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demoskopico
03-15-2008, 04:32 AM
In the xy-coordinate plane, line L and line K intersect at 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.

I reasoned in this way:

we must show whether be/ad<0
1. y=c/a and y1=k/d; ck/ad>0 where c and k are the constants. it doesn't say much
2. ck/be<0 alone insuff

together it's clear that ad and be have different signs so c is suff

anyway, it took me far more than 2 minutes to solve this one and would like to know if there is any shortcut. moreover, why does the argument tell us that the intersection point is 4,3?
thanks

JohnB
03-18-2008, 08:23 PM
In this case, the fastest way to proceed is using some geometric intuition. I'm attaching a diagram to explain. In order for the product of the slopes to be negative, one line has to go upwards from left to right, and the other has to go downwards.

Consider statement I. If the product of the x-intercepts is positive, the lines could look like A and B, so that the product of the slopes is positive. Or they could look like A and C, so that the product is negative. INSUF.

II: If the product of the y-intercepts is negative, then one line must cross the y-axis above the origin, and one below. But they could look like A and C, with negative product of the slopes, or like B and C, with positive product of slopes.

I and II together: In this case, the lines cross the y-axis on different sides of the origin, but they cross the x-axis on the same side of the origin. There's no way for this to work unless one has a negative and one has a positive slope. (Try it.) So together they're sufficient.

JohnB
03-18-2008, 08:25 PM
By the way, you're right that (4,3) is arbitrary--this question would work the same with any given point.