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# If k does not equal -1, 0 or 1, does the point of intersection of line

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If k does not equal -1, 0 or 1, does the point of intersection of line  [#permalink]

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13 Dec 2014, 02:58
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53% (02:16) correct 47% (02:35) wrong based on 128 sessions

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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 > 0
2. k > 1
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Joined: 02 Sep 2009
Posts: 61396
Re: If k does not equal -1, 0 or 1, does the point of intersection of line  [#permalink]

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13 Dec 2014, 04:13
5
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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?

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.

M19-29
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Re: If k does not equal -1, 0 or 1, does the point of intersection of line  [#permalink]

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21 Oct 2017, 03:12
Bunuel 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?

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.

M19-29

How did u get kx+b=x/k-b/k which gives x=b(k+1)/1-k^2
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Joined: 02 Sep 2009
Posts: 61396
Re: If k does not equal -1, 0 or 1, does the point of intersection of line  [#permalink]

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21 Oct 2017, 03:35
zanaik89 wrote:
Bunuel 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?

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.

M19-29

How did u get kx+b=x/k-b/k which gives x=b(k+1)/1-k^2

$$kx + b=\frac{x}{k}-\frac{b}{k}$$;

$$b+\frac{b}{k}=\frac{x}{k}-kx$$;

$$b+\frac{b}{k}=x(\frac{1}{k}-k)$$;

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

$$\frac{(bk+b)}{(1-k^2)}=x$$.
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Re: If k does not equal -1, 0 or 1, does the point of intersection of line  [#permalink]

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29 Nov 2019, 22:20
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Re: If k does not equal -1, 0 or 1, does the point of intersection of line   [#permalink] 29 Nov 2019, 22:20
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