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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 [#permalink]

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22 May 2010, 14:41
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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
[Reveal] Spoiler: OA

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

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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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22 May 2010, 15:57
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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 [ 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 [ 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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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?

Kudos [?]: 756 [0], given: 6

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

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

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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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