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 350,000 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:

Statement 1 and Statement 2 separately give an infinite number of solutions. Solving the two statements together gives you a unique solution for a and b, so you have the point (a;b). You know the line is perpendicular to y=2x and there is only one way to draw a perpendicular to a line from any given point. _________________

What is the equation of the line that is perpendicular to line \(y = 2x\) and passes through point (a, b)?

1. \(a = -b\) 2. \(a - b = 1\)

I really think that the answer should be A. Because when a = -b, it means that (-b,b). So let's say that b=5, doesn't matter whether b is positive or negative because you will still end up drawing the same line. Well, what will be its y intercept? well if b is zero, then both x and y will be zero, hence, the y intercept is at the origin. So you're gonna end up drawing the same straight line, so you can make an equation out of it.

What is the equation of the line that is perpendicular to line \(y = 2x\) and passes through point (a, b)?

1. \(a = -b\) 2. \(a - b = 1\)

I really think that the answer should be A. Because when a = -b, it means that (-b,b). So let's say that b=5, doesn't matter whether b is positive or negative because you will still end up drawing the same line. Well, what will be its y intercept? well if b is zero, then both x and y will be zero, hence, the y intercept is at the origin. So you're gonna end up drawing the same straight line, so you can make an equation out of it.

what's the OA to this problem?

My answer is A

Tarek,

please note that a and b are not variables... How can you say that b is zero ..? y=-1/2x+c a=-b b=1/2b+c b=2c From 1: you know that b=2c a=-b. there can be infinite number of solutions from this relation. and passing through (0,0) one of that solutions

I hope this is clear now. _________________

Your attitude determines your altitude Smiling wins more friends than frowning

What is the equation of the line that is perpendicular to line \(y = 2x\) and passes through point (a, b)?

1. \(a = -b\) 2. \(a - b = 1\)

I really think that the answer should be A. Because when a = -b, it means that (-b,b). So let's say that b=5, doesn't matter whether b is positive or negative because you will still end up drawing the same line. Well, what will be its y intercept? well if b is zero, then both x and y will be zero, hence, the y intercept is at the origin. So you're gonna end up drawing the same straight line, so you can make an equation out of it.

what's the OA to this problem?

My answer is A

Tarek,

please note that a and b are not variables... How can you say that b is zero ..? y=-1/2x+c a=-b b=1/2b+c b=2c From 1: you know that b=2c a=-b. there can be infinite number of solutions from this relation. and passing through (0,0) one of that solutions

I hope this is clear now.

I didn't say that b is for sure zero. All I said that no matter what the value of b is, even if it is zero, we would still be drawing the same line, no? for example, (-1,1), (0,0), (5,-5), EVEN if it is (-10, 10), wouldn't all these points fall on the same line? and we already have the perpendicular slope of -1/2, so what's wrong with this reasoning? i'm still trying to figure out what's wrong with this reasoning and sorry for being persistent but i guess it's the best way to learn here...lol

please note that a and b are not variables... How can you say that b is zero ..? y=-1/2x+c a=-b b=1/2b+c b=2c From 1: you know that b=2c a=-b. there can be infinite number of solutions from this relation. and passing through (0,0) one of that solutions

I hope this is clear now.

I didn't say that b is for sure zero. All I said that no matter what the value of b is, even if it is zero, we would still be drawing the same line, no? for example, (-1,1), (0,0), (5,-5), EVEN if it is (-10, 10), wouldn't all these points fall on the same line? and we already have the perpendicular slope of -1/2, so what's wrong with this reasoning? i'm still trying to figure out what's wrong with this reasoning and sorry for being persistent but i guess it's the best way to learn here...lol

No they are not the same. y intercept changes.. when "b" changes. -a=b=2c y=-1/2x+c for (-1,1) y intercept 2 y=-1/2x+2 --> equation (1) (0,0) y=-1/2x --> equation (2) for (5,-5) y=-1/2x+c --> -5=-1/2*5+c= -5/2 y=-1/2x+-5/2 -----> equation (3)

Does equations (1),(2) and (3) are same???? No. _________________

Your attitude determines your altitude Smiling wins more friends than frowning

please note that a and b are not variables... How can you say that b is zero ..? y=-1/2x+c a=-b b=1/2b+c b=2c From 1: you know that b=2c a=-b. there can be infinite number of solutions from this relation. and passing through (0,0) one of that solutions

I hope this is clear now.

I didn't say that b is for sure zero. All I said that no matter what the value of b is, even if it is zero, we would still be drawing the same line, no? for example, (-1,1), (0,0), (5,-5), EVEN if it is (-10, 10), wouldn't all these points fall on the same line? and we already have the perpendicular slope of -1/2, so what's wrong with this reasoning? i'm still trying to figure out what's wrong with this reasoning and sorry for being persistent but i guess it's the best way to learn here...lol

No they are not the same. y intercept changes.. when "b" changes. -a=b=2c y=-1/2x+c for (-1,1) y intercept 2 y=-1/2x+2 --> equation (1) (0,0) y=-1/2x --> equation (2) for (5,-5) y=-1/2x+c --> -5=-1/2*5+c= -5/2 y=-1/2x+-5/2 -----> equation (3)

Does equations (1),(2) and (3) are same???? No.

oh perfect! now i see where i went wrong. Thank you so much for the great explanation.