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

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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.


Answer: C.

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

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New post 21 Oct 2017, 04:12
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.


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

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New post 21 Oct 2017, 04:35
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.


Answer: C.

M19-29


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


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

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

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

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

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

New to the Math Forum?
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Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
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Kudos [?]: 129475 [0], given: 12201

Re: If k does not equal -1, 0 or 1, does the point of intersection of line   [#permalink] 21 Oct 2017, 04:35
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