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# Intercepts

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Intern
Joined: 15 Apr 2003
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03 Jun 2003, 08:41
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

This was already posted but can someone explain why its E?

I get B.

A line intercepts x-axis at "a" and y-axis at "b". Is the slope of the line negative?

1) a^2 = b^2
2) a^3 = b^3

I get B. (I'm assuming that both niether a or b= 0 otherwise this problem would be pointless.)

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Intern
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03 Jun 2003, 13:56
I dont think you can assume that a=b=0.

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Manager
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03 Jun 2003, 14:20
I also think it is B
Any number rised to an odd number maintain its sign (positive or negative), so a=b, so slope can not be negative

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Manager
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04 Jun 2003, 01:23
Why is it unreasonable to assume that a and b can equal 0???

Of course that's possible...what? you've never seen a graph that passes through the origin? E is the asnwer

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Intern
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04 Jun 2003, 02:00

A line intercepts x-axis at "a" and y-axis at "b". Is the slope of the line negative?

1) a^2 = b^2
2) a^3 = b^3

Why all you guys debate over the value "0" ?
Just try this way.
The question asked you whether the slope negative. Also it can assume that x=a and y=b Therefore, the slope can calculatyed on B/A

In statement I: if B positive, A Negative then slope is negative -- YES
if B positive, A positive then slope is positive -- NO

In statement II : it can infer from that equation and A and B MUST have the same value. Thus, no matter how you calculate the slope is always be positive. So it is enough to answer this question which is NO.

In sum, I go for B surely. for those who doubt about the value of 0 I don't know what you are thinking of ?

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Manager
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04 Jun 2003, 02:47
1
KUDOS
A line intercepts x-axis at "a" and y-axis at "b", where "a" is positive. Is the slope of the line negative?

1) a^2 = b^2
2) a^3 = b^3

the answer to this question is B, because u know that "a" and "b" will both have the same sign, because it is given that "a" is positive and so "a" cannot be zero

A line intercepts x-axis at "a" and y-axis at "b". Is the slope of the line negative?

1) a^2 = b^2
2) a^3 = b^3

in this question we don't know whether "a" is positive or negative or zero........a line passing through the origin can have a negative slope (line going from second quadrant to fourth quadrant through the origin), a positive slope (line going from third quadrant to first quadrant through the origin), zero slope (horizontal line) or infinite slope (vertical line).......so the answer should be E

I had the above concept in mind when i framed this question........

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Founder
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05 Jun 2003, 16:35
brstorewala wrote:
A line intercepts x-axis at "a" and y-axis at "b", where "a" is positive. Is the slope of the line negative?

1) a^2 = b^2
2) a^3 = b^3

the answer to this question is B, because u know that "a" and "b" will both have the same sign, because it is given that "a" is positive and so "a" cannot be zero

A line intercepts x-axis at "a" and y-axis at "b". Is the slope of the line negative?

1) a^2 = b^2
2) a^3 = b^3

in this question we don't know whether "a" is positive or negative or zero........a line passing through the origin can have a negative slope (line going from second quadrant to fourth quadrant through the origin), a positive slope (line going from third quadrant to first quadrant through the origin), zero slope (horizontal line) or infinite slope (vertical line).......so the answer should be E

I had the above concept in mind when i framed this question........

Experience....

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Intern
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19 Feb 2008, 15:12
The way I reasoned it is as follows:
In general, the two coordinates are (a, 0) and (0, b). So, the associated slope (m) = -b/a.

1. a^2 = b^2 ==> (a-b)(a+b) = 0; If a = b then m = -1. If a = -b then m = 1. So, unclear.
2. a^3 = b^3 ==> (a-b)(a^2 - ab + b^2) = 0; If a - b = 0 then m = -1. However, the (a^2 - ab + b^2) = 0 with the given info that a^3 = b^3 leads to a = b. This leads to a^2 = b^2 = 0. The slope is undefined.

So, A, B, D are out.

Now, one could argue that using 1 and 2 together, one arrives at m = -1. And thus, C is a possibility.

However, I am not sure if this process of elimination is "correct". Hence, perhaps E is a "safe" bet...
Any thoughts?

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Re: Intercepts   [#permalink] 19 Feb 2008, 15:12
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