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DS - Coordinate geometry

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Manager
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DS - Coordinate geometry [#permalink]

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New post 24 Jun 2009, 02:51
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A
B
C
D
E

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Senior Manager
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Re: DS - Coordinate geometry [#permalink]

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New post 24 Jun 2009, 09:23
IMO A.
Line with a negative slope must intesect quadrant II.

This line must be y= -x/6 + m(intrecept)
in quadrant II, x values are negative and y values are positive.

So a line that intersects quadrant 2 must have points like that

y1=-x1/6 +m>0
-x1/6>-m (as x1 is negative -x1/6 must be positive)
So if m is positive than inequation holds, if m is negative one can find an -x1 value that is greater than -m for all values of -m.
Thus, 1) is sufficient

2 is unsufficient. e.g. line k => y=-6 or for any positive value of n line k => y=-6 + nx
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Re: DS - Coordinate geometry [#permalink]

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New post 24 Jun 2009, 18:40
I think A is the answer too. Can you please confirm the OA???
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Re: DS - Coordinate geometry [#permalink]

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New post 25 Jun 2009, 03:47
sdrandom1 wrote:
I think A is the answer too. Can you please confirm the OA???


Yes, It's correct, OA is A.
Re: DS - Coordinate geometry   [#permalink] 25 Jun 2009, 03:47
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