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xy plane [#permalink] New post 02 Jan 2006, 17:26
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Re: xy plane [#permalink] New post 02 Jan 2006, 18:15
from i, m = -ve. the line can pass through (a, b) = (2, 1), (1, 2) (-2, -1), (-1, 2), (2, -1) and so on.
From ii, b>a. (a,b) could be any values for example (1, 2) (-2, -1)

E. i donot see it is C.
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Slope [#permalink] New post 02 Jan 2006, 19:00
I thought that a negative slope meant that the line had to run (for lack of a better term) from Northwest to Southeast - always down and to the right? Is that incorrect?
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Re: xy plane [#permalink] New post 02 Jan 2006, 19:59
HIMALAYA wrote:
from i, m = -ve. the line can pass through (a, b) = (2, 1), (1, 2) (-2, -1), (-1, 2), (2, -1) and so on.
From ii, b>a. (a,b) could be any values for example (1, 2) (-2, -1)

E. i donot see it is C.


C is correct. taking both statements we can determine that the any point on the line is either in fourth or 2nd quadrant. with statement 2 we can locate the point in fourth quad with -ve X and +ve Y
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Re: xy plane [#permalink] New post 02 Jan 2006, 20:08
old_dream_1976 wrote:
HIMALAYA wrote:
from i, m = -ve. the line can pass through (a, b) = (2, 1), (1, 2) (-2, -1), (-1, 2), (2, -1) and so on.
From ii, b>a. (a,b) could be any values for example (1, 2) (-2, -1)

E. i donot see it is C.


C is correct. taking both statements we can determine that the any point on the line is either in fourth or 2nd quadrant. with statement 2 we can locate the point in fourth quad with -ve X and +ve Y


oohhhhhhh thats correct.

the line passes through orgin to a,b.
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 [#permalink] New post 02 Jan 2006, 20:11
I will take E here.

From 1) Line k is of the form y = -mx+c (c being the y intercept; See attached picture)
From 2) a < b
=> b = -ma + c

Depending on the value of c, the line pass through different quandrant (See attached picture)
HTH.
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 [#permalink] New post 02 Jan 2006, 20:24
C it is.

the slope of the line = (b-0)/(a -0)= b/a
from 1, b/a < 0
from 2, we have a<b

Combine two statements, it must be that: a< 0 <b
--> b is positive --> suff

Last edited by laxieqv on 02 Jan 2006, 20:27, edited 2 times in total.
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 [#permalink] New post 02 Jan 2006, 20:25
giddi77

Remember the line passes thru the origin and therefore your figure is incorrect.

The correct answer is C. The point (a,b) is in the IInd quadrant, where the y-axis -point b- is positive.

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giddi77 wrote:
I will take E here.

From 1) Line k is of the form y = -mx+c (c being the y intercept; See attached picture)
From 2) a < b
=> b = -ma + c

Depending on the value of c, the line pass through different quandrant (See attached picture)
HTH.
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 [#permalink] New post 02 Jan 2006, 20:29
laxieqv wrote:
C it is.

the slope of the line = (b-0)/(a -0)= b/a
from 1, b/a < 0
from 2, we have a<b

Combine two statements, it must be that: a<0<b --> b is positive --> suff


laxie you are right, I missed the passing through the origin part. Thanks for the explanation.
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 [#permalink] New post 02 Jan 2006, 20:34
giddi77 wrote:
laxieqv wrote:
C it is.

the slope of the line = (b-0)/(a -0)= b/a
from 1, b/a < 0
from 2, we have a<b

Combine two statements, it must be that: a<0<b --> b is positive --> suff


laxie you are right, I missed the passing through the origin part. Thanks for the explanation.



The forum has technical errors, I typed a < 0 < b and it turns " a<0 b " :x
  [#permalink] 02 Jan 2006, 20:34
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