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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]  26 Feb 2011, 07:21
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In the xy-plane, if line k has negative slope and passes through the point (-5,r), is the x-intercept of line k positive?

(1) The slope of line K is -5
(2) r>0

OPEN DISCUSSION OF THIS QUESTION IS HERE: in-the-xy-plane-if-line-k-has-negative-slope-and-passes-135197.html
[Reveal] Spoiler: OA

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Last edited by Bunuel on 31 May 2013, 03:12, edited 2 times in total.
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Re: x-intercept [#permalink]  26 Feb 2011, 07:36
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In the xy-plane, if line k has negative slope and passes through the point (-5,r), is the x-intercept of line k positive?

This question can be done with graphic approach (just by drawing the lines) or with algebraic approach.

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), and x is the independent variable of the function y.

We are told that slope of line k is negative (m<0) and it passes through the point (-5,r): y=mx+b --> r=-5m+b.

Question: is x-intercept of line k positive? x-intercep is the value of x for y=0 --> 0=mx+b --> is x=-\frac{b}{m}>0? As we know that m<0, then the question basically becomes: is b>0?.

(1) The slope of line k is -5 --> m=-5<0. We've already known that slope was negative and there is no info about b, hence this statement is insufficient.

(2) r>0 --> r=-5m+b>0 --> b>5m=some \ negative \ number, as m<0 we have that b is more than some negative number (5m), hence insufficient, to say whether b>0.

(1)+(2) From (1) m=-5 and from (2) r=-5m+b>0 --> r=-5m+b=25+b>0 --> b>-25. Not sufficient to say whether b>0.

Answer: E.

Graphic approach:

If the slope of a line is negative, the line WILL intersect quadrants II and IV. X and Y intersects of the line with negative slope have the same sign. Therefore if X and Y intersects are positive, the line intersects quadrant I; if negative, quadrant III.

When we take both statement together all we know is that slope is negative and that it crosses some point in II quadrant (-5, r>0) (this info is redundant as we know that if the slope of the line is negative, the line WILL intersect quadrants II). Basically we just know that the slope is negative - that's all. We can not say whether x-intercept is positive or negative from this info.

Below are two graphs with positive and negative x-intercepts. Statements that the slope=-5 and that the line crosses (-5, r>0) are satisfied.

y=-5x+5:
Attachment:

1.png [ 9.73 KiB | Viewed 1486 times ]

y=-5x-20:
Attachment:

2.png [ 10.17 KiB | Viewed 1488 times ]

More on this please check Coordinate Geometry chapter of Math Book: math-coordinate-geometry-87652.html

Also discussed here: og-12-ds-question-line-concept-very-good-one-key-wrong-89300.html

Hope it helps.
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Re: x-intercept [#permalink]  26 Feb 2011, 07:31
Equation of a line passing through a point:

y = mx + c

x = -5 and y = r

r = -5m + c

x intercept is the point at which y = 0

mx + c = 0 => x = -c/m

Question: is -c/m positive?

x will be positive is c and m have opposite signs

Statement 1:
m = -5 Thus we need to see if c is positive!

Given: r = -5m + c => r = 25 + c => c = r - 25

Since we don't know the value of r it's not possible to say if c is +ve or -ve

NS

Statement 2:
r > 0 => -5m + c > 0

or c > 5m

Again, this tells nothing about the sign of c or m.

NS

Statement 1 and 2:
From 1: c = r - 25 and m < 0
From 2: r > 0

Again, we don't know the sign of m and c from the above statements.
NS

Ans: 'E'
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Re: x-intercept [#permalink]  26 Feb 2011, 07:47
GMATD11 wrote:
I used the concept :: for x- intercept put y=0

so
a) slope is -ve and point (-5,r) lies on the line

y= mx+c
r= (-)(-5) + c
we have to put y=0 to calculate x intercept. equation ill be

0 =-(-5)+c => c is -ve
Sufficient

b) NS

OA is different

The key here is to know that is y-intercept is +ve; x-intercept will be +ve because the line has -ve slope and when y-intercept is +ve, x-intercept will be +ve. We just need to know whether c(the y-intercept) is +ve or -ve.

You can't write the statement in red;
y= mx+c

It passes through (x=-5,y=r)
r = m(-5)+c
r = -5m + c

We have 3 unknowns here; r,m and c

(1)
We know m=-5
r = -5m + c
r = -5* -5 + c
r = 25+c

Still r is unknown; we can't know c.

(2)
r>0
r = -5m + c

r = (some +ve value) + c
c = r - (some +ve value)

If (some +ve value) > r; c will be -ve and x-intercept will also be -ve
If (some +ve value) < r; c will be +ve and x-intercept will also be +ve

Not Sufficient.

Combining both;
r = -5* -5 + c
r = c+25
c = 25-r and r>0

Still; if r>25; c will be -ve and x-intercept will be -ve as well.
Still; if r<25; c will be +ve and x-intercept will be +ve as well.
Not Sufficient.

Ans: "E"
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In the xy-plane, if line k has negative slope and passes [#permalink]  25 Apr 2012, 05:09
Bhai wrote:
Quote:
-5 - c/m > 0 i.e -5 -c > 0 i.e c< -5
X intercept -c/ m but we don't know whether m is + or - hence Insuff

cool_jonny009,
Can you please simplify on this.

y = mx+c and the line passes through (-5,r)
-5=mr+c == > r = (-5 -c)/m

as the stmt says r > 0 so (-5 -c)/m > 0
which means c < -5

now X intercept is when y = 0 which will be -c/m
as we know c is negative but we don't know any thing about m ...so can't conclude whether the X intercept is - ve or + ve

------------------------------------------------------------------------------------------------------------
I guess u made a mistake of taking the x cordinate when substituting for y in the Eqn Y=mX+C.
y = mx+c and the line passes through (-5,r)
Hence
r=(-5)*(-5)+c == > r = 25+c.
Hence insufficient.
Correct Answer should be E.
In the xy-plane, if line k has negative slope and passes   [#permalink] 25 Apr 2012, 05:09
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