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GMATprep1 coordinate geometry

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Source: — Data Sufficiency |
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by Jim@Grockit » Mon Sep 27, 2010 6:58 am
rb90 wrote:Please explain the working of this sum.Really need the help.Thnks.The answer is in a box in the attatchment.
A negative slope (from Statement 1) means that the line -- any line with any negative slope -- looks like this [ \ ] and there is literally no way to draw it without going into Quadrant II at some point (imagine putting an infinitely long "" anywhere on the graph pictured). Only lines with a positive slope [ / ] or a slope of 0 [ -- ] and a negative y-intercept avoid Quadrant II.

Statement 2 gives us a negative y-intercept, but a line that's [ / ] or [ -- ] could avoid quadrant II from there.
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by rb90 » Sat Oct 09, 2010 4:15 am
For statement2,
y=mx+c
6=0+c
=>c=6.

For y=0,
0=mx+6
=> mx= -6
Now either m<0 or x<0 , but in either case, the line is going through QII.
Howver, this would make D the answer,which isnt correct, but please explain where am i going wrong.
Need help.Thanks
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