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crackgmat007
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Recall that a line can be expressed in the form y = mx + b where m is the slope and b is the y-intercept. We will use this form, but label the y-intercept as c since the point (a,b) is used in this problem.
So the line k is of the form y = mx + c.
The line passes through (0,0) and (a,b). So the slope m = b/a. So k is of the form y = (b/a)x + c. plug in either point to find that c = 0.
So line k is expressed as y = (b/a)x
(1). Slope is negative. Then b/a < 0. Choose a = 1, b = -2. The slope is negative and b is negative. Now choose a = -1, b = 2. The slope is negative and b is positive. This is insufficient.
(2). a < b. Choose a = 1, b = 2. a<b and b is positive. Now choose. a = -2, b = -1. Then a<b and b is negative. This is insufficient.
(1)+(2). since ab does NOT equal 0, b does NOT equal 0. So b > 0 or b < 0.
Case b > 0: Slope is negative and a<b tells us that a<0 and b > 0 so both conditions are met.
Case b < 0: Since a < b and b < 0, a < 0. a and b both being negative gives us that the slope b/a > 0. But this does not satisfy condition 1. So b cannot be less than zero.
Therefore b > 0 and C is the answer.
