
(2) ax > 0 tells us nothing about the sign of y NOT SUFF

(T) If (-a,b) is in quadrant ii, -a < 0 and b > 0, Thus a > 0 , as is x (as per (2)). Therefore, y > 0 positive, as per (1). Thus (-x,y) is in quadrant ii as well. Likewise, (-x,y) is in quadrant iv if and only if (-a,b) is in quadrant iv. SUFF
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