Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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.

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

Show Tags

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 _________________

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

Show Tags

08 Aug 2012, 10:37

1

This post received KUDOS

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 _________________

I've failed over and over and over again in my life and that is why I succeed--Michael Jordan Kudos drives a person to better himself every single time. So Pls give it generously Wont give up till i hit a 700+

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

Show Tags

08 Aug 2012, 11:16

2

This post received KUDOS

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]

Show Tags

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.

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

Show Tags

09 Aug 2012, 03:41

Expert's post

2

This post was BOOKMARKED

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]

Show Tags

22 Aug 2014, 06:06

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

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.

So, my final tally is in. I applied to three b schools in total this season: INSEAD – admitted MIT Sloan – admitted Wharton – waitlisted and dinged No...

HBS alum talks about effective altruism and founding and ultimately closing MBAs Across America at TED: Casey Gerald speaks at TED2016 – Dream, February 15-19, 2016, Vancouver Convention Center...

By Libby Koerbel Engaging a room of more than 100 people for two straight hours is no easy task, but the Women’s Business Association (WBA), Professor Victoria Medvec...