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

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Does line Ax + By + C = 0 (A is not 0) intersect the x-axis [#permalink] New post 02 Dec 2007, 03:59
Does line Ax + By + C = 0 (A is not 0) intersect the x-axis on the negative side?

1. BA < 0
2. AC > 0

Please explain your answer.
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Re: 18.13 X axis [#permalink] New post 02 Dec 2007, 07:06
bmwhype2 wrote:
Does line Ax + By + C = 0 (A is not 0) intersect the x-axis on the negative side?

1. BA <0> 0

Please explain your answer.


Getting E.

Ax + By + C = 0
By = -Ax - C (cannot divide by B just yet since B could be 0)

Stat 1:
Tells us that B is not 0 and that A and B have the same sign.
y = - (A/B)x - C
To find x, y = 0:
-(A/B)x = C
x = -(B/A) * C
B/A will have the same sign therefore -(B/A) will be negative which makes me think that the answer to the stem is yes. However, what if C = 0? The answer to the stem is no. Insuff.

Stat 2:
Tells us that A & C have opposite signs. I don't think that this alone helps us in determine the answer. Insuff.

Together:
If A is +ve and C is -ve then x intercept is +ve
If A is -ve and C is +ve then x intercept is -ve
Insuff.
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Re: 18.13 X axis [#permalink] New post 02 Dec 2007, 08:40
AX + BY + C = 0

Y = -AX/B – C/B

Assume that this line intersects x axis at (-n, 0), where n is a +ve integer. As it is given that the given equation interest the x-axis on –ve side.

Then
0 = AN/B – C/B

AN = C
A/C = N here n is +ve integer
So either A> 0 & C > 0 or C< 0 & A<0> 0

So 2 alone is sufficient. Ans is B
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 [#permalink] New post 02 Dec 2007, 09:29
im also getting B.

Equation ends up y = (-Ax-C)/B

We are interested in the x intercept, so set the equation above to 0, and you end up with:

C=-Ax

Statement 1: tells us either B<0 or A<0. If A<0>0, so either A and C are both positive, or A and C are both negative

If you consider either case, and plug into x = -(C/A), you always end up with a negative x.

Sufficient.
  [#permalink] 02 Dec 2007, 09:29
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Does line Ax + By + C = 0 (A is not 0) intersect the x-axis

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