A line passes through the point (1,P). Is the slope of this line>0?
1) Point (p, 13) lies on the line
2) Ponts (0, 1) lies on the line
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309) is slope negative
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mohit11
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P could be positive or negative.
A: Possible points
(1, P) (P, 13) - P is positive
(1, -P) (-P 13) - P is negative
Slope - Y2 -Y1/X2 - X1
1st case - 13 - P/ P - 1 if P =1 , then slope is infinity
If P>1 and less than 13 , then slope is positive
if P> 1 and greater than 13 , then the slope is negative.
Doing the same approximation for second set of points is not required, since we cannot answer the question using A.
Possible correction answers : B C E
B : Possible points
(1, P) (0,1)
(1 -P) (0,1)
1st case slope - 1-P/0-1 .. If P =1 , then Slope = 0
if P >1, then slope is negative
Again, approximation for second case is not required since using B we cannot answer the question
Possible answers - C, E
If we use both A, B
Then Line passes through, (1,P), (P,13) and (0,1)
Again depending on whether P is positive or negative, Slope could be positive/negative or 0
Hence Answer is E
Please post the OA
A: Possible points
(1, P) (P, 13) - P is positive
(1, -P) (-P 13) - P is negative
Slope - Y2 -Y1/X2 - X1
1st case - 13 - P/ P - 1 if P =1 , then slope is infinity
If P>1 and less than 13 , then slope is positive
if P> 1 and greater than 13 , then the slope is negative.
Doing the same approximation for second set of points is not required, since we cannot answer the question using A.
Possible correction answers : B C E
B : Possible points
(1, P) (0,1)
(1 -P) (0,1)
1st case slope - 1-P/0-1 .. If P =1 , then Slope = 0
if P >1, then slope is negative
Again, approximation for second case is not required since using B we cannot answer the question
Possible answers - C, E
If we use both A, B
Then Line passes through, (1,P), (P,13) and (0,1)
Again depending on whether P is positive or negative, Slope could be positive/negative or 0
Hence Answer is E
Please post the OA
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KFreudenberg
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To determine the slope you need to use y1-y2/x1-x2
For statement 1:
(1, P) = (x1, y1) and (P, 13) = (x2, y2) so (P-13)/(1-P) = slope
if P is 1 then there is a zero in the denominator which deems the slope undefined
You could test other numbers to see that the slope could be positive, negative or zero, but you already know one combination that deems the slope undefined so statement 1 is insufficient
This leaves us with B, C, and E
For statement 2:
(0, 1) = (x2, y2)
(P-1)/(1-0) = slope
Using this statement the denominator is already defined for us as positive (1).
If P > 1, then slope is positive
If P < 1, then slope is negative
With two possible values, statement 2 is insufficient
This leaves us with C and E.
For answer option C I used all the points to do equation y = mx +b to see if I could get a specific value for p.
13 = p * (p-1)/(1) + 1
12 = p^2 - p which means P could be 4 or -3
Looks like answer E is correct.
I think this is the most thorough way, but being thorough has really cost me time in some of my practice GMATs if there are quicker "tricks" that bring you to E, please let me know!
For statement 1:
(1, P) = (x1, y1) and (P, 13) = (x2, y2) so (P-13)/(1-P) = slope
if P is 1 then there is a zero in the denominator which deems the slope undefined
You could test other numbers to see that the slope could be positive, negative or zero, but you already know one combination that deems the slope undefined so statement 1 is insufficient
This leaves us with B, C, and E
For statement 2:
(0, 1) = (x2, y2)
(P-1)/(1-0) = slope
Using this statement the denominator is already defined for us as positive (1).
If P > 1, then slope is positive
If P < 1, then slope is negative
With two possible values, statement 2 is insufficient
This leaves us with C and E.
For answer option C I used all the points to do equation y = mx +b to see if I could get a specific value for p.
13 = p * (p-1)/(1) + 1
12 = p^2 - p which means P could be 4 or -3
Looks like answer E is correct.
I think this is the most thorough way, but being thorough has really cost me time in some of my practice GMATs if there are quicker "tricks" that bring you to E, please let me know!
