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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 [#permalink] New post 14 Mar 2012, 00:04
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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] 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] 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] 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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Re: In the xy-plane, if line k has negative slope, is the [#permalink] New post 15 Mar 2012, 08:15
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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] New post 09 Aug 2014, 03:03
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In the xy-plane, if line k has negative slope, is the [#permalink] 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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Re: In the xy-plane, if line k has negative slope, is the [#permalink] New post 12 Aug 2014, 07:30
Expert's post
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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RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: In the xy-plane, if line k has negative slope, is the [#permalink] 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.
Re: In the xy-plane, if line k has negative slope, is the   [#permalink] 13 Aug 2014, 06:49
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