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

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Q: In a rectangular coordinate system, does the line k [#permalink] New post 18 Mar 2007, 15:27
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A
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C
D
E

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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
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 [#permalink] New post 18 Mar 2007, 15:38
(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, 16:24, edited 1 time in total.
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 [#permalink] New post 18 Mar 2007, 15:58
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.
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 [#permalink] New post 18 Mar 2007, 15:59
BTW, your answer is correct. It is A and you are a genius!!! :)
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 [#permalink] New post 18 Mar 2007, 16:15
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
Fig1_Family of - x div 6.gif [ 2.94 KiB | Viewed 896 times ]

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 [#permalink] New post 18 Mar 2007, 16:17
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 ;)...
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 [#permalink] New post 24 Mar 2007, 21:21
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 ?
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 [#permalink] New post 25 Mar 2007, 00:51
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.
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 [#permalink] New post 25 Mar 2007, 02:39
Excellent discussion grad_mba and suithink :)

suithink brought already all the light on the question raised by grad_mba :) Nothing to add :D :)
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 [#permalink] New post 25 Mar 2007, 03:11
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
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 [#permalink] New post 25 Mar 2007, 06:19
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.

Hence the answer is 'A'
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 [#permalink] New post 25 Mar 2007, 11:12
hey thanks suithink :)
  [#permalink] 25 Mar 2007, 11:12
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