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# In the xy-coordinate plane, line l and line k intersect at

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In the xy-coordinate plane, line l and line k intersect at [#permalink]

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25 Jul 2009, 11:23
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In the xy-coordinate plane, line l and line k intersect at the point (4, 3). Is the product of their slopes negative?

(1) The product of the x-intercepts of lines l and k is positive.
(2) The product of the y-intercepts of lines l and k is negative.

Source: GMATPrep
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25 Jul 2009, 12:44
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Statment (1) is not sufficient ....plz refer image attached......possibility 2 will lead us to -ve product where as other two are +ve products of slopes....hence insufficient

Statment (2) Sufficient ...see image attached

Hence B....lemme know if its right

If u think this is right....Kudos plz...trying to access tests
Attachments

Stat 2.GIF [ 5.32 KiB | Viewed 909 times ]

Stat 1.GIF [ 8.88 KiB | Viewed 913 times ]

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Bhushan S.
If you like my post....Consider it for Kudos

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25 Jul 2009, 21:38
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Thanks for your post. You have explained really well, but unfortunately OA is C. Anyway kudos for such a great explanation , after going through your post I was able to solve the problem.

Just posting my explanation for your reference:
Attachment:

Untitled.gif [ 9.1 KiB | Viewed 896 times ]

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26 Jul 2009, 05:43
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Good question...what is the source?

But I am not sure if this kind of question comes on real GMAT.
Because if you combine 1 and 2 then there is one case which is not possible at all !! i.e. if both x-intercepts are negative then both y-intercepts SHOULD definitely be positive. Hence these two conditions CANNOT be combined.

Only if the two x-intercepts are positive, we get multiple options of which the only possibility is to have one negative slope and one positive slope.
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26 Jul 2009, 06:24
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Economist wrote:
Good question...what is the source?

But I am not sure if this kind of question comes on real GMAT.
Because if you combine 1 and 2 then there is one case which is not possible at all !! i.e. if both x-intercepts are negative then both y-intercepts SHOULD definitely be positive. Hence these two conditions CANNOT be combined.

Only if the two x-intercepts are positive, we get multiple options of which the only possibility is to have one negative slope and one positive slope.

Source GMATPrep, OA is C
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26 Jul 2009, 07:21
The above graphic has left out one possible case when considering Statement 2 here. One line certainly must have a negative y-intercept, and therefore a positive slope. The other line has a positive y-intercept. If that intercept is high, as in the diagram above, this line will have a negative slope. However, if the y-intercept is close to the origin (say at y=1), the line can have a positive slope. From S2 alone, we can't determine whether the product of the slopes will be positive or negative.

edit - sorry, I missed bigoyal's solution above, which is perfect.
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28 Jul 2009, 07:52
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there is a easy solution for this, with out graphs..

considering the equations
Y1 = m1X1 + K1 --> 1
Y2 = m2X2 + K2 ---> 2

(1) The product of the x-intercepts of lines l and k is positive.
X-intercept y =0
hence $$K1K2/M1M2 > 0$$ NOT SUFF

(2) The product of the y-intercepts of lines l and k is negative.
K1K2 < 0

NOT SUFF

1& 2
sure M1M2 < 0 so hence SUFF, I dint use the point (4,3) though.
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28 Jul 2009, 11:04
That sounds very precise.
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28 Jul 2009, 16:02
This is one crazy question man! Kudos to all of you who tackled it.
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28 Jul 2009, 20:43
whatthehell wrote:
This is one crazy question man! Kudos to all of you who tackled it.

ya, I agree. And moreover since its from GMATPrep, we cannot ignore questions of this complexity
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Re: product of slopes   [#permalink] 28 Jul 2009, 20:43
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