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by btgyes » Sat Feb 19, 2011 10:58 pm
In the rectangular coordinate system, line k has equation y = mx + b, where m and b are constants. does the line k intersect quadrant II?

1) Slope of k is -1/6

2) The y-intercept of k is -6
Source: — Data Sufficiency |

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by Geva@EconomistGMAT » Sat Feb 19, 2011 11:18 pm
btgyes wrote:In the rectangular coordinate system, line k has equation y = mx + b, where m and b are constants. does the line k intersect quadrant II?

1) Slope of k is -1/6

2) The y-intercept of k is -6
One thing people miss about this question is that a line stretches infinitely in both directions - if a question wanted to limit the line, it would use the term "line segment".

With this in mind, all you really need to know is whether the line will pass through quadrant II (negative x, positive y quadrant) at SOME point.

(1) As long as the line has a negative slope (doesn't matter if it's a negative fraction or negative number smaller than -1), the line will pass through quadrant II somewhere - the line declines from left to right (think of a car driving on the line from left to right - with a negative slope, that car is driving downhill), so at some point the line will ht Quad II.
You need the y intercept (in stat. (2) to determine whether that point of intersection with quad II will happen left or right of the y-axis, but not that the intersection will happen. Thu, stat. (1) is sufficient to answer the question on its own.

(2) alone, this is insufficiient, since you do not know the slope.
If the slope of the line is negative, then the line will hit the second quadrant somewhere to the left of the y axis.\
If the slope of the line is positive, or zero (parallel to the x axis), the the line will never hit the second quad.

Thus, (2) is insufficient. Answer is A.
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by btgyes » Mon Feb 21, 2011 11:04 pm
Geva@MasterGMAT wrote:
btgyes wrote:In the rectangular coordinate system, line k has equation y = mx + b, where m and b are constants. does the line k intersect quadrant II?

1) Slope of k is -1/6

2) The y-intercept of k is -6
One thing people miss about this question is that a line stretches infinitely in both directions - if a question wanted to limit the line, it would use the term "line segment".

With this in mind, all you really need to know is whether the line will pass through quadrant II (negative x, positive y quadrant) at SOME point.

(1) As long as the line has a negative slope (doesn't matter if it's a negative fraction or negative number smaller than -1), the line will pass through quadrant II somewhere - the line declines from left to right (think of a car driving on the line from left to right - with a negative slope, that car is driving downhill), so at some point the line will ht Quad II.
You need the y intercept (in stat. (2) to determine whether that point of intersection with quad II will happen left or right of the y-axis, but not that the intersection will happen. Thu, stat. (1) is sufficient to answer the question on its own.

(2) alone, this is insufficiient, since you do not know the slope.
If the slope of the line is negative, then the line will hit the second quadrant somewhere to the left of the y axis.\
If the slope of the line is positive, or zero (parallel to the x axis), the the line will never hit the second quad.

Thus, (2) is insufficient. Answer is A.
Hi Geva,

Could you plz explian why a line having negative slope will go through 2nd quardent.... mathematically...


as this is not the individual ques... there could be several scenario... ques can be maneuver... such as 3td quardent, forth quardent so it wont be sufficient to cram that LINE HAVING NEGATIVE SLOPE WILL PASS THROUGH 2ND QUADENT...

so plz prove mathematically....

THANKS....!

Thanks...!

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by sanju09 » Tue Feb 22, 2011 12:40 am
btgyes wrote:In the rectangular coordinate system, line k has equation y = mx + b, where m and b are constants. does the line k intersect quadrant II?

1) Slope of k is -1/6

2) The y-intercept of k is -6

(1) A negative slope suggests that the intercepts are either both positive or both negative, and in either case the line definitely intersects quadrant II. Sufficient

(2) A negative y-intercept could take the line to quadrant II only if its x-intercept is also negative, otherwise it can't. With no info about the x-intercept, this statement is [spoiler]Insufficient


A
[/spoiler]
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by Geva@EconomistGMAT » Tue Feb 22, 2011 12:44 am
forget mathematically, go visually. :)


Image

Play around with the declining line in the figure above - if you move it to the left or right, a "gentler" or "steeper" declining slope, the line will always pass through the 2nd quadrant at some point (assuming, as stated before, that the line stretches infinitely in both directions, rather than a finite line segment). It is possible to draw the line so that it does not go through the 1st or 3rd quadrants (depending on whether the line intersects the y axis above below or above the origin, respectively), but the line cannot help move through the 2nd and 4th quadrants.

In order to miss the 2 quad, the line must have a slope of zero (parallel to the x axis) or a positive slope ("rising line"), and even then the y intercept must be negative.
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