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If k does not equal -1, 0 or 1, does the point of intersecti

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by sachin_yadav » Sat Apr 30, 2016 11:06 am
Please if anyone can reply to this question
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by GMATGuruNY » Sat Apr 30, 2016 2:25 pm
sachin_yadav 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
Line 1: y = kx + b
Line 2, rephrased with y isolated: y = (1/k)x - (1/k)b
At the point of intersection, the two lines must have the same y-coordinate, implying that the portions in blue are equal:
kx + b = (1/k)x - (1/k)b

kx - (1/k)x = -b - (1/k)b

Multiplying both sides by k, we get:
k²x - x = -bk - b

x(k² - 1) = -b(k+1)

x(k+1)(k-1) = -b(k+1)

x(k-1) = -b

x = -b/(k-1)

The value of x will be negative if b and k-1 have the SAME SIGN.
Question stem, rephrased:
Do b and k-1 have the same sign?

Statement 1:
Case 1: b=1 and k=2, implying that k-1 = 2-1 = 1.
In this case, b and k-1 have the same sign.
Case 2: b=1 and k=1/2, implying that k-1 = 1/2 - 1 = -1/2.
In this case, b and k-1 have different signs.

Since the answer is YES in Case 1 but NO in Case 2, INSUFFICIENT.

Statement 2:
Since k>1, k-1 > 0.
No information about b.
INSUFFICIENT.

Statements combined:
Since bk>0 and k>1, b>0.
Since k>1, k-1>0.
Thus, b and k-1 have the same sign.
SUFFICIENT.

The correct answer is C.
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by sachin_yadav » Wed May 04, 2016 10:25 pm
Mitch,

Thanks for your explanation. I have understood to what you have just explained algebraically; However i am more looking for a graphical solution. I don't have any simulation otherwise i would have made the graph and uploaded it. So that i can ask you where i am making the mistake. The graph that i am making is giving me OA E, which is wrong.
Also, how should i decide whether to solve graphically or algebraically.

Regards
Sachin
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by GMATGuruNY » Wed May 11, 2016 3:25 am
sachin_yadav 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
One way to plot a line:

1. Determine the y-intercept by plugging in x=0 and solving for y
2. Determine the x-intercept by plugging in y=0 and solving for x
3. Draw a line through the two intercepts

Statement 1: kb > 0
Test one case that also satisfies Statement 2.
Case 1: k=2, b=1
Here, the two lines are y=2x+1 and x=2y+1.
The following graph is yielded:
Image

Test one case that does NOT also satisfy Statement 2.
Case 2: k=1/2, b=1
Here, the two lines are y= (1/2)x + 1 and x = (1/2)y + 1.
The following graph is yielded:
Image

Since the point of intersection has a negative x-coordinate in Case 1 but a nonnegative x-intercept in Case 2, INSUFFICIENT.

Statement 2: k>1
Case 1 also satisfies Statement 2.

Case 3: k=2, b=0
Here, the two lines are y=2x and x=2y.
The following graph is yielded:
Image

Since the point of intersection has a negative x-coordinate in Case 1 but a nonnegative x-intercept in Case 3, INSUFFICIENT.

Statements combined:
Case 1 satisfies both statements.

Case 4: k=5, b=10
Here, the two lines are y=5x+10 and x=5y+10.
The following graph is yielded:
Image

Cases 1 and 4 illustrate that -- if kb>0 and k>1 -- the point of intersection must have a negative x-coordinate.
SUFFICIENT.

The correct answer is C.
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