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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, 02:59
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D
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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.

M18-13
[Reveal] Spoiler: OA

Last edited by Bunuel on 13 Nov 2013, 01:03, edited 2 times in total.
Renamed the topic, edited the question, added the OA and moved to DS forum.
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Re: 18.13 X axis [#permalink] New post 02 Dec 2007, 06:06
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, 07:40
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, 08:29
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.
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Re: Does line Ax + By + C = 0 (A is not 0) intersect the x-axis [#permalink] New post 12 Nov 2013, 14:38
i was doing this on the gmatclub tests and i cannot figure out why all we need is x = -c/a

"So, the x-intercept of line ax+by+c=0 is x=−c/a."
I plugged in 0 so ax+ by+ c = 0
then y = ( -ax - c ) / b

was i supposed to think of this question like this
ax + b (0) + c = 0
ax + c = 0
x = -c/a

then use that equation to figure out what x is???

(when i was doing this before viewing the solution, i assumed that we would need a/b to solve because -a/b * x.)
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Re: Does line Ax + By + C = 0 (A is not 0) intersect the x-axis [#permalink] New post 12 Nov 2013, 22:23
@laserglare

Yes, to find the x-intercept of a line, the point where it intersects the x-axis, we set the y-coordinate to 0. You correctly replaced y as 0 in the equation Ax+By+C=0, which gave you the x-intercept of -C/A.

Here 2 alone is sufficient because if AC>0, then we have either both A and C are positive or both A and C are negative, in both scenarios -C/A is negative, meaning the x-intercept is negative or intersects the x-axis to the left of the origin.

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Re: Does line Ax + By + C = 0 (A is not 0) intersect the x-axis [#permalink] New post 12 Nov 2013, 22:29
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laserglare wrote:
i was doing this on the gmatclub tests and i cannot figure out why all we need is x = -c/a

"So, the x-intercept of line ax+by+c=0 is x=−c/a."
I plugged in 0 so ax+ by+ c = 0
then y = ( -ax - c ) / b

was i supposed to think of this question like this
ax + b (0) + c = 0
ax + c = 0
x = -c/a

then use that equation to figure out what x is???

(when i was doing this before viewing the solution, i assumed that we would need a/b to solve because -a/b * x.)


Given Ax + By + C = 0 is the equation of a line. You need to figure out whether it intersects x axis on the negative side i.e. in the second quadrant. You want to know that when the line crosses the x axis (if it does), is x co-ordinate negative there? When does a line cross the x axis? When its y co-ordinate is 0. So how will you know the point where the line crosses the x axis?
You put y = 0.
Ax + B*0 + C = 0
x = -C/A
So when y = 0, x = -C/A

We want to know whether this x cor-ordinate (-C/A) is negative. It will be negative when C/A is positive i.e. both C and A will have the same sign (either both positive or both negative)
Statement 2 tells you that C and A have the same sign (since their product is positive). Hence it is enough alone.

Answer (B)
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Re: Does line Ax + By + C = 0 (A is not 0) intersect the x-axis [#permalink] New post 13 Nov 2013, 01:04
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bmwhype2 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.

M18-13


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: Does line Ax + By + C = 0 (A is not 0) intersect the x-axis   [#permalink] 13 Nov 2013, 01:04
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