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

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SVP
Joined: 21 Jul 2006
Posts: 1510
In the xy-coordinate plane, line l and line k intersect at [#permalink]

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28 Oct 2008, 13:43
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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.

Thanks
Current Student
Joined: 28 Dec 2004
Posts: 3357
Location: New York City
Schools: Wharton'11 HBS'12

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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
Senior Manager
Joined: 31 Jul 2008
Posts: 296

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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 )
VP
Joined: 05 Jul 2008
Posts: 1408

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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??
Senior Manager
Joined: 21 Apr 2008
Posts: 269
Location: Motortown

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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
Current Student
Joined: 28 Dec 2004
Posts: 3357
Location: New York City
Schools: Wharton'11 HBS'12

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28 Oct 2008, 18:17
what if there is no X-axis intercept and you have 2 lines with positive slopes, with one having a negative y intercept and the other a straight line..
SVP
Joined: 21 Jul 2006
Posts: 1510

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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 x-intercept is positive. this means that both lines are touching the x-axes 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 y-intercept, while the second line could be going downward from above the y-intercept 4. So not suff.

(1&2) when both the lines have a positive x-intercept, while one of the y-intercept 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.

SVP
Joined: 29 Aug 2007
Posts: 2473

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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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Senior Manager
Joined: 21 Apr 2008
Posts: 269
Location: Motortown

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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
Re: DS: Geometry   [#permalink] 29 Oct 2008, 04:46
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