akpareek wrote:in the xy plane, line L passes through point (1,p). Does L have a positive slope ?
1. L passes through point (-p,13)
2. L passes through point (0,1)
yep, this one is wacky.
what's the source of this problem?
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as proved by other posters, each of the individual statements is insufficient.
if you take the statements together, then you now have
three slopes, ALL of which must be equal to each other.
let's call the points A(1, p), B(-p, 13), and C(0, 1).
... then you've got three slopes:
slope AB = (p - 13)/(1 + p)
slope BC = (13 - 12)/(-p - 0) = -12/p
slope AC = (p - 1)/(1 - 0) = p - 1
you can set any two of these three equal to give an equation, so you may as well go with the easiest ones, i.e., the last two.
this gives -12/p = p - 1.
multiply out to -12 = p^2 - p.
this gives p^2 - p + 12 = 0, which is the same weird equation (no real solutions) that other posters got above.
therefore, you get
no such points at all, meaning that the combination of the two statements is actually a mathematical impossibility.
the gmat won't do that to you, so this problem is a bad, bad problem.
