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# Does the point of intersection of line y=Kx+ B and line

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CEO
Joined: 21 Jan 2007
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Does the point of intersection of line y=Kx+ B and line [#permalink]  13 Nov 2007, 06:04
Does the point of intersection of line y=Kx+ B and line x=Kx+B have a negative coordinate?

1. k > 0, b>0
2. k > 1

Manager
Joined: 01 Nov 2007
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since the slope of the lines is same dont you think the lines will be parallal.

and does the question asks that both the x,y cordinate will be negative or any one of them.?
VP
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Re: intersecting lines [#permalink]  14 Nov 2007, 10:42
bmwhype2 wrote:
Does the point of intersection of line y=Kx+ B and line x=Kx+B have a negative coordinate?

1. k > 0, b>0
2. k > 1

I think the answer is C. i remember doing it. if it is the right answer i might post the explanation
Director
Joined: 13 Dec 2006
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Is there any typo error?

I think the second equation should be X= KY + B...

if so then 1 is sufficent, hence answer is A

On solving both the equation we will have X = -B((K^2 + 1)/(K^2 - 1))

and Y = B[1- (K^2 +1)/(K^2-1)], since both K and B are greater than 0, X and Y will be negative coordinate. If K=1, K^2 - 1 = 0 and hence X will have infinite negative value. So Case 2 can be ignored.

Amar
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Re: intersecting lines [#permalink]  15 Nov 2007, 04:44
bmwhype2 wrote:
bmwhype2 wrote:
Does the point of intersection of line y=Kx+ B and line x=KY+B have a negative coordinate?

1. k > 0, b>0
2. k > 1

sorry.

Actually, if there is no error and the 2nd graph is vertical line, then the answer is C.

Let's see :

Functions are : y = kx+b, x = b/(1-k)

1 is not sufficient because if b and k both positive then :

y = kx + b passes I, III, IV quadrants.
x = b/(1-k) when k <1>1 then it is vertical to the left of Y and the intersection will be in either quadrant III or IV and 1 or 2 coordinates will be negative.

2 is not sufficient because it gives us obviousely not enough information.

Thus the unswer is C
Re: intersecting lines   [#permalink] 15 Nov 2007, 04:44
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