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# Q: In a rectangular coordinate system, does the line k

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Senior Manager
Joined: 20 Feb 2007
Posts: 256
Q: In a rectangular coordinate system, does the line k [#permalink]

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18 Mar 2007, 16:27
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Q: In a rectangular coordinate system, does the line k intersect quadrant II?

(1) the slope of k is -1/6
(2) y-intercept of k is -6
SVP
Joined: 01 May 2006
Posts: 1796

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18 Mar 2007, 16:38
(A) for me

From1
A negative slope forces a line to pass by the cadran II (y>0 and x<0).

We have : y = -1/6*x + b... If x < 6*b and x < 0, y > 0 and we are in the cadran II.

SUFF.

From2
We cannot conclude. All depend on the slope again.

o If the slope is equal to 0, then y = -6 is the line equation and we never pass in the cadran II
o If the slope is equal to -1, then if x = -7 , we are in the cadran II

INSUFF.

Last edited by Fig on 18 Mar 2007, 17:24, edited 1 time in total.
Senior Manager
Joined: 20 Feb 2007
Posts: 256

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18 Mar 2007, 16:58
Fig, I am not very good at coordinate geometry. Could you please explain this, especially the part written in red?

A negative slope for a line to pass by the cadran II (y>0 and x<0).

We have : y = -1/6*x + b... If x < 6*b and x <0> 0 and we are in the cadran II.

SUFF.

Thanks.
Senior Manager
Joined: 20 Feb 2007
Posts: 256

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18 Mar 2007, 16:59
BTW, your answer is correct. It is A and you are a genius!!!
SVP
Joined: 01 May 2006
Posts: 1796

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18 Mar 2007, 17:15
Summer3 wrote:
Fig, I am not very good at coordinate geometry. Could you please explain this, especially the part written in red?

A negative slope for a line to pass by the cadran II (y>0 and x<0).

We have : y = -1/6*x + b... If x < 6*b and x <0> 0 and we are in the cadran II.

SUFF.

Thanks.

I prefer to draw an XY plan ...

All lines sharing a similar slope are parallel to one another and called a family of parallel lines. The most simple line of a family is of a kind : y = a*x. Here, we have y = -1/6 * x... Look at the Fig 1... The line passes by the cadran II. Imagine any parallel line to it, it must pass by the cadran II

And, to respond to your question, we can rethink of the definition of the cadran II: y>0 and x<0... Let see if x could be negative when y is postive.
o y>0
<=> -1/6*x + b > 0 as y = -1/6*x + b, the equation of the line
<=> 1/6*x < b
<=> x < 6 * b >>>>> So, yes, x could be negative when y > 0.
Attachments

Fig1_Family of - x div 6.gif [ 2.94 KiB | Viewed 1354 times ]

SVP
Joined: 01 May 2006
Posts: 1796

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18 Mar 2007, 17:17
Summer3 wrote:
BTW, your answer is correct. It is A and you are a genius!!!

The bold part is overstated and it is the wrong answer choice actually ...
Director
Joined: 13 Mar 2007
Posts: 543
Schools: MIT Sloan

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24 Mar 2007, 22:21
well, i got stuck in the same question, probably over thought,

for any line with a negative slope, it should intersect quadrant 2 - agreed

but only if the line extends infinitely on both directions !!

the line can entirely lay on quad4 and still have negative slope !
(any line parallel to a line with slope -1/6 )

so guess the question should have told that the line is infinite or it should have given some end points ..

since its a DS problem, i read too much into it and chose E !

any inputs on the above ?
Manager
Joined: 16 Dec 2005
Posts: 60

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25 Mar 2007, 01:51
From Wikipedia,
J
A line can be described as an infinitely thin, infinitely long, perfectly straight curve (the term curve in mathematics includes "straight curves")

In geometry, a line segment is a part of a line that is bounded by two end points, and contains every point on the line between its end points.

Line in xy plane normally refers to line NOT line segment.
SVP
Joined: 01 May 2006
Posts: 1796

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25 Mar 2007, 03:39

Manager
Joined: 26 Feb 2007
Posts: 92

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25 Mar 2007, 04:11
Q: In a rectangular coordinate system, does the line k intersect quadrant II?

(1) the slope of k is -1/6
this means equation of line is y=-x/6+c where c is y intercept
if y=0 then x=6c means x can be positive or negative for y=0 so it is not sufficient
(2) y-intercept of k is -6

similary 2 alone is not sufficient

but if we combine 1 and 2, then line equation is y=-x/6-6
points of this line are (-36,0) and(0,-6) which intercepts quadrant III not II

so C is the answer for me
Director
Joined: 14 Jan 2007
Posts: 774

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25 Mar 2007, 07:19
kumarajeet06 wrote:
Q: In a rectangular coordinate system, does the line k intersect quadrant II?

(1) the slope of k is -1/6
this means equation of line is y=-x/6+c where c is y intercept
if y=0 then x=6c means x can be positive or negative for y=0 so it is not sufficient
(2) y-intercept of k is -6

similary 2 alone is not sufficient

but if we combine 1 and 2, then line equation is y=-x/6-6
points of this line are (-36,0) and(0,-6) which intercepts quadrant III not II

so C is the answer for me

If you carefully look at the stem- we have to find out whether the line with slope -1/6 passes thru the second quadrnt.
the equation of the line will be y =(-1/6)x + c where c =y intercept (when x=0)
we don't know about c. The possible values of c are zero, -ve and +ve numbers. For all values of c, the line passes through the quad II.
Hence statement 1 is sufficient.

Statement 2 is insuff as we don't know the slope of the line so not possible to know where the line will pass through from.

Director
Joined: 13 Mar 2007
Posts: 543
Schools: MIT Sloan

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25 Mar 2007, 12:12
hey thanks suithink
25 Mar 2007, 12:12
Display posts from previous: Sort by
The passage provided is a discussion about a coordinate geometry problem from a GMAT prep context. The problem asks whether a line \(k\) intersects the second quadrant in a rectangular coordinate system, given two pieces of information: the slope of \(k\) and the y-intercept of \(k\).

### Content Structure:
- The passage starts with a timestamp and a reference to GMAT Club Daily Prep.
- It then presents a coordinate geometry problem and two statements providing information about the line \(k\).

### Analysis:
single_question_with_answer

- the problem asks whether the line \(k\) intersects the second quadrant.
- statement 1 provides the slope of \(k\), which is \(-\frac{1}{6}\).
- statement 2 provides the y-intercept of \(k\), which is \(-6\).

### explanation:
1. **statement 1:**
   - the slope of \(k\) is \(-\frac{1}{6}\).
   - a negative slope means the line is decreasing from left to right.
   - for a line with a negative slope, it will always intersect the second quadrant (where \(y > 0\) and \(x < 0\)) because it will eventually cross the y
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
It is currently 26 Jun 2017, 04:04

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# Q: In a rectangular coordinate system, does the line k

Author Message
Senior Manager
Joined: 20 Feb 2007
Posts: 256
Q: In a rectangular coordinate system, does the line k [#permalink]

### Show Tags

18 Mar 2007, 16:27
00:00

Difficulty:

(N/A)

Question Stats:

100% (00:00) correct 0% (00:00) wrong based on 2 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Q: In a rectangular coordinate system, does the line k intersect quadrant II?

(1) the slope of k is -1/6
(2) y-intercept of k is -6
SVP
Joined: 01 May 2006
Posts: 1796

### Show Tags

18 Mar 2007, 16:38
(A) for me

From1
A negative slope forces a line to pass by the cadran II (y>0 and x<0).

We have : y = -1/6*x + b... If x < 6*b and x < 0, y > 0 and we are in the cadran II.

SUFF.

From2
We cannot conclude. All depend on the slope again.

o If the slope is equal to 0, then y = -6 is the line equation and we never pass in the cadran II
o If the slope is equal to -1, then if x = -7 , we are in the cadran II

INSUFF.

Last edited by Fig on 18 Mar 2007, 17:24, edited 1 time in total.
Senior Manager
Joined: 20 Feb 2007
Posts: 256

### Show Tags

18 Mar 2007, 16:58
Fig, I am not very good at coordinate geometry. Could you please explain this, especially the part written in red?

A negative slope for a line to pass by the cadran II (y>0 and x<0).

We have : y = -1/6*x + b... If x < 6*b and x <0> 0 and we are in the cadran II.

SUFF.

Thanks.
Senior Manager
Joined: 20 Feb 2007
Posts: 256

### Show Tags

18 Mar 2007, 16:59
BTW, your answer is correct. It is A and you are a genius!!!
SVP
Joined: 01 May 2006
Posts: 1796

### Show Tags

18 Mar 2007, 17:15
Summer3 wrote:
Fig, I am not very good at coordinate geometry. Could you please explain this, especially the part written in red?

A negative slope for a line to pass by the cadran II (y>0 and x<0).

We have : y = -1/6*x + b... If x < 6*b and x <0> 0 and we are in the cadran II.

SUFF.

Thanks.

I prefer to draw an XY plan ...

All lines sharing a similar slope are parallel to one another and called a family of parallel lines. The most simple line of a family is of a kind : y = a*x. Here, we have y = -1/6 * x... Look at the Fig 1... The line passes by the cadran II. Imagine any parallel line to it, it must pass by the cadran II

And, to respond to your question, we can rethink of the definition of the cadran II: y>0 and x<0... Let see if x could be negative when y is postive.
o y>0
<=> -1/6*x + b > 0 as y = -1/6*x + b, the equation of the line
<=> 1/6*x < b
<=> x < 6 * b >>>>> So, yes, x could be negative when y > 0.
Attachments

Fig1_Family of - x div 6.gif [ 2.94 KiB | Viewed 1354 times ]

SVP
Joined: 01 May 2006
Posts: 1796

### Show Tags

18 Mar 2007, 17:17
Summer3 wrote:
BTW, your answer is correct. It is A and you are a genius!!!

The bold part is overstated and it is the wrong answer choice actually ...
Director
Joined: 13 Mar 2007
Posts: 543
Schools: MIT Sloan

### Show Tags

24 Mar 2007, 22:21
well, i got stuck in the same question, probably over thought,

for any line with a negative slope, it should intersect quadrant 2 - agreed

but only if the line extends infinitely on both directions !!

the line can entirely lay on quad4 and still have negative slope !
(any line parallel to a line with slope -1/6 )

so guess the question should have told that the line is infinite or it should have given some end points ..

since its a DS problem, i read too much into it and chose E !

any inputs on the above ?
Manager
Joined: 16 Dec 2005
Posts: 60

### Show Tags

25 Mar 2007, 01:51
From Wikipedia,
J
A line can be described as an infinitely thin, infinitely long, perfectly straight curve (the term curve in mathematics includes "straight curves")

In geometry, a line segment is a part of a line that is bounded by two end points, and contains every point on the line between its end points.

Line in xy plane normally refers to line NOT line segment.
SVP
Joined: 01 May 2006
Posts: 1796

### Show Tags

25 Mar 2007, 03:39

Manager
Joined: 26 Feb 2007
Posts: 92

### Show Tags

25 Mar 2007, 04:11
Q: In a rectangular coordinate system, does the line k intersect quadrant II?

(1) the slope of k is -1/6
this means equation of line is y=-x/6+c where c is y intercept
if y=0 then x=6c means x can be positive or negative for y=0 so it is not sufficient
(2) y-intercept of k is -6

similary 2 alone is not sufficient

but if we combine 1 and 2, then line equation is y=-x/6-6
points of this line are (-36,0) and(0,-6) which intercepts quadrant III not II

so C is the answer for me
Director
Joined: 14 Jan 2007
Posts: 774

### Show Tags

25 Mar 2007, 07:19
kumarajeet06 wrote:
Q: In a rectangular coordinate system, does the line k intersect quadrant II?

(1) the slope of k is -1/6
this means equation of line is y=-x/6+c where c is y intercept
if y=0 then x=6c means x can be positive or negative for y=0 so it is not sufficient
(2) y-intercept of k is -6

similary 2 alone is not sufficient

but if we combine 1 and 2, then line equation is y=-x/6-6
points of this line are (-36,0) and(0,-6) which intercepts quadrant III not II

so C is the answer for me

If you carefully look at the stem- we have to find out whether the line with slope -1/6 passes thru the second quadrnt.
the equation of the line will be y =(-1/6)x + c where c =y intercept (when x=0)
we don't know about c. The possible values of c are zero, -ve and +ve numbers. For all values of c, the line passes through the quad II.
Hence statement 1 is sufficient.

Statement 2 is insuff as we don't know the slope of the line so not possible to know where the line will pass through from.

Director
Joined: 13 Mar 2007
Posts: 543
Schools: MIT Sloan

### Show Tags

25 Mar 2007, 12:12
hey thanks suithink
25 Mar 2007, 12:12
Display posts from previous: Sort by
user
The passage provided is a discussion about a coordinate geometry problem from a GMAT prep context. The problem asks whether a line \(k\) intersects the second quadrant in a rectangular coordinate system, given two pieces of information: the slope of \(k\) and the y-intercept of \(k\).

### Content Structure:
- The passage starts with a timestamp and a reference to GMAT Club Daily Prep.
- It then presents a coordinate geometry problem and two statements providing information about the line \(k\).

### Analysis:
TYPE: SINGLE_QUESTION_WITH_ANSWER

- The problem asks whether the line \(k\) intersects the second quadrant.
- Statement 1 provides the slope of \(k\), which is \(-\frac{1}{6}\).
- Statement 2 provides the y-intercept of \(k\), which is \(-6\).

### Explanation:
1. **Statement 1:**
   - The slope of \(k\) is \(-\frac{1}{6}\).
   - A negative slope means the line is decreasing from left to right.
   - For a line with a negative slope, it will always intersect the second quadrant (where \(y > 0\) and \(x < 0\)) because it will eventually cross the y