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If k does not equal -1, 0 or 1, does the point of intersection of line 13 Dec 2014, 03:58
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C
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85% (hard)

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54% (02:32) correct 46% (02:37) wrong based on 263 sessions

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


(1) \(kb \gt 0\)

(2) \(k \gt 1\)

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C
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If k does not equal -1, 0 or 1, does the point of intersection of line 13 Dec 2014, 05:13
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If \(k\) is not equal to -1, 0, or 1, does the point of intersection of the line \(y = kx + b\) and the 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 \(x = ky + b\)). Set them equal to obtain the x-coordinate of the intersection point: \(kx + b=\frac{x}{k}-\frac{b}{k}\), which simplifies to \(x=\frac{b(k+1)}{1-k^2}=\frac{b(k+1)}{(1-k)(1+k)}=\frac{b}{1-k}\).

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

(1) \(kb \gt 0\).

This statement indicates that \(k\) 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) \(k \gt 1\).

The above implies that the denominator of \(x=\frac{b}{1-k}\) is negative, but we lack information about \(b\). Not sufficient.

(1)+(2) From statement (2) we know that \(k\) is positive, and from statement (1) we infer that \(k\) and \(b\) share the same sign, thus \(b\) must also be positive. Hence, the numerator (\(b\)) is positive and the denominator (\(1-k\)) is negative (as \(k > 1\)), indicating that \(x=\frac{b}{1-k}\) is negative. Sufficient.


Answer: C

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If k does not equal -1, 0 or 1, does the point of intersection of line 21 Oct 2017, 04:12
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Bunuel
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.


Answer: C.

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How did u get kx+b=x/k-b/k which gives x=b(k+1)/1-k^2
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If k does not equal -1, 0 or 1, does the point of intersection of line 21 Oct 2017, 04:35
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Bunuel
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.


Answer: C.

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

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

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

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

\(\frac{(bk+b)}{(1-k^2)}=x\).
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If k does not equal -1, 0 or 1, does the point of intersection of line 23 May 2020, 08:26
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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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Asked: 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?
Intersection of line y and line x:
\(y = kx + b = \frac{(x-b)}{k} = \frac{x}{k} - \frac{b}{k}\)
\((k-\frac{1}{k})x + (b+\frac{b}{k}) = 0\)
\(x = b(1+\frac{1}{k})/(\frac{1}{k} - k) ; 1+\frac{1}{k} = \frac{(k+1)}{k} & \frac{1}{k}-k = \frac{(1-k^2)}{k} =\frac{ (1+k)(1-k)}{k}\)
\(x = \frac{b}{(1-k)}; y=\frac{kb}{(1-k)} + b = \frac{(kb + b-bk)}{(1-k)} = \frac{b}{(1-k)}\)
x - coordinate of the point of intersection of line y=kx + b and line x=ky+b = \(\frac{b}{1-k}\)

1. kb > 0
k & b have same sign
NOT SUFFICIENT

2. k > 1
1-k <0
But sign of b is unknown
NOT SUFFICIENT

(1) + (2)
1. kb > 0
k & b have same sign
2. k > 1>0
1-k <0
Since k>0 ; b>0
\(x = \frac{b}{1-k} < 0\)
SUFFICIENT

IMO C
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If k does not equal -1, 0 or 1, does the point of intersection of line 05 Jul 2023, 01:26
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