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

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In the xy-plane, if line k has negative slope and passes through the point (s, -2), is the x-intercept of the line k positive?

(1) s = 0
(2) The y-intercept of line k is negative
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Re: Is the x-intercept of the line k positive? [#permalink]

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New post 22 May 2010, 15:11
Can someone help me on how to approach and solve such problems. Thanks in advance.

In the xy-plane, if line k has negative slope and passes through the point (s, -2), is the x-intercept of the line k positive?

(1) s = 0
(2) The y-intercept of line k is negative

y=mx+c
given that m is -ve & the line passes through (s,-2)

the equation of the line will be
y= -mx+c ( substitute (s,-2) in the equation )
therefore,
y+2 =-m(x-s)
y+2=-mx+ms

question is x intercept of line is +ve.,
to find x intercept, put y=0. so the equation will be,x=(ms-2)/m;
Condition 1) if S=0; then x will be -ve

Condition 2) The y-intercept of line k is negative
we cant deducew any thing from it.


Ans - A)

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Re: Is the x-intercept of the line k positive? [#permalink]

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seekmba wrote:
Can someone help me on how to approach and solve such problems. Thanks in advance.

In the xy-plane, if line k has negative slope and passes through the point (s, -2), is the x-intercept of the line k positive?

(1) s = 0
(2) The y-intercept of line k is negative


Drawing method:

(1) Line k passes through the point (0, -2) and has negative slope. Line k can be something like this:
Attachment:
1.png
1.png [ 9.16 KiB | Viewed 3457 times ]

So you can see that x intercept must be negative. Sufficient.

(2) The y-intercept of line k is negative and line has negative slope. Line k can be something like this:
Attachment:
2.png
2.png [ 9.87 KiB | Viewed 3457 times ]

Again you can see that when y-intercept is negative and slope is negative, then x-intercept must be negative too. Sufficient.

Answer: D.

Algebraic approach:

Equation of a line in point intercept form is \(y=mx+b\), where: \(m\) is the slope of the line; \(b\) is the y-intercept of the line (the value of \(y\) for \(x=0\)); \(x\) is the independent variable of the function \(y\).

X-intercept is the value of \(x\) for \(y=0\) --> \(x_{intercept}=-\frac{b}{m}\).

Question: is \(-\frac{b}{m}>0\). As given that slope is negative (\(m<0\)), basically the question becomes is \(b>0\)?

(1) Line k passes through the point (0, -2) --> \(-2=0*m+b\) --> \(b=-2<0\). Answer to the question is NO. Sufficient.

(2) The y-intercept of line k is negative --> as y-intercept is value of \(b\), so we are directly told that \(b<0\). Answer to the question is NO. Sufficient.

Answer: D.

For more on this issue refer to the Coordinate Geometry chapter of the Math book (link in my signature).

Hope it helps.
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Re: Is the x-intercept of the line k positive? [#permalink]

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New post 24 May 2010, 06:37
Thanks so much Bunuel. I cannot believe how easily you explained it. I have some doubts in the graphs solution but totally understood the algebraic approach.

one question that I have though is:

In the graphs, I see the functions: f(x) = -x-2 and f(x) = -3x-1. How did you derive these formulas and how did you use them towards the question?

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Re: Is the x-intercept of the line k positive? [#permalink]

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New post 24 May 2010, 06:58
seekmba wrote:
Thanks so much Bunuel. I cannot believe how easily you explained it. I have some doubts in the graphs solution but totally understood the algebraic approach.

one question that I have though is:

In the graphs, I see the functions: f(x) = -x-2 and f(x) = -3x-1. How did you derive these formulas and how did you use them towards the question?


These lines are just examples of what actual line k could be look like, \(f(x) = -x-2\) and \(f(x) = -3x-1\) do not represent the exact equations of the line k. I just needed to draw lines for statements 1 and 2 (each line satisfying the given statement) to illustrate the point that x-intercept must be negative.

\(f(x)=-x-2\) is just one of the infinite lines with negative slope (\(m=-1<0\)) and with y-intercept equal to -2 (line passes through the pint \((0,-2)\)).

\(f(x)=-3x-1\) is just one of the infinite lines with negative slope (\(m=-3<0\)) and with negative y-intercept (\(y_{intercept}=b=-1<0\)).

Hope it's clear.
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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

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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

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New post 13 Oct 2016, 07:11
seekmba wrote:
In the xy-plane, if line k has negative slope and passes through the point (s, -2), is the x-intercept of the line k positive?

(1) s = 0
(2) The y-intercept of line k is negative


took a minute to solve...but definitely can be cracked in under 30s...

1. if s=0,then -2 is the y interceptor. we know for sure that line k doesn't pass through 1st quadrant, because the slope is negative. therefore, it can't be possible to have an x-intercept of the line positive.
B, C, and E - out.

2. y intercept of line is negative. it means that the line doesn't pass through the quadrant 1. it can't be possible to have an x intercept that is positive.
sufficient.

D is the answer.

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In the xy-plane, if line k has negative slope and passes   [#permalink] 13 Oct 2016, 07:11
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In the xy-plane, if line k has negative slope and passes

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