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# In the rectangular coordinate system, does the line K (not

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In the rectangular coordinate system, does the line K (not [#permalink]

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10 Dec 2005, 19:33
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In the rectangular coordinate system, does the line K (not shown) intersect the second quadrant?

a) The slope of K is -1/6
b)The Y intercept of K is -6.
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10 Dec 2005, 20:41
amy_v wrote:
In the rectangular coordinate system, does the line K (not shown) intersect the second quadrant?

a) The slope of K is -1/6
b)The Y intercept of K is -6.

BTW, "rectangular coordinate system" , I assume it means the Cartesian plane
1. quadrant 2: when x <0, y>0
line K : y= -1/6 x + b ( we don't know b yet)
of coz, there'll be negative values of x which produce positive y with any b
----> line K intersects 2nd quadrant
---->suff

2. - 6 : see my illustration with the two red lines.
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11 Dec 2005, 05:50
laxieqv wrote:
amy_v wrote:
In the rectangular coordinate system, does the line K (not shown) intersect the second quadrant?

a) The slope of K is -1/6
b)The Y intercept of K is -6.

BTW, "rectangular coordinate system" , I assume it means the Cartesian plane
1. quadrant 2: when x <0, y>0
line K : y= -1/6 x + b ( we don't know b yet)
of coz, there'll be negative values of x which produce positive y with any b
----> line K intersects 2nd quadrant
---->suff

Hi Laxie, I am trying to rephrase your explanation for statement 1 . Pls let me know if I got it correct..

we are asked to find if the x<0andY>0 satisfies the equation for the line

y=-1/6x+b

Now if x<0 then -1/6x becomes positive and Y is positive. that makes B positive. Since the y intercept when x=o is positive , the line intercects the II quadrant...correct?
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Joined: 24 Sep 2005
Posts: 1890
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11 Dec 2005, 07:06
amy_v wrote:
laxieqv wrote:
amy_v wrote:
In the rectangular coordinate system, does the line K (not shown) intersect the second quadrant?

a) The slope of K is -1/6
b)The Y intercept of K is -6.

BTW, "rectangular coordinate system" , I assume it means the Cartesian plane
1. quadrant 2: when x <0, y>0
line K : y= -1/6 x + b ( we don't know b yet)
of coz, there'll be negative values of x which produce positive y with any b
----> line K intersects 2nd quadrant
---->suff

Hi Laxie, I am trying to rephrase your explanation for statement 1 . Pls let me know if I got it correct..

we are asked to find if the x<0andY>0 satisfies the equation for the line

y=-1/6x+b

Now if x<0 then -1/6x becomes positive and Y is positive. that makes B positive. Since the y intercept when x=o is positive , the line intercects the II quadrant...correct?

Well, like this....
in case:
+ b>=0: for every x<0 , we have -1/6 x + b > 0 coz the two terms are positive. ----> y> 0 -----> the line surely intersects quadrant 2
+ b<0 : when x< 0 , we have -1/6 x > 0 ....in some cases when the absolute value of b > -1/6x , y < 0 ....but there're certainly values of x which make the absolute value of -1/6 x larger than that of b ----> y is positive. Since a line is infinitive, there're values of x large enough to outweigh the absolute value of this negative b ----> the line must intersect quadrant 2

A little abstract, hope you understand my explanation, if you're still confused, i'll try my better best to explain to you
Manager
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11 Dec 2005, 12:32
laxieqv wrote:
amy_v wrote:
laxieqv wrote:
amy_v wrote:
In the rectangular coordinate system, does the line K (not shown) intersect the second quadrant?

a) The slope of K is -1/6
b)The Y intercept of K is -6.

BTW, "rectangular coordinate system" , I assume it means the Cartesian plane
1. quadrant 2: when x <0, y>0
line K : y= -1/6 x + b ( we don't know b yet)
of coz, there'll be negative values of x which produce positive y with any b
----> line K intersects 2nd quadrant
---->suff

Hi Laxie, I am trying to rephrase your explanation for statement 1 . Pls let me know if I got it correct..

we are asked to find if the x<0andY>0 satisfies the equation for the line

y=-1/6x+b

Now if x<0 then -1/6x becomes positive and Y is positive. that makes B positive. Since the y intercept when x=o is positive , the line intercects the II quadrant...correct?

Well, like this....
in case:
+ b>=0: for every x<0 , we have -1/6 x + b > 0 coz the two terms are positive. ----> y> 0 -----> the line surely intersects quadrant 2
+ b<0 : when x< 0 , we have -1/6 x > 0 ....in some cases when the absolute value of b > -1/6x , y < 0 ....but there're certainly values of x which make the absolute value of -1/6 x larger than that of b ----> y is positive. Since a line is infinitive, there're values of x large enough to outweigh the absolute value of this negative b ----> the line must intersect quadrant 2

A little abstract, hope you understand my explanation, if you're still confused, i'll try my better best to explain to you

gotcha!!!! as usual perfect!!
Re: DS:Coordiante Geometry-GMATPREP   [#permalink] 11 Dec 2005, 12:32
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