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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) \(k \gt 1\)
(1) \(kb \gt 0\)
(2) \(k \gt 1\)
16 Sep 2014, 01:07
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Official Solution:
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
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
