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In the xycoordinate plane, line l and line k intersect at [#permalink]
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28 Oct 2008, 13:43
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In the xycoordinate plane, line l and line k intersect at the point (4,3). Is the product of their slopes negative?
(1) The product of the xintercepts of lines l and k is positive
(2) The product of the yintercepts of lines l and k is negative.
Please explain your answer! Thanks



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Re: DS: Geometry [#permalink]
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28 Oct 2008, 14:00
i get C..for a while i was debating if it should be E..
basically just draw the graph out..i think with X intercepts positive and y intercept negative, that means on the lines has a negative slope



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Re: DS: Geometry [#permalink]
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28 Oct 2008, 14:40
B ?
if the product of the y intercept is ve then the slopes will be opposite (for eg. one can consider them to be perpendicular for which m1*m2=1 )



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Re: DS: Geometry [#permalink]
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28 Oct 2008, 15:50
I am not sure if we have to use the point (4,3)
L is y1=m1x1 + b1 and K is y2= m2x2 + b2
(1) is saying b1b2/m1m2 is +ve. dont know if both are ve or +ve. Insuff
(2) is saying b1b2 is ve. Insuff about m1m2
together
b1b2/m1m2 = +ve and b1b2= ve. Hence m1m2 = ve
C??



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Re: DS: Geometry [#permalink]
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28 Oct 2008, 16:44
B
L is y1=m1x1 + b1 and K is y2= m2x2 + b2 Substituting (4,3) 3 = 4m1+ b1, 3 = 4m2+b2 b1=3/4m1, b2 = 3/4m2
1. X intercept for the equations is calculated when y = 0 b1/c1 * b2/c2 = 1  In Sufficient
2. Y intercept for the equations is calculated when x = 0 3/4m1* 3/4m2 = 1 m1*m2 = 16/9
Hence proves that the slopes are negative  Sufficient



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Re: DS: Geometry [#permalink]
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28 Oct 2008, 18:17
what if there is no Xaxis intercept and you have 2 lines with positive slopes, with one having a negative y intercept and the other a straight line..



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Re: DS: Geometry [#permalink]
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28 Oct 2008, 20:11
The OA here is C. The easiest way to solve this problem is by drawing a graph. When the question asks whether the product of the slope is negative, what the question is really asking is whether one line is going upward and the second line is going downward through the intersection of (4,3):
1) the product of the xintercept is positive. this means that both lines are touching the xaxes in quadrant 1. On the one hand, 1 line could be going upward through (4,3), while the second line goes downward through (4,3). On the other hand, both lines could be going upward through the point (4,3) not suff.
2) product of the y intercepts is negative. On the one hand, both lines could be going upward through the point (4,3), but 1 of the lines has the y intercept in the negative side, while the second line could still be going upward from just above the origin. On the other hand, 1 line could be going upward from the negative yintercept, while the second line could be going downward from above the yintercept 4. So not suff.
(1&2) when both the lines have a positive xintercept, while one of the yintercept is negative while the second is positive, then obviously, 1 is going upward, and the second is going downward, so the product of the slope is negative, and therefore Suff.
Answer is C



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Re: DS: Geometry [#permalink]
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28 Oct 2008, 20:35
LiveStronger wrote: B
L is y1=m1x1 + b1 and K is y2= m2x2 + b2 Substituting (4,3) 3 = 4m1+ b1, 3 = 4m2+b2
b1=3/4m1, b2 = 3/4m2
1. X intercept for the equations is calculated when y = 0 b1/c1 * b2/c2 = 1  In Sufficient
2. Y intercept for the equations is calculated when x = 0 3/4m1* 3/4m2 = 1 m1*m2 = 16/9
Hence proves that the slopes are negative  Sufficient How did you get the marked above?
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Re: DS: Geometry [#permalink]
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29 Oct 2008, 04:46
GMAT TIGER wrote: LiveStronger wrote: B
L is y1=m1x1 + b1 and K is y2= m2x2 + b2 Substituting (4,3) 3 = 4m1+ b1, 3 = 4m2+b2
b1=3/4m1, b2 = 3/4m2
1. X intercept for the equations is calculated when y = 0 b1/c1 * b2/c2 = 1  In Sufficient
2. Y intercept for the equations is calculated when x = 0 3/4m1* 3/4m2 = 1 m1*m2 = 16/9
Hence proves that the slopes are negative  Sufficient How did you get the marked above? I know I made a stupid mistake, but all I can do right now is laugh










