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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

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

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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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 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.