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M19-29

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M19-29  [#permalink]

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New post 16 Sep 2014, 00:06
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Question Stats:

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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) \(kb \gt 0\)

(2) \(k \gt 1\)

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M19-29  [#permalink]

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New post 16 Sep 2014, 00:07
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Official Solution:


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: \(kx + b=\frac{x}{k}-\frac{b}{k}\), which gives \(x=\frac{b(k+1)}{1-k^2}=\frac{b(k+1)}{(1-k)(1+k)}=\frac{b}{1-k}\).

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

(1) \(kb \gt 0\). This statement tells that \(k\) and \(b\) have the same sign. Now, if \(b \gt 0\) and \(k=2\) then the answer is YES but if \(b \gt 0\) and \(k=\frac{1}{2}\) then the answer is NO. Not sufficient.

(2) \(k \gt 1\). So, the 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.


Answer: C.
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Re: M19-29  [#permalink]

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New post 20 Jul 2016, 11:16
Hi Bunuel,

Is there a way to resolve this using graphical method? I tried but I am facing deadlock in statement 1.
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Re: M19-29  [#permalink]

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New post 17 Aug 2016, 05:06
Bunuel,

how do you get from kx+b= (x - b) / k to x = b(k+1)/1−k2 ? the "1-k2" in the denominator i just cant figure out how you get that.
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Re: M19-29  [#permalink]

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New post 17 Aug 2016, 06:05
lydennis8 wrote:
Bunuel,

how do you get from kx+b= (x - b) / k to x = b(k+1)/1−k2 ? the "1-k2" in the denominator i just cant figure out how you get that.


\(kx+b= \frac{(x - b)}{k}\)

Multiply by k: \(k^2x + bk = x - b\)

Rearrange: \(bk + b = x - k^2x\)

Factor out b and x: \(b(k + 1) = x(1 - k^2)\)

\(\frac{b(k + 1)}{(1 - k^2)} = x\).

Hope it's clear.
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Re: M19-29  [#permalink]

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New post 24 Aug 2016, 05:08
I know this is a silly question: how did you get y=x/k-b/k?
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Re: M19-29  [#permalink]

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New post 24 Aug 2016, 05:10
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Re: M19-29  [#permalink]

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New post 13 Dec 2017, 10:45
hey!

would it be correct to solve this question this way?

ky+b=kx+b
x=0, y=b ; y=0 x=b
(1) both negative or both positive - not suf
(2) not suf
(1+2) if k is positive then according to (1) b also positive - then x positive.

am i right?

thanks!
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Re: M19-29 &nbs [#permalink] 13 Dec 2017, 10:45
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