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# A straight line passes through (1,p). Does it have a

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A straight line passes through (1,p). Does it have a [#permalink]

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23 Nov 2005, 22:24
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

A straight line passes through (1,p). Does it have a positive slope?

1). It passes through (-p, 13)
2). It passes through (0,1)

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Re: Slope of a line [#permalink]

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23 Nov 2005, 23:40
C. from i and ii, individually, we can do nothing. but from i and ii, togather, it can be said thet the slope is +ve. if p=-ve, the three points cannot make a straight line. only if p=+ve makes the line a straigt. so, it is C.

will revise, if any..
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Re: Slope of a line [#permalink]

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23 Nov 2005, 23:42
getzgetzu wrote:
A straight line passes through (1,p). Does it have a positive slope?

1). It passes through (-p, 13)
2). It passes through (0,1)

1. the slope is (1+p)/ (p-13) ...can't tell if it's negative or positive
in case p=13 ( the fraction doesn't exist) then the line is parallel to x-axis and pass point (1,13) --->insuff

2. similar to 1 ---> insuff

1 and 2:
to see if there's case in which the line passes the three point. Let make the two slopes equal
(1+p)/(p-13) = 1/(p-1) ---> p^2-p+12=0 , this equation has no solution ---> the three points can't be in the line together
in case p=13 and 1, the same.
-->insuff
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23 Nov 2005, 23:46
OA is E.... Laxi u just rock! Amazing explaination
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Re: Slope of a line [#permalink]

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23 Nov 2005, 23:47
laxieqv wrote:
1. the slope is (1+p)/ (p-13) ...can't tell if it's negative or positive
in case p=13 ( the fraction doesn't exist) then the line is parallel to x-axis and pass point (1,13) --->insuff
2. similar to 1 ---> insuff
1 and 2:
to see if there's case in which the line passes the three point. Let make the two slopes equal
(1+p)/(p-13) = 1/(p-1) ---> p^2-p+12=0 , this equation has no solution ---> the three points can't be in the line together
in case p=13 and 1, the same. -->insuff

laxi, the information says that it is a straight line and passes through (1,p), (-p,13) and (0,1), then it must have +ve slop otherwise it is not a st line. and given information is that it is a st line. so C.
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23 Nov 2005, 23:54
A straight line passes through (1,p). Does it have a positive slope?

1). It passes through (-p, 13)
2). It passes through (0,1)

slope = (y1-y2)/(x1-x2)

1) slope = (p-13)/(1+p). insufficient
2) slope = (p-1). insufficient

1&2) slope = (p-13)/(1+p) = (p-1)
(p-13) = -(1+p)(1-p) = p^2-1
p^2-p+12=0
p = [1+root(-47)]/2
p = [1-root(-47)]/2

Insufficient.

(E)
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Re: Slope of a line [#permalink]

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23 Nov 2005, 23:56
HIMALAYA wrote:
laxieqv wrote:
1. the slope is (1+p)/ (p-13) ...can't tell if it's negative or positive
in case p=13 ( the fraction doesn't exist) then the line is parallel to x-axis and pass point (1,13) --->insuff
2. similar to 1 ---> insuff
1 and 2:
to see if there's case in which the line passes the three point. Let make the two slopes equal
(1+p)/(p-13) = 1/(p-1) ---> p^2-p+12=0 , this equation has no solution ---> the three points can't be in the line together
in case p=13 and 1, the same. -->insuff

laxi, the information says that it is a straight line and passes through (1,p), (-p,13) and (0,1), then it must have +ve slop otherwise it is not a st line. and given information is that it is a st line. so C.

As i proved this straight line can't pass the three points no matter if slope is + or - ...
I simply proved that two statements can't go together ----> not to mention whether they both can supply sufficient information for answer or not ...
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24 Nov 2005, 00:06
gamjatang wrote:
A straight line passes through (1,p). Does it have a positive slope?

1). It passes through (-p, 13)
2). It passes through (0,1)

slope = (y1-y2)/(x1-x2)

1) slope = (p-13)/(1+p). insufficient
2) slope = (p-1). insufficient

1&2) slope = (p-13)/(1+p) = (p-1)
(p-13) = -(1+p)(1-p) = p^2-1
p^2-p+12=0
p = [1+root(-47)]/2
p = [1-root(-47)]/2

Insufficient.

(E)

o bba, you can't take roots of negative numbers, except you're considering imaginary numbers
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Re: Slope of a line [#permalink]

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24 Nov 2005, 00:15
laxieqv wrote:
HIMALAYA wrote:
laxieqv wrote:
1. the slope is (1+p)/ (p-13) ...can't tell if it's negative or positive
in case p=13 ( the fraction doesn't exist) then the line is parallel to x-axis and pass point (1,13) --->insuff
2. similar to 1 ---> insuff
1 and 2:
to see if there's case in which the line passes the three point. Let make the two slopes equal
(1+p)/(p-13) = 1/(p-1) ---> p^2-p+12=0 , this equation has no solution ---> the three points can't be in the line together
in case p=13 and 1, the same. -->insuff

laxi, the information says that it is a straight line and passes through (1,p), (-p,13) and (0,1), then it must have +ve slop otherwise it is not a st line. and given information is that it is a st line. so C.

As i proved this straight line can't pass the three points no matter if slope is + or - ...
I simply proved that two statements can't go together ----> not to mention whether they both can supply sufficient information for answer or not ...

you are proving that the line cannot passes the three points. but it is not the question. there is no dispute whether the line passes or not the three points. the line passes the three points as the information given in the question and statement says that the line passes through the three points. so if the line passes the three points, then it must have +ve slop.
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24 Nov 2005, 00:18
laxieqv wrote:
gamjatang wrote:
A straight line passes through (1,p). Does it have a positive slope?

1). It passes through (-p, 13)
2). It passes through (0,1)

slope = (y1-y2)/(x1-x2)

1) slope = (p-13)/(1+p). insufficient
2) slope = (p-1). insufficient

1&2) slope = (p-13)/(1+p) = (p-1)
(p-13) = -(1+p)(1-p) = p^2-1
p^2-p+12=0
p = [1+root(-47)]/2
p = [1-root(-47)]/2

Insufficient.

(E)

o bba, you can't take roots of negative numbers, except you're considering imaginary numbers

I was trying to say that NO SUCH LINE EXISTS.

How do we know if the slope of the line is positive or negative when the line doesn't even exist?

Thus (E).
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Re: Slope of a line [#permalink]

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24 Nov 2005, 00:24
HIMALAYA wrote:
C. from i and ii, individually, we can do nothing. but from i and ii, togather, it can be said thet the slope is +ve. if p=-ve, the three points cannot make a straight line. only if p=+ve makes the line a straigt. so, it is C.

will revise, if any..

uhm, can you prove the bold part?! ...if i missed sth here...anyway, OA is E ..
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24 Nov 2005, 00:26
gamjatang wrote:
laxieqv wrote:
gamjatang wrote:
A straight line passes through (1,p). Does it have a positive slope?

1). It passes through (-p, 13)
2). It passes through (0,1)

slope = (y1-y2)/(x1-x2)

1) slope = (p-13)/(1+p). insufficient
2) slope = (p-1). insufficient

1&2) slope = (p-13)/(1+p) = (p-1)
(p-13) = -(1+p)(1-p) = p^2-1
p^2-p+12=0
p = [1+root(-47)]/2
p = [1-root(-47)]/2

Insufficient.

(E)

o bba, you can't take roots of negative numbers, except you're considering imaginary numbers

I was trying to say that NO SUCH LINE EXISTS.

How do we know if the slope of the line is positive or negative when the line doesn't even exist?

Thus (E).

you're right!
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Re: Slope of a line [#permalink]

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24 Nov 2005, 00:30
laxieqv wrote:
HIMALAYA wrote:
C. from i and ii, individually, we can do nothing. but from i and ii, togather, it can be said thet the slope is +ve. if p=-ve, the three points cannot make a straight line. only if p=+ve makes the line a straigt. so, it is C.
will revise, if any..

uhm, can you prove the bold part?! ...if i missed sth here...anyway, OA is E ..

i might be wrong, but we should not be going after OA rather we should always endevor to find what is right/correct. i am sending you a PM.
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Re: Slope of a line [#permalink]

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24 Nov 2005, 00:39
HIMALAYA wrote:
laxieqv wrote:
HIMALAYA wrote:
C. from i and ii, individually, we can do nothing. but from i and ii, togather, it can be said thet the slope is +ve. if p=-ve, the three points cannot make a straight line. only if p=+ve makes the line a straigt. so, it is C.
will revise, if any..

uhm, can you prove the bold part?! ...if i missed sth here...anyway, OA is E ..

i might be wrong, but we should not be going after what is given OE. we should always endevor to find what is right and correct. i am sending u PM.

hik, i am not distracted by the OA ...it came after my post, didn't it?! ...Also, OA may be wrong at times ...I just hope you prove the bold part ...it's easy for one to ignore a point ..I do so often..that's why i just want you to clarify the bold part so that i can know what i missed.
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24 Nov 2005, 03:09
unless you know the value of P you can never calculate the slope.

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24 Nov 2005, 20:09
I agree that this question is a little ambiguous. If a line doesn't exist, then does it have a positive slope? I would say No.
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24 Nov 2005, 20:09
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# A straight line passes through (1,p). Does it have a

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