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In the XY-Plane, is the slope of the line k equal to 0 ? [#permalink]
 08 Aug 2012, 09:18
Question Stats:
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57% (00:22) wrong based on 2 sessions
In the XY-Plane, is the slope of the line k equal to 0 ? (1) The X-intercept of k is 0 (2) The Y-intercept 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 x-intercept =0; it means that m could be either infinity or a non-zero. 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 y-intercept = 0=> c=0 and c/m=0 => m is not equal to zero.
It's crazy, but I am a bit lost....
Thoughts?
Last edited by Bunuel on 13 Aug 2012, 06:51, edited 5 times in total.
Moved topic to DS subforum and edited the question.
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Re: In the XY-Plane , is the slope of the line k equal to 0 ? [#permalink]
08 Aug 2012, 09:40
voodoochild wrote: In the XY-Plane , is the slope of the line k equal to 0 ?
(1) The X- intercept of k is 0
(2) The y-intercept 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 x-intercept =0; it means that m could be either infinity or a non-zero. 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 y-intercept = 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 x-intercept and b is y-intercept. 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 XY-Plane , is the slope of the line k equal to 0 ? [#permalink]
08 Aug 2012, 10:37
There is no need of the y=mx+c equation here. A line has the slope zero if it is parallel to X-axis or lies on the X-axis 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 XY-Plane , is the slope of the line k equal to 0 ? [#permalink]
08 Aug 2012, 11:16
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voodoochild wrote: In the XY-Plane , is the slope of the line k equal to 0 ?
(1) The X- intercept of k is 0
(2) The y-intercept 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 x-intercept =0; it means that m could be either infinity or a non-zero. 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 y-intercept = 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: m23-74149-20.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 XY-Plane , is the slope of the line k equal to 0 ? [#permalink]
08 Aug 2012, 11:25
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SOURH7WK wrote: voodoochild wrote: In the XY-Plane , is the slope of the line k equal to 0 ?
(1) The X- intercept of k is 0
(2) The y-intercept 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 x-intercept =0; it means that m could be either infinity or a non-zero. 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 y-intercept = 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 x-intercept and b is y-intercept.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 non-zero, 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 XY-Plane, is the slope of the line k equal to 0 ? [#permalink]
09 Aug 2012, 03:41
In the XY-Plane, is the slope of the line k equal to 0 ? Equation of a line in point-intercept form is y=mx+b, where m is the slope of that line. Now, if m=0, then the equation becomes y=b, which means that y does not change as x increases, so the line in exactly horizontal. The slope of any horizontal line is always zero.(1) The X-intercept of k is 0. No sufficient. (2) The Y-intercept of k is 0. Not sufficient. (1)+(2) From above we know that line k passes through the origin but we can not say whether it's horizontal (X-axis) or not. Not sufficient. Answer: E. For more check Coordinate Geometry chapter of Math Book: math-coordinate-geometry-87652.htmlHope it helps.
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Re: In the XY-Plane, is the slope of the line k equal to 0 ? [#permalink]
01 Dec 2012, 06:36
Bunuel wrote: In the XY-Plane, is the slope of the line k equal to 0 ? Equation of a line in point-intercept form is y=mx+b, where m is the slope of that line. Now, if m=0, then the equation becomes y=b, which means that y does not change as x increases, so the line in exactly horizontal. The slope of any horizontal line is always zero.(1) The X-intercept of k is 0. No sufficient. (2) The Y-intercept of k is 0. Not sufficient. (1)+(2) From above we know that line k passes through the origin but we can not say whether it's horizontal (X-axis) or not. Not sufficient. Answer: E. For more check Coordinate Geometry chapter of Math Book: math-coordinate-geometry-87652.htmlHope it helps. 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 XY-Plane, is the slope of the line k equal to 0 ?
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01 Dec 2012, 06:36
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