In the xy-plane, the line k passes through the origin and through the point (a,b), where ab does not = 0. Is b positive?
(1) The slope of line k is negative
(2) a<b
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GMAT Prep2?? (Slope)
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from 1) b can lie in II or IV quadrant. Not Sufficient
from 1) and 2) (a,b) will lie in second quadrant only. since in IV a will always be greater than b.
Answer should be C
from 1) and 2) (a,b) will lie in second quadrant only. since in IV a will always be greater than b.
Answer should be C
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Another approach to solve this question:-
We know that line passes through two points: O (0, 0) and P (a, b).
Now, by the equation of line,
(x - x1)/(y-y1) = (x2 - x1)/(y2-y1)
x/y = a/b
ay = bx
y = bx/a
Now by condition 1: Slope is negative.
Any equation line could be: y = mx + c (where c is constant)
and our line equation is y = bx/a
This means c = 0 and m = b/a
Now as m is negative, therefore b/a < 0
Now if b is negative then b < a otherwise a < b
It yields two different results, hence Condition 1 is not alone sufficient,
Condition (2) also alone INSUFFICIENT, as it said a < b, but doesn't give any clues of their signs.
Now when we take both statements together, we have already proved that
of a < b then b is positive.
So both statements together are SUFFICIENT.
We know that line passes through two points: O (0, 0) and P (a, b).
Now, by the equation of line,
(x - x1)/(y-y1) = (x2 - x1)/(y2-y1)
x/y = a/b
ay = bx
y = bx/a
Now by condition 1: Slope is negative.
Any equation line could be: y = mx + c (where c is constant)
and our line equation is y = bx/a
This means c = 0 and m = b/a
Now as m is negative, therefore b/a < 0
Now if b is negative then b < a otherwise a < b
It yields two different results, hence Condition 1 is not alone sufficient,
Condition (2) also alone INSUFFICIENT, as it said a < b, but doesn't give any clues of their signs.
Now when we take both statements together, we have already proved that
of a < b then b is positive.
So both statements together are SUFFICIENT.