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Point (r,s) lies on the line L. Is the line s intercept with

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Point (r,s) lies on the line L. Is the line s intercept with [#permalink]

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Point (r,s) lies on the line L. Is the line’s intercept with axis-x greater than r?

(1) L has a negative slope.
(2) s<0
[Reveal] Spoiler: OA

Last edited by Bunuel on 19 Oct 2012, 03:31, edited 1 time in total.
Edited the question.

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Re: Is the line’s intercept with axis-x greater than r? [#permalink]

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This question can be easily solved if you understand the concept of slope and x intercept.

Negative slope means that the line is going down from left to right.
x intercept is the point at which the line crosses the x axis.

You know that (r, s) lies on the line and want to find if x intercept is greater than r.
Statement 1: If line has negative slope, (r, s) could still be above x axis or below x axis on the line. If it is above x axis, r < x intercept. If it is below x axis, r > x intercept. See the 1st diagram below.
Statement 2: If s < 0, it means the point (r, s) lies below x axis (s is the y co-ordinate and it is negative below x axis). But if the slope of the line is positive, r < x intercept and if the slope is negative, r > x intercept. See the 2nd diagram below.
Attachment:
Ques.jpg
Ques.jpg [ 17.47 KiB | Viewed 4076 times ]


When both the statements are combined, slope of line is negative and (r, s) lies below the x axis, then (r, s) has to lie to the right of the x intercept. Hence r will always be greater than x intercept.
Attachment:
Ques1.jpg
Ques1.jpg [ 9.6 KiB | Viewed 4072 times ]


So we can conclusively answer the question with 'No. Line's x intercept is not greater than r.' Answer (C).
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Re: Is the line’s intercept with axis-x greater than r? [#permalink]

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dimitri92 wrote:
Point (r,s) lies on the line L. Is the line’s intercept with axis-x greater than r?
1). L has a negative slope.
2). s<0


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 \(y\).

Point (r,s) lies on the line L --> \(s=mr+b\).

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

Q: is \(-\frac{b}{m}>r\)? --> substitute \(b\) and simplify --> \(-\frac{s-mr}{m}>r\)? --> \(-\frac{s}{m}+r>r\)? --> \(\frac{s}{m}<0\)?

So the question became: is \(\frac{s}{m}<0\)? we need to know signs of \(s\) and \(m\).

(1) \(m<0\). No info about \(s\). Not sufficient.

(2) \(s<0\). No info about \(m\). Not sufficient.

(1)+(2) Sufficient.

Answer: C.
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Re: Is the line’s intercept with axis-x greater than r? [#permalink]

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New post 19 May 2011, 03:03
a+b

x intercept in the 3rd quadrant has the least value.
any value of r in the 3rd quadrant will be > than the intercept value of x.

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Re: Is the line’s intercept with axis-x greater than r? [#permalink]

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New post 19 May 2011, 12:59
Very good explanation Bunuel!

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Re: Point (r,s) lies on the line L. Is the line s intercept with [#permalink]

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Re: Point (r,s) lies on the line L. Is the line s intercept with [#permalink]

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Re: Point (r,s) lies on the line L. Is the line s intercept with [#permalink]

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New post 31 Dec 2015, 19:09
oh damn..did the classic DS mistake..took the info from A, and answered that B is sufficient...

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Re: Point (r,s) lies on the line L. Is the line s intercept with [#permalink]

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New post 11 Jan 2017, 01:13
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Re: Point (r,s) lies on the line L. Is the line s intercept with   [#permalink] 11 Jan 2017, 01:13
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