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M18#13

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M18#13 [#permalink] New post 07 Feb 2009, 05:22
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Does line Ax + By + C = 0 (A is not 0) intersect the x-axis on the negative side?

1. BA < 0
2. AC > 0

[Reveal] Spoiler: OA
B

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Re: M18#13 [#permalink] New post 01 Jan 2011, 07:16
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321kumarsushant wrote:
Hey all,

my answer is different from all.
i ll go with E.
here is my explanation:
ax+by+c=0;
or, y= -(c+ax)/b; or, y= -(c^2+acx) / bc ; (on multiplying C )
statement 1 : BA <0 ;not applicable
statement 2 : AC >0 ; applicable;
in the eq, c^2 is +ve
ac is +ve
bc not known;
on comparing with standard (y=mx+c)
m = -ac/bc ; since bc is unknown, you cant predict the slope of line ,
hence you cant say, whether it will cut Y-Axis or not.


next;

ax+by+c=0;
or, y= -(c+ax)/b; or, y= -(bc+bax) / b^2 ; (on multiplying B )
statement 1 : BA <0 ; applicable
statement 2 : AC >0 ; not applicable;
in the eq, b^2 is +ve
ab is -ve
bc not known;
on comparing with standard (y=mx+c)
m = -ba/b^2 ; since b^2 is +ve, the slope of line is +ve .
constant (i.e c in standard eq) = bc; not known
since, bc is not known, whether it will be on +ve or -ve Y-axis.
so, you cant say that whether this line will cut the X-axis or not.


please make me correct if m wrong.


hence you cant say, whether it will cut Y-Axis or not.


Official answer is B, not E.

Does line Ax + By + C = 0 (A is not 0) intersect the x-axis on the negative side?

ax+by+c=0 is equation of a line. Note that the line won't have interception with x-axis when a=0 (and c\neq{0}): in this case the line will be y=-\frac{c}{b} and will be parallel to x -axis.

Now, in other cases (when a\neq{0}) x-intercept of a line will be the value of x when y=0, so the value of x=-\frac{c}{a}. Question basically asks whether this value is negative, so question asks is -\frac{c}{a}<0? --> is \frac{c}{a}>0? --> do c and a have the same sign?

(1) BA < 0. Not sufficient as we can not answer whether c and a have the same sign.
(2) AC > 0 --> c and a have the same sign. Sufficient.

Answer: B.

Check more on this topic here: math-coordinate-geometry-87652.html

Hope it helps.
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Re: M16#13 [#permalink] New post 07 Feb 2009, 05:24
I solved this as follows:

Ax+By+C=0

therefore y = -(A/B) x + (C/B)

When y = 0, x = (BC)/A

therefore you need both (1) and (2) to check if x is negative.
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Re: M16#13 [#permalink] New post 10 Feb 2009, 05:02
I'm pretty sure that's wrong..... (A/B)X + (C/B) = Y
Y=0 --> X = -C/A

Then we need only (b) to answer the question...

Who is right here?
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Re: M16#13 [#permalink] New post 10 Feb 2009, 10:51
tzvister wrote:
I'm pretty sure that's wrong..... (A/B)X + (C/B) = Y
Y=0 --> X = -C/A

Then we need only (b) to answer the question...

Who is right here?


absolutely you are correct,

the general form of equn of line is x/a + y/b = 1 ( a and b are intercept of x and y axis respectively) comparing it to the given equn a = -C/A, hence from stmt 2 itself v can answer the question
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Re: M16#13 [#permalink] New post 11 Feb 2009, 06:53
topmbaseeker wrote:
I solved this as follows:

Ax+By+C=0

therefore y = -(A/B) x + (C/B)

When y = 0, x = (BC)/A
therefore you need both (1) and (2) to check if x is negative.


your approach is correct.. you did mistake in the equation.
Y= -(A/B)x+(-C/B) =0

--> -Ax-C =0 -->x intercept = -C/A

B is the answer.
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Re: M18#13 [#permalink] New post 24 Dec 2009, 06:17
X=(-By-C)/A
when y=0, X=-C/A, B is sufficient.

This question is not hard, just be careful x-intersection means y=0.


topmbaseeker wrote:
Does line Ax + By + C = 0 (A is not 0) intersect the x-axis on the negative side?

1. BA < 0
2. AC > 0

* Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient
* Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient
* BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient
* EACH statement ALONE is sufficient
* Statements (1) and (2) TOGETHER are NOT sufficient

[Reveal] Spoiler: OA
B

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Re: M18#13 [#permalink] New post 01 Jan 2011, 02:28
Hey all,

my answer is different from all.
i ll go with E.
here is my explanation:
ax+by+c=0;
or, y= -(c+ax)/b; or, y= -(c^2+acx) / bc ; (on multiplying C )
statement 1 : BA <0 ;not applicable
statement 2 : AC >0 ; applicable;
in the eq, c^2 is +ve
ac is +ve
bc not known;
on comparing with standard (y=mx+c)
m = -ac/bc ; since bc is unknown, you cant predict the slope of line ,
hence you cant say, whether it will cut Y-Axis or not.


next;

ax+by+c=0;
or, y= -(c+ax)/b; or, y= -(bc+bax) / b^2 ; (on multiplying B )
statement 1 : BA <0 ; applicable
statement 2 : AC >0 ; not applicable;
in the eq, b^2 is +ve
ab is -ve
bc not known;
on comparing with standard (y=mx+c)
m = -ba/b^2 ; since b^2 is +ve, the slope of line is +ve .
constant (i.e c in standard eq) = bc; not known
since, bc is not known, whether it will be on +ve or -ve Y-axis.
so, you cant say that whether this line will cut the X-axis or not.


please make me correct if m wrong.


hence you cant say, whether it will cut Y-Axis or not.
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Re: M18#13 [#permalink] New post 07 Jan 2012, 16:40
I missed step 1 which is put it in slope form...messed me up.
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Re: M18#13 [#permalink] New post 06 Jan 2013, 14:55
Answer=B
I hope this helps
The question is if x intercept is negative, which can be found by solving for x, assuming y=0, x= -(C/A)
situation 1 says BA<0 - no relevance to whether x intercept is -ve or =ve - not sufficient
2) AC>0, would mean both a and c are either =ve or -ve in either situation X intercept is -ve therefore B is sufficient,
also A not equal to zero would mean C cannot be zero as well. so this covers all grounds i guess.

Correct answer is B
Re: M18#13   [#permalink] 06 Jan 2013, 14:55
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