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Re: DS - Coordinate geometry [#permalink]
24 Jun 2009, 09:23
IMO A. Line with a negative slope must intesect quadrant II.
This line must be y= -x/6 + m(intrecept) in quadrant II, x values are negative and y values are positive.
So a line that intersects quadrant 2 must have points like that
y1=-x1/6 +m>0 -x1/6>-m (as x1 is negative -x1/6 must be positive) So if m is positive than inequation holds, if m is negative one can find an -x1 value that is greater than -m for all values of -m. Thus, 1) is sufficient
2 is unsufficient. e.g. line k => y=-6 or for any positive value of n line k => y=-6 + nx