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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.

WE 1: 7 Yrs in Automobile (Commercial Vehicle industry)

Re: In the XY-Plane, is the slope of the line k equal to 0 ? [#permalink]

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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]

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08 Aug 2012, 10:37

1

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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 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]

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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 _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Last edited by EvaJager on 08 Aug 2012, 11:45, edited 1 time in total.

Re: In the XY-Plane, is the slope of the line k equal to 0 ? [#permalink]

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08 Aug 2012, 11:25

1

This post received KUDOS

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.
_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

In the xy-plane is the slope of line k equal to 0?

The questions basically asks whether line k is a horizontal line (the slope of any horizontal line is always zero. For more check here: math-coordinate-geometry-87652.html).

(1) The x-intercept of k is 0. Now, I'm not a verbal expert, but the x-intercept implies that there is only one point of interception with x-axis, which means that we can eliminate y=0 line. So, we have that line k is not y=0 and has x-intercept, thus it cannot be horizontal --> the slope does not equal to 0. Sufficient.

(2) The y-intercept of k is 0. Clearly insufficient.

Re: In the XY-Plane, is the slope of the line k equal to 0 ? [#permalink]

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22 Aug 2014, 06:06

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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?

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.

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?

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.