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Re: DS co-ordinate geometry question [#permalink]
14 Mar 2012, 01:07

thanks , i also got the following explanation and dont understand the logic behind their deduction from 1 ,

Explanation

If a line has negative slope, the intercepts will have the same sign. So if we can find the sign of the x-intercept, we can answer the question.

Statement (1) is insufficient. It's possible that both intercepts are negative, for instance if the x-intercept is -4, the y-intercept could be -2. This is a relatively flat slope--as it turns out, it's true if the slope is greater than -1. It's also possible that both intercepts are positive. For instance, if the x-intercept is 3, the y-intercept could be 5. The negative slope here is steeper--in general, less than -1.

Statement (2) is also insufficient. Such a slope is relatively steep, but it could result in positive or negative intercepts--the slope of the line doesn't determine the location of the line.

Taken together, the statements are sufficient. In (1), we learned that if the slope is less than -1, both intercepts are positive. Since the slope is less than -2, both intercepts must be positive. Choice (C) is correct.

Re: DS co-ordinate geometry question [#permalink]
14 Mar 2012, 19:11

vdadwal wrote:

thanks , i also got the following explanation and dont understand the logic behind their deduction from 1 ,

Explanation

If a line has negative slope, the intercepts will have the same sign. So if we can find the sign of the x-intercept, we can answer the question.

Statement (1) is insufficient. It's possible that both intercepts are negative, for instance if the x-intercept is -4, the y-intercept could be -2. This is a relatively flat slope--as it turns out, it's true if the slope is greater than -1. It's also possible that both intercepts are positive. For instance, if the x-intercept is 3, the y-intercept could be 5. The negative slope here is steeper--in general, less than -1.

Statement (2) is also insufficient. Such a slope is relatively steep, but it could result in positive or negative intercepts--the slope of the line doesn't determine the location of the line.

Taken together, the statements are sufficient. In (1), we learned that if the slope is less than -1, both intercepts are positive. Since the slope is less than -2, both intercepts must be positive. Choice (C) is correct.

Can you help ?

thanks

proceed graphically and check the slope, 1) when the intercepts are in first quadrant, you will see the slope should be less than tan(135) i.e. less than -1 to satisfy the condition y>x intercept. (at -1 you will see x=y intercept) similarly, when in third quadrant slope should be greater than tan (135) i.e. -1

insufficient

2) insufficient

both 1 and 2 slope less than -2 i.e. less than -1 hence both intercept are positive.

hope this clarifies _________________

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Re: In the xy-plane, if line k has negative slope, is the [#permalink]
15 Mar 2012, 08:15

1

This post received KUDOS

Expert's post

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In the xy-plane, if line k has negative slope, is the y-intercept of line k positive?

Equation of a line in point intercept form is \(y=mx+b\), where: \(m\) is the slope of the line and \(b\) is the y-intercept of the line (the value of \(y\) for \(x=0\)). So, basically we are asked whether \(b>0\).

(1) The x-intercept of line k is less than the y-intercept of line k --> x-intercept is value of \(x\) for \(y=0\), so it's \(-\frac{b}{m}\). The statement says that: \(-\frac{b}{m}<b\) --> multiply by negative \(m\) and flip the sign of the inequality: \(-b>bm\) --> \(b(m+1)<0\). Now, in order \(b>0\) to be true \(m+1\) should be negative, so the question becomes: is \(m+1<0\)? --> is \(m<-1\). We don't know that. Not sufficient.

(2) The slope of line k is less than -2. Insufficient on its own.

(1)+(2) From (1) the question became: "is \(m<-1\)?" and (2) says that \(m<-2\). Sufficient.

Re: In the xy-plane, if line k has negative slope, is the [#permalink]
09 Aug 2014, 03:03

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Re: In the xy-plane, if line k has negative slope, is the [#permalink]
27 Aug 2015, 20:47

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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