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# DS - Coordinate geometry

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Manager
Joined: 12 Apr 2006
Posts: 218
Location: India
Followers: 1

Kudos [?]: 24 [0], given: 17

DS - Coordinate geometry [#permalink]  24 Jun 2009, 02:51
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
Attachment:

DS.jpg [ 13.07 KiB | Viewed 772 times ]
Senior Manager
Joined: 23 Jun 2009
Posts: 355
Location: Turkey
Schools: UPenn, UMich, HKS, UCB, Chicago
Followers: 5

Kudos [?]: 96 [0], given: 63

Re: DS - Coordinate geometry [#permalink]  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
Manager
Joined: 30 May 2009
Posts: 218
Followers: 3

Kudos [?]: 66 [0], given: 0

Re: DS - Coordinate geometry [#permalink]  24 Jun 2009, 18:40
I think A is the answer too. Can you please confirm the OA???
Manager
Joined: 12 Apr 2006
Posts: 218
Location: India
Followers: 1

Kudos [?]: 24 [0], given: 17

Re: DS - Coordinate geometry [#permalink]  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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