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:

We don't need to calculate anything to answer this question.

If L and K are lines in xy- plane, is the product of the slope of l and k equal to -1?

The product of two lines equal to -1 if these lines are perpendicular to each other. So, the question basically asks whether L and K are perpendicular to each other.

Now, if we knew the equations of each line we would be able to answer the question. What do we need to find equation of a line? Since a line is defined by two points, then if we knew any two points of a line then we could get its equatiom

(1) Line L passes through the origin and the point (1,2). We have two points of line L: (0, 0) and (1, 2). So, we can get the equation of line L, though we know nothing about line K. Not sufficient.

(2) Line K has x-intercept 4 and y-intercept 2. We have two points of line K: (4, 0) and (0, 2). So, we can get the equation of line K, though we know nothing about line L. Not sufficient.

(1)+(2) We know both equation, hence we can answer whether the lines are perpendicular to each other. Sufficient.

Answer: C.

venmic wrote:

Can anyone give a better explanation on this one please

Hovv does K have a slope of -1/2 out of novvhere at all

Please any expert

Given two points \((x_1,y_1)\) and \((x_2,y_2)\) on a line, the slope \(m\) of the line is:

\(m=\frac{y_2-y_1}{x_2-x_1}\).

Since x-intercept of line K is 4, then it passes through point (4, 0) and since y-intercept of line K is 2, the it passes through point (0, 2). Hence it's slope is \(m=\frac{2-0}{0-4}=-\frac{1}{2}\).

If l and K are lines in xy- plane, is the product of the slope of l and k equal to -1?

1. Line l passes through the origin and the point (1,2). 2. Line K has x-intercept 4 and y-intercept 2.

C

1: we dont know anything about the other line but we can find the slope of line l = 2

2: again we dont know the other line's slope but k's is -1/2

Together 2*-1/2= -1

But you could just realize that 1 and 2 are suff together b/c we know the slopes we are going to get a YES OR A NO NOT BOTH. so u could save some time.

If l and K are lines in xy- plane, is the product of the slope of l and k equal to -1?

1. Line l passes through the origin and the point (1,2). 2. Line K has x-intercept 4 and y-intercept 2.

C

1: we dont know anything about the other line but we can find the slope of line l = 2

2: again we dont know the other line's slope but k's is -1/2

Together 2*-1/2= -1

But you could just realize that 1 and 2 are suff together b/c we know the slopes we are going to get a YES OR A NO NOT BOTH. so u could save some time.

Its not clear if the x-intercept is going to be a (4,0). All it said is that x-intercept is 4 and y-intercept is 2. Even a coordinate of (-4,0) equals an x-intercept of 4.

So the x coordinates can be (4,0) or (-4,0) and y coordinates can be (0,2) or (0,-2)

So there are two possible lines that are perpendicular and the other 2 are not.

my mistake if 1) is true then 2) will have a specific set of coordinates.

C is correct...

Need to be more cautious.

what do u mean by 2 has to have a specific set of cordinates .. form what i see i agree to your earlier assumption ...when u say line k has 4 as x cordinate and 2 as y cordinate u can very well say that it is a cordinate of the same point (4,2) then u cannot say anything about K's slope as it just passes through one point .. please explain as i still feel it should be E

Re: If L and K are lines in xy- plane, is the product of the [#permalink]

Show Tags

20 Oct 2012, 14:46

read 2) carefully - line K has an x-intercept of 4 (meaning it hits the x axis at 4--- so the point would therefore be 4,0) and y-intercept of 2 (meaning it hits the y axis at 2---so the point would therefore be 0,2). So we now have (4,0) and (0,2) as our two points for line K and we can plug it into our slope formula of y2 - y1/x2 - x1 --> y2 = 2 y1 = 0 x2 - 0 y1 = 4 --> 2-0/0-4 ---> 2/-4 = -1/2.

Re: If L and K are lines in xy- plane, is the product of the [#permalink]

Show Tags

27 Jul 2015, 13:37

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.
_________________

Re: If L and K are lines in xy- plane, is the product of the [#permalink]

Show Tags

28 Nov 2016, 21:56

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.
_________________