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In the xy plane, line l and line k intersect at the point

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Senior Manager
Joined: 08 Aug 2005
Posts: 251
In the xy plane, line l and line k intersect at the point [#permalink]

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27 Apr 2006, 23:56
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

In the xy plane, line l and line k intersect at the point (4,3). Is the product of their slopes negative?

1) The product of x intercepts of lines l and k is positive.

2) The product of y intercepts of lines l and k is negative.
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Joined: 20 Mar 2005
Posts: 201
Location: Colombia, South America
Re: Gmat Prep xy plane [#permalink]

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28 Apr 2006, 02:12
getzgetzu wrote:
In the xy plane, line l and line k intersect at the point (4,3). Is the product of their slopes negative?

1) The product of x intercepts of lines l and k is positive.

2) The product of y intercepts of lines l and k is negative.

from the main statement we know that the lines pass through the first quadrant

then what? I think is easier to solve if you draw a cartesian plane

from (1) the intercepts are either both positive or both negative

in that case if both -ve, then slopes would be positive
in that case if both +ve, then impossible to know, if they are at the right of point (4,3) would be -ve, at the left of that point positive, but one can at the right, the other one at the left, so I'd say (1) is insuff.

from (2) product of y intercepts are negative, so one of the intercepts is bigger than one and the other one smaller
for the one has negative intercept slope is always +ve, but we don't know for sure for the one that has +ve intercept, it depends if is lower or higher than point (4,3)

so insuff.

taking (1) and (2)

intercepts of x must be both positive,
the line with -ve y intercept would have a positive slope
the line with +ve y intercept would have a negative slope

=> I would say the answer is C

sorry, too wordy, a little graph would work much better
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Joined: 23 Jan 2006
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28 Apr 2006, 06:55
OA is A?
say line l has x-intercept of (1,0)
say line k has x-intercept of (350,0)
slope of l = (3-0)/(4-1) = 3
slope of k = (3-0)/(4-350) = -3/346
their products would be negative
now say k has x intercept of (2,0)
slope of k = (3-0)/(4-2) = 3/2
their products would be positive.

Is say E on this one....

actually, I change my answer to C. lines passng thru the point 4,3 will have y-intercept, x-intercept, slope
+,-,+
+,+,-
-,+,+

the only way to have the product of both intercepts both be positive, is to have the same combination of y-int, x-int, and slope, and the product of the slopes will always be positive.
C it is...

Last edited by kook44 on 28 Apr 2006, 07:12, edited 1 time in total.
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Joined: 23 Mar 2006
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28 Apr 2006, 06:56
Why do you say E sm176811
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Joined: 29 Apr 2003
Posts: 1403

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28 Apr 2006, 10:10
Loner wrote:
Why do you say E sm176811

I need to draw it out. I could draw lines where both the conditions satisfied, yet the slopes were of the same signs!
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Joined: 05 Jan 2006
Posts: 381

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28 Apr 2006, 22:14
I have seen this question discussed before, someone has very nice explaination of using y=mx+c

y1=m1x1+c1
y2=m2x2+c2

is m1*m2 negative

x intercept = -c/m
y intercept = c

1) product of x intercept positive so (-c1/m1)*(-c2/m2) > 0 can not really tell anything about m1 and m2

2) product of y intercept positive c1*c2<0 can not say anything about m1 and m2

but 1 and 2 togather c1c2/m1m2 > 0 and c1c2<0 so m1*m2<0

hence C is sufficient...
28 Apr 2006, 22:14
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