M19-29 : Retired Discussions [Locked]
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# M19-29

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Joined: 07 Sep 2010
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07 Jun 2013, 23:12
If k does not equal -1, 0 or 1, does the point of intersection of line $$y=kx+b$$ and line $$x=ky+b$$ have a negative x-coordinate?
(1)$$kb>0$$

(2) $$k>1$$

Help will be appreciated.
Thanks
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08 Jun 2013, 03:29
imhimanshu wrote:
If k does not equal -1, 0 or 1, does the point of intersection of line $$y=kx+b$$ and line $$x=ky+b$$ have a negative x-coordinate?
(1)$$kb>0$$

(2) $$k>1$$

Help will be appreciated.
Thanks

m19 q29

If k does not equal -1, 0 or 1, does the point of intersection of line y = kx + b and line x = ky + b have a negative x-coordinate?

We have equations of two lines: $$y = kx + b$$ and $$y=\frac{x}{k}-\frac{b}{k}$$ (from $$x = ky + b$$). Equate to get the x-coordinate of the intersection point: $$kx + b=\frac{x}{k}-\frac{b}{k}$$ --> $$x=\frac{b(k+1)}{1-k^2}=\frac{b(k+1)}{(1-k)(1+k)}=\frac{b}{1-k}$$.

So, the question basically asks whether $$x=\frac{b}{1-k}$$ is negative.

(1) kb > 0 --> $$k$$ and $$b$$ have the same sign. Now, if $$b>0$$ and $$k=2$$ then the answer is YES but if $$b>0$$ and $$k=\frac{1}{2}$$ then the answer is NO. Not sufficient.

(2) k > 1 --> denominator of $$x=\frac{b}{1-k}$$ is negative, but we have no info about $$b$$. Not sufficient.

(1)+(2) Since from (2) $$k$$ is positive and from (1) $$k$$ and $$b$$ have the same sign, then $$b$$ is positive too. So, numerator ($$b$$) is positive and denominator ($$1-k$$) is negative, which means that $$x=\frac{b}{1-k}$$ is negative. Sufficient.

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Re: M19-29   [#permalink] 08 Jun 2013, 03:29
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# M19-29

Moderator: Bunuel

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