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In the xy-plane, if line k has negative slope, is the

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

(1) The x-intercept of line k is less than the y-intercept of line k.

(2) The slope of line k is less than -2.

It is a DS question, can you help and explain the answer?
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Re: DS co-ordinate geometry question [#permalink]

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New post 14 Mar 2012, 00:51
vdadwal wrote:
In the xy-plane, if line k has negative slope, is the y-intercept of line k positive?

(1) The x-intercept of line k is less than the y-intercept of line k.

(2) The slope of line k is less than -2.

It is a DS question, can you help and explain the answer?


let me try:
we have line k say: y=mx+c
we need to find if c>0

1) x-intercept, i.e. y=0
x=-c/m
y-intercept, i.e. x=0
y=c

hence -c/m<c
=> c*((1/m)+1)>0

i.e. for different value of "m", "c" can be both positive and negative

hence insufficient

2) cant infer anything about c
insufficient

1+2

if m<-2

c*(m+1)<0 (m<0 hence sign change)
as m<-2
hence m+1<-1
i.e. negative
i.e. c>0

Sufficient

hence C

hope it helps..!!!
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Re: DS co-ordinate geometry question [#permalink]

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New post 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.

Can you help ?

thanks
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Re: DS co-ordinate geometry question [#permalink]

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New post 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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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.

Answer: C.
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Re: In the xy-plane, if line k has negative slope, is the [#permalink]

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In the xy-plane, if line k has negative slope, is the [#permalink]

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New post 11 Aug 2014, 04:13
Bunuel,

I dont understand

-b/m<b --> multiply by negative m and flip the sign of the inequality: -b>bm --> b(m+1)<0... can you explain?

IF -b/m<b, then -b<bm....b(m+1)>0...Can you explain how b(1+m) < 0?
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New post 12 Aug 2014, 07:30
saikrishna123 wrote:
Bunuel,

I dont understand

-b/m<b --> multiply by negative m and flip the sign of the inequality: -b>bm --> b(m+1)<0... can you explain?

IF -b/m<b, then -b<bm....b(m+1)>0...Can you explain how b(1+m) < 0?


When you multiply by a negative value you must flip the sign of the inequality.


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

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New post 13 Aug 2014, 06:49
-b/m<b --> If we multiply by negative m, we have to flip the sign as well as multiply by -m on both sides. Isnt this correct?

If I multiply (left side equation) -b/m by -m => -b/m*-m => b
If I multiply (right side equation) b by -m => -b*m
If I flip the sign,

(Left side) b > -b*m (right side) => b(1+m) > 0....Where did I go wrong? Please clarify my concept.
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New post 01 Nov 2015, 02:02
hi bunuel plz explain, why the sing of m is not considered here( at x intersept) y = mx +b , x = -b/m, why its not x = b/m(taking -m, negative slope)
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New post 04 Nov 2015, 07:19
Hi Bunuel,

I have the same doubt as vipulgoel, however i couldn't understand your follow-up explanation. Since we know m is negative, shouldnt we take the sign into consideration ? Could you please explain what do you mean by "you do not substitute a variable say, x by -x" ?

Thanks.
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Swaroopdev wrote:
Hi Bunuel,

I have the same doubt as vipulgoel, however i couldn't understand your follow-up explanation. Since we know m is negative, shouldnt we take the sign into consideration ? Could you please explain what do you mean by "you do not substitute a variable say, x by -x" ?

Thanks.


Say it's given that x=a, and you know that x is negative do you substitute x by -x in this case? No.
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Re: In the xy-plane, if line k has negative slope, is the [#permalink]

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New post 06 Nov 2015, 21:08
hi, Let me try , y = mx+ c is a general form, irrespective of slope, first just write x intercept (without considering - ve slope), now as Bunuel did multiply with -m(negative slope on both sides, that's how -ve slope comes in picture)
Re: In the xy-plane, if line k has negative slope, is the   [#permalink] 06 Nov 2015, 21:08
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