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by itsmebharat » Thu May 26, 2011 3:42 am
In xy coordinate plane, line l and line k intersect at point (4,3). Is product of the slopes negative.

I. product of x intercepts of lines l and k are positive
II. product of y intercepts of lines l and k is negative.
I am not an Expert, please feel free to suggest if there is an error.
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by irock » Thu May 26, 2011 3:54 am
line L: y1 = m1x1 + b1; x-intercept = -b1/m1; y-intercept = b1
line K: y2 = m2x2 + b2; x-intercept = -b2/m2; y-intercept = b2

(4,3) lies on both.
3 = 4m1 + b1 = 4m2 + b2

Is m1m2 < 0?

1. (-b1/m1)*(-b2/m2) > 0
b1b2/m1m2 > 0
If b1b2 > 0, m1m2 > 0
If b1b2 < 0, m1m2 < 0
NOT SUFFICIENT.

2. b1b2 < 0
(3-4m1)(3-4m2) < 0
NOT SUFFICIENT.

Together, b1b2 < 0 and m1m2 < 0.
SUFFICIENT. Answer is C.
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by BellTheGMAT » Thu May 26, 2011 5:04 am
IMO C
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by Frankenstein » Thu May 26, 2011 9:01 am
Hi,
Equation of a line with x-intercept 'a' and y-intercept 'b' is (x/a)+(y/b)=1 and its slope is (-b/a)
Let the lines l and k be (x/a1)+(y/b1)=1 and (x/a2)+(y/b2)=1.
Product of slopes is (-b1/a1).(-b2/a2) =b1b2/a1a2
From (1) a1.a2>0. No info. about b1,b2
Not sufficient
From (1) b1.b2<0. No info. about a1,a2
Not sufficient
Combining we get b1b2/a1a2 < 0
Sufficient

Hence, answer C

Cheers!
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by amar66 » Thu May 26, 2011 11:08 am
IMO C
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