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Re: DS- line k intersect quadrant II? [#permalink]
24 Jul 2008, 05:26
nmohindru wrote:
gmatcraze wrote:
Q. In the rectangular co-ordinate system shown does the line k intersect quadrant II?
1. The slope of k is -1/6 2. The y-intercept of k is -6
Answer is C)
For line equation you need to know slope and y or x intercept.
but here we don't need to find the line equation, we only need to determine if the line k intersects quadrant II .... how can we determine that? From (1), we know that the line has a negative slope and therefore intersects either quadrant II or quadrant IV. Hence not SUFF. From (2), we know that the point is (0,-6). And this lies in quadrant IV. Hence B.
Re: DS- line k intersect quadrant II? [#permalink]
24 Jul 2008, 05:43
See what I wrote below. Its bold and in red.
gmatcraze wrote:
nmohindru wrote:
gmatcraze wrote:
Q. In the rectangular co-ordinate system shown does the line k intersect quadrant II?
1. The slope of k is -1/6 2. The y-intercept of k is -6
Answer is C)
For line equation you need to know slope and y or x intercept.
but here we don't need to find the line equation, we only need to determine if the line k intersects quadrant II .... how can we determine that? From (1), we know that the line has a negative slope and therefore intersects either quadrant II or quadrant IV[we don't know this because we don't know the y-intersect. It could potentially go through all quandrants, but not at the same time of course. Max 3 quandrants at the same time]. Hence not SUFF. From (2), we know that the point is (0,-6). And this lies in quadrant IV [something that is actually on the x-axis or y-axis is not IN any quandrant...it is on the axis. Furthermore, with the y-intercept at -6, we still don't know which direction the line goes. We only know where it begins. The line could be a y=-6 and the line be parallel to the x-axis, or the line could have slope of -1 and it does go through Quadrant II. Hence B.
Just wondering if my reasoning is correct?
_________________
------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.
Re: DS- line k intersect quadrant II? [#permalink]
24 Jul 2008, 06:04
nmohindru wrote: gmatcraze wrote: Q. In the rectangular co-ordinate system shown does the line k intersect quadrant II?
1. The slope of k is -1/6 2. The y-intercept of k is -6
Answer is C)
For line equation you need to know slope and y or x intercept.
but here we don't need to find the line equation, we only need to determine if the line k intersects quadrant II .... how can we determine that? From (1), we know that the line has a negative slope and therefore intersects either quadrant II or quadrant IV. Hence not SUFF. A line with negative slope will intersect either quadrant I or III. From (2), we know that the point is (0,-6). And this lies in quadrant IV. Hence B.
Re: DS- line k intersect quadrant II? [#permalink]
02 Dec 2008, 20:52
question : does the line k intersect quadrant II ? answer from gmatcraze : From (1), we know that the line has a negative slope and therefore intersects either quadrant II or quadrant IV. The best answer choice is A ( according to GMAt prep)
Re: DS- line k intersect quadrant II? [#permalink]
04 Oct 2010, 02:30
1
This post was BOOKMARKED
citing Bunuel from this website.
SLOPE AND QUADRANTS: 1. If the slope of a line is negative, the line WILL intersect quadrants II and IV. X and Y intersects of the line with negative slope have the same sign. Therefore if X and Y intersects are positive, the line intersects quadrant I; if negative, quadrant III. 2. If the slope of line is positive, line WILL intersect quadrants I and III. Y and X intersects of the line with positive slope have opposite signs. Therefore if X intersect is negative, line intersects the quadrant II too, if positive quadrant IV.
Re: DS- line k intersect quadrant II? [#permalink]
15 Oct 2010, 02:02
2
This post received KUDOS
Expert's post
6
This post was BOOKMARKED
gmatcraze wrote:
Q. In the rectangular co-ordinate system shown does the line k intersect quadrant II?
1. The slope of k is -1/6 2. The y-intercept of k is -6
1. If the slope of line is negative, line WILL intersect quadrants II and IV. X and Y intersects of the line with negative slope have the same sign. Therefore if X and Y intersects are positives, line intersects the quadrant I too, if negative quadrant III.
2. If the slope of line is positive, line WILL intersect quadrants I and III. Y and X intersects of the line with positive slope have opposite signs. Therefor if X intersect is negative, line intersects the quadrant II too, if positive quadrant IV.
3. Every line (but the one crosses origin or parallel to X or Y axis) crosses three quadrants. Only the line which crosses origin (0,0) OR is parallel of either of axis crosses two quadrants.
4. The line with slope 0 is parallel to X-axis and crosses quadrant I and II, if the Y intersect is positive OR quadrants III and IV, if the Y intersect is negative.
If you draw the couple of lines with different slopes you'll understand this better.
BACK TO THE ORIGINAL QUESTION:
Statement (1) says that slope is negative (case 1) so the line will intersect the quadrants II and IV (line goes from up to down). Sufficient.
Statement (2) provides us with y-intercept -6, now if line has positive slope then the line goes from down to up and thus won't intersect quadrant II but if the slope is negative then the line goes from up to down thus it will intersect quadrant II. Two different answers. Not sufficient.
Answer: A.
For more on this topic check Coordinate Geometry chapter of Math Book (link in my signature).
Re: rectangular co-ordinate system [#permalink]
20 Mar 2011, 03:16
As you know the equation of a line is y=kx+b II quadrant means that for x<0 y>0 (1) says that the line is y=-1/6x+b. Whatever b you have (positive or negative), this is some finite number. Taking x as low as you can (going to -infinity), you always could get -1/6*x as much positive as you need such that the sum -1/6x+b is positive. In other words, solving -1/6*x+b=0 you get the for any x<6b y is positive, obviously for some negative values of x this will also hold. (2) says that y(0)=-6 which means that b=-6. Is the slope is positive, then such line will never intersect II quadrant. For example y=-6+x will never intersect it.
So (1) is sufficient, (2) is not. The answer is (A) _________________
If my post is useful for you not be ashamed to KUDO me! Let kudo each other!
Re: rectangular co-ordinate system [#permalink]
20 Mar 2011, 04:30
1
This post received KUDOS
The answer is A, I think you can visualize this without any calculation.
Also, please check MATH book, and Bunuel's tutorials where it's explained what type of lines with what signs pass from which quadrants. _________________
Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)
Re: rectangular co-ordinate system [#permalink]
20 Mar 2011, 06:13
1
This post received KUDOS
It's A. 1. A line with a negative slope always passes through second and fourth quadrant. Thus sufficient Likewise A line with positive slope always passes through First and Third Quadrant.
2. this is insufficient, If the slope of the line is negative, then the line will touch the second quadrant 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 touch the second quad.
Re: In the rectangular co-ordinate system shown does the line k [#permalink]
25 Nov 2014, 10:18
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Re: In the rectangular co-ordinate system does the line k [#permalink]
28 Sep 2015, 04:08
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________
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