Mo2men 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?
1. kb > 0
2. k > 1
Can we solve this question be drawing lines???
Mitch has beautifully explained the graphical solution, here is the Algebraic one.
We wish to know the nature of x-coordinate.
We have two intersecting line: y = kx+b and x = ky+b;
By plugging in the value of y from the first line into the second, we get,
x = k(kx+b)+b
=> x=k^2x+bk+b
=> x(1-k^2) = b(k+1)
=> x(1+k)(1-k)=b(k+1)
=> x=b/(1-k)
Signs of b and k will determine whether x is negative/positive.
S1: kb > 0
kb > 0 implies that kb is positive. Thus, either both are positive or both are negative.
Case 1: b and k are positive.
x=b/(1-k) may or may not be negative.
(a) If 0 < k < 1, x = b / |(1-k)| => x is not negative.
(b) If k > 1, x = b / (-|(1-k)|) => x is negative. No unique answer.
There is no need to discuss case 2 (b and k are negative.) as we already concluded that S1 itself in not sufficient.
S2: k > 1
There is no information about the sign of b. Insufficient.
S1 and S2:
From S1, we know that b and k are of the same sign and from S2, we know that k is positive thus, b and k each is positive.
We saw in the Case 1 analysis that even if b and k each is positive, there is no unique answer. However, if k > 1, we have do not have Case 1.a. Thus, x is negative. Sufficient.
The correct answer:
C
Relevant book:
Manhattan Review GMAT Coordinate Geometry Guide
Hope this helps!
-Jay
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