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178 Does straight line have a negative slope?

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by ern5231 » Wed Oct 07, 2009 5:19 pm
In rectangular coordinate system, a straight line has a point (c,d) in the first quadrant. Does the line have a negative slope?
1) the straight line also crosses (c+2, d-3)
2) this straight line's x-intercept is greater than y-intercept
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Source: — Data Sufficiency |
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by sunnyjohn » Wed Oct 07, 2009 6:22 pm
IMO:D
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by crackgmat007 » Wed Oct 07, 2009 9:38 pm
By POE, I could get it to C or E. I would guess E.

1. Gives many possibilities
2. Picking numbers, if X intercept is 4 & Y is 3, line has a negative slope. If Y is -3, line has a positive slope.
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by rohittk » Thu Oct 08, 2009 1:56 am
Answer is D as slope = (y2 - y1)/ (x2 - x1).
Interpolate values into this equation and you know why the answer is D.
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by ershovici » Thu Oct 08, 2009 3:27 am
I think the answer is A.
Just drow a scetch and you will see that the line which passes through these two points can not have positive slope.
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by crackgmat007 » Fri Oct 09, 2009 7:34 am
rohittk wrote:Answer is D as slope = (y2 - y1)/ (x2 - x1).
Interpolate values into this equation and you know why the answer is D.
Can you explain how B is sufficient?
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Re: 178 Does straight line have a negative slope?

by Brent@GMATPrepNow » Fri Oct 09, 2009 10:42 am
ern5231 wrote:In rectangular coordinate system, a straight line has a point (c,d) in the first quadrant. Does the line have a negative slope?
1) the straight line also crosses (c+2, d-3)
2) this straight line's x-intercept is greater than y-intercept
(1) If the line passes through (c,d) and (c+2, d-3), we can apply the slope formula to see that the slope is -3/2 --> SUFF

(2) INSUFF
Case a) x-intercept=6, y-intercept=1 --> negative slope
Case b) x-intercept=6, y-intercept=-1 --> positive slope
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