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20 Jan 2008, 23:03
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Line L has a slope that is greater than -2. Is the x-intercept of line L negative?
1.L passes through the point (3,-7)
2.L passes through the point (5,-9)
1.L passes through the point (3,-7)
2.L passes through the point (5,-9)
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A
we can solve it by drawing or by algebra.
L: \(y=ax+b\) and a e \((-2,+\infty)\)
the x-intercept of line L is b
1. \(-7=a*3+b\) ==> \(b=-3*a-7\) b e \((-\infty,-1)\) suff.
2. \(-9=a*5+b\) ==> \(b=-5*a-9\) b e \((-\infty,1)\) insuff.
we can solve it by drawing or by algebra.
L: \(y=ax+b\) and a e \((-2,+\infty)\)
the x-intercept of line L is b
1. \(-7=a*3+b\) ==> \(b=-3*a-7\) b e \((-\infty,-1)\) suff.
2. \(-9=a*5+b\) ==> \(b=-5*a-9\) b e \((-\infty,1)\) insuff.
21 Jan 2008, 03:28
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marcodonzelli
I think C is correct
x-intercept of L is under control not only of b, but also a.
x= - b/a
1. x = (-3a -7)/a =-3 - 7/a, x-intercept wil (-) if a>0 and will (+) if -2<a<0 --> not sufficient
2. x = - 5 - 9/a --> not sufficient
1 and 2 combined: a = 1, b = - 13, so x - intercept negative --> suff.
C
