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Re: The y intercept of a line L is 4. If the slope of L is [#permalink]

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12 Nov 2007, 20:29

Balvinder wrote:

The y intercept of a line L is 4. If the slope of L is negative, which of the fillowing could be the x intercept of L. I -1 II 0 III 6

A) I only B) II only C) III only D) I and II E) I and III

The stem tells us that the line is downward sloping (from left to right) and intercepts the y-axis at 4. Therefore, the all possible points of intersection on the x-axis will be positive. III is the only positive option. Answer C.

equation of a line y = mx+c. If y intercept of line l is 4, then one of the points is (0,4) - by making x zero.

We know that the slope 'm' is negative. To find the x intercept of the line, make y zero in the above line equation. We get x = -c/m. Since the value of c is 4 and the slope is negative. x has to be positive, and not negative or 0.

Re: The y intercept of line l is 4. If the slope of line l is [#permalink]

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22 Jul 2012, 18:48

Hi Bunuel, I am not convinced with the answer for the above question. Let me tell you how I approached it. The equation is y=mx+4, where 4 is the y-intercept (given) and since it has got a negative slope, the equation becomes, y=(-m)x+4. Now, to determine x-intercept, we consider y=0, then, -mx+4=0, hence mx=4 and since m has to be negative (it is given that slope is negative for the line), I chose x=-1, so that m=-4. According to the above calculations, I get the answer as A. But the answer is C. Please help me get the concept right.

Re: The y intercept of line l is 4. If the slope of line l is [#permalink]

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23 Jul 2012, 00:47

The answer is C. If the slope is negative of the line then either it passes through the 1 st quadrant <starting from II towards IV> or passes through the third quadrant <starting from II towards IV>.. Thus, the intercepts are either both positive or both negative. Given in the question, its a positive intercept, the other one will be a positive as well. Thus, 6.

Re: The y intercept of line l is 4. If the slope of line l is [#permalink]

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23 Jul 2012, 02:16

1

This post received KUDOS

sharmila79 wrote:

Hi Bunuel, I am not convinced with the answer for the above question. Let me tell you how I approached it. The equation is y=mx+4, where 4 is the y-intercept (given) and since it has got a negative slope, the equation becomes, y=(-m)x+4. Now, to determine x-intercept, we consider y=0, then, -mx+4=0, hence mx=4 and since m has to be negative (it is given that slope is negative for the line), I chose x=-1, so that m=-4. According to the above calculations, I get the answer as A. But the answer is C. Please help me get the concept right.

No. You are mixing it up. the slope is negative Hence you can write y=(-m)x+4 since -m is negative, in this case m must be positive again your slope is -m, not m!!

So later if you say mx=4 x is necessarily positive since m is positive.

Re: The y intercept of line l is 4. If the slope of line l is [#permalink]

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23 Jul 2012, 02:18

Also, even though the answer is C, this problem is not very well written IMO since an infinite negative slope could cause the x intercept to be 0 :D (though "The intercept" can imply that there is only one y intercept)

Re: The y intercept of a line L is 4. If the slope of L is [#permalink]

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22 Jul 2014, 19:44

slope(Intercept form) = - y(intercept)/ x (intercept) we are also given that slope should be negative. we are told that "y intercept of a line L is 4".

s = -4/x;

x has to be positive to maintain a negative slope .

The y intercept of a line L is 4. If the slope of L is negative, which of the following could be the x intercept of L.

I. -1 II. 0 III. 6

A. I only B. II only C. III only D. I and II E. I and III

Slope of a line = – (y intercept)/(x intercept) Since slope is given to be negative, and y intercept is positive, we can say that x intercept must be positive too. So x intercept can be only III.

Re: The y intercept of line l is 4. If the slope of line l is [#permalink]

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27 Aug 2015, 06:52

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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