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In the XYPlane, is the slope of the line k equal to 0 ? [#permalink]
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08 Aug 2012, 09:18
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In the XYPlane, is the slope of the line k equal to 0 ? (1) The Xintercept of k is 0 (2) The Yintercept of k is 0. Logically, OA E makes sense.
However, I need to understand what's wrong with the algebraic method. Please don't reply with an intuitive answer. There is no doubt about OA.
A Let the equation of line be y=mx+c
1) X intercept => 0=mx+c => x=c/m; Now if the xintercept =0; it means that m could be either infinity or a nonzero. I believe that "c/m = 0/0" is undefined. Hence, m will not be equal to Zero. No. This doesn't make sense. What am I missing here?
2) Y intercept => y= c =0; therefore, y=mx; Not sufficient.
1 and 2) x intercept = 0 and yintercept = 0=> c=0 and c/m=0 => m is not equal to zero.
It's crazy, but I am a bit lost....
Thoughts? OPEN DISCUSSION OF THIS QUESTION IS HERE: inthexyplaneistheslopeoflinekequalto146084.html
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Last edited by Bunuel on 22 Aug 2014, 10:37, edited 7 times in total.
Moved topic to DS subforum and edited the question.



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Re: In the XYPlane, is the slope of the line k equal to 0 ? [#permalink]
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08 Aug 2012, 09:40
voodoochild wrote: In the XYPlane , is the slope of the line k equal to 0 ?
(1) The X intercept of k is 0
(2) The yintercept of k is 0.
Logically, OA E makes sense.
However, I need to understand what's wrong with the algebraic method. Please don't reply with an intuitive answer. There is no doubt about OA.
A Let the equation of line be y=mx+c
1) X intercept => 0=mx+c => x=c/m; Now if the xintercept =0; it means that m could be either infinity or a nonzero. I believe that "c/m = 0/0" is undefined. Hence, m will not be equal to Zero. No. This doesn't make sense. What am I missing here?
2) Y intercept => y= c =0; therefore, y=mx; Not sufficient.
1 and 2) x intercept = 0 and yintercept = 0=> c=0 and c/m=0 => m is not equal to zero.
It's crazy, but I am a bit lost....
Thoughts? I think u should try the intercept equation x/a+y/b =1, where a is xintercept and b is yintercept. the equation u used for validating St 1 is for: If the slope 'm' and y intercept 'c' of a line are given the equation of line is Y= mX +C
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Re: In the XYPlane, is the slope of the line k equal to 0 ? [#permalink]
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08 Aug 2012, 10:37
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There is no need of the y=mx+c equation here. A line has the slope zero if it is parallel to Xaxis or lies on the Xaxis itself. Statement1 X intercept=0 can be a line passing through the origin. Statement 2 y intercept zero means the line is either parallel to y axis or passing through the origin. Both the statements are not sufficient! Hope this helps
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Re: In the XYPlane, is the slope of the line k equal to 0 ? [#permalink]
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08 Aug 2012, 11:16
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voodoochild wrote: In the XYPlane , is the slope of the line k equal to 0 ?
(1) The X intercept of k is 0
(2) The yintercept of k is 0.
Logically, OA E makes sense.
However, I need to understand what's wrong with the algebraic method. Please don't reply with an intuitive answer. There is no doubt about OA.
A Let the equation of line be y=mx+c
1) X intercept => 0=mx+c => x=c/m; Now if the xintercept =0; it means that m could be either infinity or a nonzero. I believe that "c/m = 0/0" is undefined. Hence, m will not be equal to Zero. No. This doesn't make sense. What am I missing here?
2) Y intercept => y= c =0; therefore, y=mx; Not sufficient.
1 and 2) x intercept = 0 and yintercept = 0=> c=0 and c/m=0 => m is not equal to zero.
It's crazy, but I am a bit lost....
Thoughts? (1) x intercept is 0 means y=0 when x=0. Then from m*0+c=0 we obtain c=0, regardless to the value of m, which can never be infinity. m must be a real number, doesn't matter how large or small. Did you see an equation of a straight line written as \(y = \infty{x}+1\)? The given line passes through the origin, and its equation is of the form y = mx, or it can even be x = 0 (the y axis itself). In fact, the general equation of a straight line is Ax+By+C=0, where A,B, are not simultaneously 0. If \(B\neq0\), then we can express y from the given equation and get an expression of the form y=mx+n. If B=0, and \(A\neq{0}\) we have a vertical line. The equation of a vertical line is in fact of the form x = c. Doesn't matter what the value of y, x is the same. So, the slope of such a line is not defined, as the "rise"(change in y) would be divided by the "run" (change in x) which is 0, and division by 0 has no sense. Both statements tell you the same things: that the line passes through the origin, nothing more. That's why you cannot tell anything about the slope of the line. Also, take a look at the short discussion of a linear equation of the form \(Ax=B\): m237414920.html#p1109547
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Last edited by EvaJager on 08 Aug 2012, 11:45, edited 1 time in total.



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Re: In the XYPlane, is the slope of the line k equal to 0 ? [#permalink]
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08 Aug 2012, 11:25
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SOURH7WK wrote: voodoochild wrote: In the XYPlane , is the slope of the line k equal to 0 ?
(1) The X intercept of k is 0
(2) The yintercept of k is 0.
Logically, OA E makes sense.
However, I need to understand what's wrong with the algebraic method. Please don't reply with an intuitive answer. There is no doubt about OA.
A Let the equation of line be y=mx+c
1) X intercept => 0=mx+c => x=c/m; Now if the xintercept =0; it means that m could be either infinity or a nonzero. I believe that "c/m = 0/0" is undefined. Hence, m will not be equal to Zero. No. This doesn't make sense. What am I missing here?
2) Y intercept => y= c =0; therefore, y=mx; Not sufficient.
1 and 2) x intercept = 0 and yintercept = 0=> c=0 and c/m=0 => m is not equal to zero.
It's crazy, but I am a bit lost....
Thoughts? I think u should try the intercept equation x/a+y/b =1, where a is xintercept and b is yintercept.the equation u used for validating St 1 is for: If the slope 'm' and y intercept 'c' of a line are given the equation of line is Y= mX +C This equation can be used when both the x and the y intercept are nonzero, which here is not the case. a and b are explicitly the two intercepts, and in order to appear in the denominator, neither can be 0.One should rather use the general form of the linear equation for a straight line: Ax + By + C = 0.
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Re: In the XYPlane, is the slope of the line k equal to 0 ? [#permalink]
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09 Aug 2012, 03:41



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Re: In the XYPlane, is the slope of the line k equal to 0 ? [#permalink]
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01 Dec 2012, 06:36
I'd like to congratulate GMAC on finally producing a question that is absurd/pathetic, and ambiguous at best.
If it says "the X intercept"  in standard english this statement definitely means there is only ONE intercept.
This is definitely new for me: X axis' intercept on the X axis is 'only' 0. Wow!



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Re: In the XYPlane, is the slope of the line k equal to 0 ? [#permalink]
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Re: In the XYPlane, is the slope of the line k equal to 0 ? [#permalink]
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22 Aug 2014, 06:26
voodoochild wrote: In the XYPlane, is the slope of the line k equal to 0 ? (1) The Xintercept of k is 0 (2) The Yintercept of k is 0. Logically, OA E makes sense.
However, I need to understand what's wrong with the algebraic method. Please don't reply with an intuitive answer. There is no doubt about OA.
A Let the equation of line be y=mx+c
1) X intercept => 0=mx+c => x=c/m; Now if the xintercept =0; it means that m could be either infinity or a nonzero. I believe that "c/m = 0/0" is undefined. Hence, m will not be equal to Zero. No. This doesn't make sense. What am I missing here?
2) Y intercept => y= c =0; therefore, y=mx; Not sufficient.
1 and 2) x intercept = 0 and yintercept = 0=> c=0 and c/m=0 => m is not equal to zero.
It's crazy, but I am a bit lost....
Thoughts? a line will have 0 slope if it is parallel to the x axis or the x axis itself. The x intercept is 0 means either the line makes an angle with x axis and passes through the origin in this case slope is not zero or the line is Y axis itself , in which case also slope is not 0 , its undefined. Can anybody show how a line having x intercept 0 will have 0 slope ? I was of the opinion that the x axis itself has equation y = 0 , and the x intercept of the X axis cannot be 0 , x will vary according to the position of a point in the x axis .Hence X intercept = 0 cannot be the x axis.



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Re: In the XYPlane, is the slope of the line k equal to 0 ? [#permalink]
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22 Aug 2014, 10:38
qlx wrote: voodoochild wrote: In the XYPlane, is the slope of the line k equal to 0 ? (1) The Xintercept of k is 0 (2) The Yintercept of k is 0. Logically, OA E makes sense.
However, I need to understand what's wrong with the algebraic method. Please don't reply with an intuitive answer. There is no doubt about OA.
A Let the equation of line be y=mx+c
1) X intercept => 0=mx+c => x=c/m; Now if the xintercept =0; it means that m could be either infinity or a nonzero. I believe that "c/m = 0/0" is undefined. Hence, m will not be equal to Zero. No. This doesn't make sense. What am I missing here?
2) Y intercept => y= c =0; therefore, y=mx; Not sufficient.
1 and 2) x intercept = 0 and yintercept = 0=> c=0 and c/m=0 => m is not equal to zero.
It's crazy, but I am a bit lost....
Thoughts? a line will have 0 slope if it is parallel to the x axis or the x axis itself. The x intercept is 0 means either the line makes an angle with x axis and passes through the origin in this case slope is not zero or the line is Y axis itself , in which case also slope is not 0 , its undefined. Can anybody show how a line having x intercept 0 will have 0 slope ? I was of the opinion that the x axis itself has equation y = 0 , and the x intercept of the X axis cannot be 0 , x will vary according to the position of a point in the x axis .Hence X intercept = 0 cannot be the x axis. Check here: inthexyplaneistheslopeoflinekequalto146084.html
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