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It's much easier to work toward an explanation when you know the right answer, heh.

From the first part of the question, we can be sure that the sign (+/-) of x has to equal the sign of y since (-x,y) and (-y,x) are in the same quadrant.

(1) Tells us that whatever the signs of s and t are, they have to be same (otherwise > 0 doesn't hold). But this means (-s,t) could be in one of two quadrants. (A is out, thus so is D)

(2) Tells us that the sign of t and x must be the same. Taken alone, there's not much you can do because you don't know anything about s. (B is out)

Taken together: Since sign of s = t and sign of x = t: sign of t = s = x, and furthermore sign of y = x = t = s (from the question) All the signs of the variables are the same. So (-s,t) will be in the same quadrant of (-x,y), and therefore in the same quadrant of (-y, x) (as would (-t, s). --- Answer: C

Without the OA I probably would taken much longer to work that out and maybe gotten wrong, so thanks!

If xy does not equal 0 and points (-x,y) ans (-y,x) are in the same quadrant of the xy-plane, is point (-s, t) in the same quadrant?

(1) st > 0 (2) xt > 0

Answer is C , However I prefer a different approach to solve this problem.

Use actual number to solve the probelm.

Given premise claim that sign of x and y must be same. So consider X= 5 and Y = 4. (-5,-4) and (-4,5) are in II quadrant. Consider X = -7 and Y = -8, (7,-8) and (8,-7) are in IV quadrant.

Statement 1 : Sign of s and t are same. Consider S = 2 , t = 3 , (-2, 3) is in II quadrant Consider S = -2, t = -3 ,( 2, -3) is in IV quadrant.

Statement 2: Sign of x and t are same. When X is + , t must be +. When X in -, t must be -.

Togather, we can say (-x,y) , (-y,x) and (-s, t) are in same quadrant : II or IV

Re: If xy does not equal 0 and points (-x,y) ans (-y,x) are in [#permalink]

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13 Oct 2013, 15:53

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jlgdr wrote:

marcusaurelius wrote:

If xy does not equal 0 and points (-x,y) ans (-y,x) are in the same quadrant of the xy-plane, is point (-s, t) in the same quadrant?

(1) st > 0 (2) xt > 0

OA= C

Please don't leave OA naked. Then it makes it hard for us to solve such a nice problem in real conditions. Cheers, J

If ab different from 0 and points (-a,b) and (-b,a) are in the same quadrant of the xy-plane, is point (-x,y) in the same quadrant?

The fact that points \((-a,b)\) and \((-b,a)\) are in the same quadrant means that \(a\) and \(b\) have the same sign. These points can be either in II quadrant, in case \(a\) and \(b\) are both positive, as \((-a,b)=(-,+)=(-b,a)\) OR in IV quadrant, in case they are both negative, as \((-a,b)=(+,-)=(-b,a)\) ("=" sign means here "in the same quadrant").

Now the point \((-x,y)\) will be in the same quadrant if \(x\) has the same sign as \(a\) (or which is the same with \(b\)) AND \(y\) has the same sign as \(a\) (or which is the same with \(b\)). Or in other words if all four: \(a\), \(b\), \(x\), and \(y\) have the same sign.

Note that, only knowing that \(x\) and \(y\) have the same sign won't be sufficient (meaning that \(x\) and \(y\) must have the same sign but their sign must also match with the sign of \(a\) and \(b\)).

(1) \(xy>0\) --> \(x\) and \(y\) have the same sign. Not sufficient. (2) \(ax>0\) --> \(a\) and \(x\) have the same sign. But we know nothing about \(y\), hence not sufficient.

(1)+(2) \(x\) and \(y\) have the same sign AND \(a\) and \(x\) have the same sign, hence all four \(a\), \(b\), \(x\), and \(y\) have the same sign. Thus point \((-x,y)\) is in the same quadrant as points \((-a,b)\) and \((-b,a)\). Sufficient.

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