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
OA is C
How this question can be solved graphically.
Sachin
If k does not equal -1, 0 or 1, does the point of intersecti
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- sachin_yadav
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Line 1: y = kx + bsachin_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 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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- sachin_yadav
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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
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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One way to plot a line: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
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:
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:
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:
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:
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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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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