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

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Manager
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Line L passes through point (1,4) and the product of its [#permalink] New post 04 May 2006, 08:30
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)

I got confused and selected E.
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Re: quadrant [#permalink] New post 04 May 2006, 09:06
kuristar 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)


wow, good one.... B . (5, 7)
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 [#permalink] New post 04 May 2006, 09:10
Is the answer B .?? (5,7)

i used different approach so not sure of the answer

say the intercept of x axis be a
the intercept of y axia be b

so x/a + y/b =1

slope = -b/a --- equation 1

but given a * b = -1

so a = -1/b substitute this value in equation 1

u get slope = b^2 which mean slope will always be positive

now from the point (1,4 ) find the slope with each of the no given and only
option B (5,7) has a positive slope rest of them gives -ive slope .

Do let me know the answer , as i've mentioned above i never solved using the above way , just wanted to solve it fast and thought of the above way .i didnt want to solve it in 10 min but just around 2 min .I'm curious to know the answer . Thanks.
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 [#permalink] New post 04 May 2006, 09:10
Sorry to be the 'dumb blonde' but could you explain?

Thank you!
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 [#permalink] New post 04 May 2006, 09:31
You are right, the oa is (5,7).

Now I have to try to get it right!
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 [#permalink] New post 05 May 2006, 02:17
I dont see why I did that, but still I took the slope for each of the cases,

ie. (y2 - y1)/ (x2 - x1)

For all except B, I got a negative slope.

Since B was the only different one, I selected B.

But can someone please explain why the slope has to be positive?
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 [#permalink] New post 05 May 2006, 02:44
remgeo,

if you see my post above i have derived why slope has to be positive .

the slope has to be B^2 and obviously that will always be positive .

Thanks.
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 [#permalink] New post 05 May 2006, 23:49
I selected B.

I just drew out an xy graph and put a dot on (1,4). From there, I knew that the product of the x intercept and y intercept needed to be negative.

So I went down through the answer choices and looked for a line that crossed either the x or the y axis at a negative point.

(A) Doesn't work because when you draw the line, the x and y intercepts are both positive. Hence when you multiple them, the number will be positive. If you want to solve it out, the x intercept here would be (5,0). The y intercept would be (0,5). When you multiply, the ans is positive.
(B) When you plot this point and draw a line that goes through (1,4), you will see a negative slope. The y intercept is positive, but if you keep tracing the line, it crosses the x axis at a negative point. When you multiply the two products, the number will be negative. B is correct.
(C) Same situation as A
(D) Same situation as A. Despite the negative number, the x intercept is still positive.
(E) Same situation as A. Despite the negative number, the x intercept is still positive.
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 [#permalink] New post 05 May 2006, 23:57
Answer is B...

I found drawing the graph easier than solving it using y=mx+c!
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 [#permalink] New post 06 May 2006, 11:16
Agree with B :wink:
  [#permalink] 06 May 2006, 11:16
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