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Line L passes through point (1,4), and the product of its

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Line L passes through point (1,4), and the product of its [#permalink] New post 19 Mar 2006, 13:08
00:00
A
B
C
D
E

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0% (00:00) correct 100% (00:39) wrong based on 0 sessions
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Line L passes through point (1,4), and the product of its intercepts with axis-x and axis-y is negative. Which of the following point is on the line L?


A. (2,3)
B. (5,7)
C. (4,2)
D. (4,-1)
E. (5, -1)
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 [#permalink] New post 19 Mar 2006, 13:13
B (5,7)
We can deduce from the conditions in the question that slope of the line > 0. From there onwards, it is POE.
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 [#permalink] New post 19 Mar 2006, 13:22
I do understand that B leads to a positive slope.. but what I dont understand is why we need a positive slope?? because the product of the intecepts is negative?
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 [#permalink] New post 19 Mar 2006, 13:26
Yes...
Let us take the equation of the line to be y = mx + c. (We do not know the nature of the numbers yet).
x-intercept (when y = 0) = -c/m
y-intercept (when x = 0) = c
Product of intercepts = -(c^2)/m < 0 as per the question.
=> m < 0 since c^2 is always positive.
Hope this helps.
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 [#permalink] New post 19 Mar 2006, 13:31
chuckle wrote:
Yes...
Let us take the equation of the line to be y = mx + c. (We do not know the nature of the numbers yet).
x-intercept (when y = 0) = -c/m
y-intercept (when x = 0) = c
Product of intercepts = -(c^2)/m < 0 as per the question.
=> m < 0 since c^2 is always positive.
Hope this helps.


Going by your logic, Only C has -ve slope (= -2/3). So C should be the answer.
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 [#permalink] New post 19 Mar 2006, 13:39
Sorry giddi...
I made a mistake. m should be > 0.
Also, all the answer choices except B give negative slope...
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 [#permalink] New post 19 Mar 2006, 14:34
chuckle wrote:
Sorry giddi...
I made a mistake. m should be > 0.
Also, all the answer choices except B give negative slope...


Thanks! I also missed the -ve sign :twisted:

And no need for sorry pal. Your solution is really neat :beer
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 [#permalink] New post 20 Mar 2006, 09:49
Yes, indeed very neat solution :)

My approach to this problem was slightly different.

I visualized the line ;), if product of y intercept & x intercept is < 0,
it means the line cuts the in such a way that the slope is > 0.

Only (5,7) satisfies the requirement!
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Re: PS + Coordinate geometry [#permalink] New post 20 Mar 2006, 11:56
bewakoof wrote:
Line L passes through point (1,4), and the product of its intercepts with axis-x and axis-y is negative. Which of the following point is on the line L?


A. (2,3)
B. (5,7)
C. (4,2)
D. (4,-1)
E. (5, -1)


Correct me if I am wrong, but shouldn't the question be worded:
"Which of the following points could be on the line L?"

Good question, nonetheless. Where is it from?
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 [#permalink] New post 21 Mar 2006, 09:05
It's obvious when you make a drawing. Line L can be only in the direction of red lines (y+, x- or y-, x+). The only point that meets the criteria is point B ... green line is line L.
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 [#permalink] New post 21 Mar 2006, 14:55
general eqn of the line:
x/a+y/b=1.
m = -a/b
we have ab < 0.
so m>0,

therefore, angle made at the x axis is acute.
only (B) does that.

Thanks,
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Re: PS + Coordinate geometry [#permalink] New post 21 Mar 2006, 18:21
kook44 wrote:
bewakoof wrote:
Line L passes through point (1,4), and the product of its intercepts with axis-x and axis-y is negative. Which of the following point is on the line L?


A. (2,3)
B. (5,7)
C. (4,2)
D. (4,-1)
E. (5, -1)


Correct me if I am wrong, but shouldn't the question be worded:
"Which of the following points could be on the line L?"

Good question, nonetheless. Where is it from?


i got this question from some other forum.. I dont remember, from where..
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Re: PS + Coordinate geometry   [#permalink] 21 Mar 2006, 18:21
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