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:

2)If P and Q are two points on the line 3x + 4y = -15 such that OP and OQ= 9 units, The area of triangle POQ is given by

a) 18* sqrt(2) b) 3* sqrt(2) c) 6* sqrt(2) d) 15*sqrt(2)

Look at the the diagram below:

OC is perpendicular to AB (PQ), so it's height of AOB and POQ;

\(OA=5\) and \(OB=\frac{15}{4}\) (points A and B are intersection of the line \(3x+4y=-15\) with the X and Y axis respectively) --> \(AB=hypotenuse=\sqrt{OA^2+OB^2}=\frac{25}{4}\);

As AOB and OCB are similar then \(\frac{OC}{OB}=\frac{OA}{AB}\) --> \(OC=3\);

Bunuel, what makes us think that point o is Origin in this case???

Well, we must know the coordinates of the point O to solve the question. As origin in coordinate geometry usually marked with letter O (as the center of a circle) then we can assume that O is origin in this case. Though I think on GMAT it would be explicitly stated.
_________________

2)If P and Q are two points on the line 3x + 4y = -15 such that OP and OQ= 9 units, The area of triangle POQ is given by

a) 18* sqrt(2) b) 3* sqrt(2) c) 6* sqrt(2) d) 15*sqrt(2)

Look at the the diagram below:

Attachment:

graph.php.png

OC is perpendicular to AB (PQ), so it's height of AOB and POQ;

\(OA=5\) and \(OB=\frac{15}{4}\) (points A and B are intersection of the line \(3x+4y=-15\) with the X and Y axis respectively) --> \(AB=hypotenuse=\sqrt{OA^2+OB^2}=\frac{25}{4}\);

As AOB and OCB are similar then \(\frac{OC}{OB}=\frac{OA}{AB}\) --> \(OC=3\);

Bunuel, what makes us think that point o is Origin in this case???

Well, we must know the coordinates of the point O to solve the question. As origin in coordinate geometry usually marked with letter O (as the center of a circle) then we can assume that O is origin in this case. Though I think on GMAT it would be explicitly stated.

2)If P and Q are two points on the line 3x + 4y = -15 such that OP and OQ= 9 units, The area of triangle POQ is given by

a) 18* sqrt(2) b) 3* sqrt(2) c) 6* sqrt(2) d) 15*sqrt(2)

Look at the the diagram below:

Attachment:

graph.php.png

OC is perpendicular to AB (PQ), so it's height of AOB and POQ;

\(OA=5\) and \(OB=\frac{15}{4}\) (points A and B are intersection of the line \(3x+4y=-15\) with the X and Y axis respectively) --> \(AB=hypotenuse=\sqrt{OA^2+OB^2}=\frac{25}{4}\);

As AOB and OCB are similar then \(\frac{OC}{OB}=\frac{OA}{AB}\) --> \(OC=3\);

2)If P and Q are two points on the line 3x + 4y = -15 such that OP and OQ= 9 units, The area of triangle POQ is given by

a) 18* sqrt(2) b) 3* sqrt(2) c) 6* sqrt(2) d) 15*sqrt(2)

Look at the the diagram below:

Attachment:

graph.php.png

OC is perpendicular to AB (PQ), so it's height of AOB and POQ;

\(OA=5\) and \(OB=\frac{15}{4}\) (points A and B are intersection of the line \(3x+4y=-15\) with the X and Y axis respectively) --> \(AB=hypotenuse=\sqrt{OA^2+OB^2}=\frac{25}{4}\);

As AOB and OCB are similar then \(\frac{OC}{OB}=\frac{OA}{AB}\) --> \(OC=3\);

Re: If P and Q are two points on the line 3x + 4y =-15 such that [#permalink]

Show Tags

01 Mar 2014, 12:16

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.
_________________

Bunuel could you please elaborate a little bit more on the relation between similar triangles? How could you get OC/OB = OA/AB, I'm not able to see it.

Bunuel could you please elaborate a little bit more on the relation between similar triangles? How could you get OC/OB = OA/AB, I'm not able to see it.

Could you please explain how we can derive SQRT (OA^2 + OB^2) = 25/4?

AB is the hypotenuse of right triangle AOB, hence \(AB=hypotenuse=\sqrt{OA^2+OB^2}\), where \(OA=5\) and \(OB=\frac{15}{4}\) (points A and B are intersection of the line \(3x+4y=-15\) with the X and Y axis respectively).

2)If P and Q are two points on the line 3x + 4y = -15 such that OP and OQ= 9 units, The area of triangle POQ is given by

a) 18* sqrt(2) b) 3* sqrt(2) c) 6* sqrt(2) d) 15*sqrt(2)

Look at the the diagram below:

Attachment:

graph.php.png

OC is perpendicular to AB (PQ), so it's height of AOB and POQ;

\(OA=5\) and \(OB=\frac{15}{4}\) (points A and B are intersection of the line \(3x+4y=-15\) with the X and Y axis respectively) --> \(AB=hypotenuse=\sqrt{OA^2+OB^2}=\frac{25}{4}\);

As AOB and OCB are similar then \(\frac{OC}{OB}=\frac{OA}{AB}\) --> \(OC=3\);

2)If P and Q are two points on the line 3x + 4y = -15 such that OP and OQ= 9 units, The area of triangle POQ is given by

a) 18* sqrt(2) b) 3* sqrt(2) c) 6* sqrt(2) d) 15*sqrt(2)

Look at the the diagram below:

Attachment:

graph.php.png

OC is perpendicular to AB (PQ), so it's height of AOB and POQ;

\(OA=5\) and \(OB=\frac{15}{4}\) (points A and B are intersection of the line \(3x+4y=-15\) with the X and Y axis respectively) --> \(AB=hypotenuse=\sqrt{OA^2+OB^2}=\frac{25}{4}\);

As AOB and OCB are similar then \(\frac{OC}{OB}=\frac{OA}{AB}\) --> \(OC=3\);

Re: If P and Q are two points on the line 3x + 4y =-15 such that [#permalink]

Show Tags

07 Mar 2015, 09:04

Bunuel wrote:

kobinaot wrote:

help pls!

2)If P and Q are two points on the line 3x + 4y = -15 such that OP and OQ= 9 units, The area of triangle POQ is given by

a) 18* sqrt(2) b) 3* sqrt(2) c) 6* sqrt(2) d) 15*sqrt(2)

Look at the the diagram below:

Attachment:

graph.php.png

OC is perpendicular to AB (PQ), so it's height of AOB and POQ;

\(OA=5\) and \(OB=\frac{15}{4}\) (points A and B are intersection of the line \(3x+4y=-15\) with the X and Y axis respectively) --> \(AB=hypotenuse=\sqrt{OA^2+OB^2}=\frac{25}{4}\);

As AOB and OCB are similar then \(\frac{OC}{OB}=\frac{OA}{AB}\) --> \(OC=3\);

Re: If P and Q are two points on the line 3x + 4y =-15 such that [#permalink]

Show Tags

21 Mar 2016, 11:01

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.
_________________

There’s something in Pacific North West that you cannot find anywhere else. The atmosphere and scenic nature are next to none, with mountains on one side and ocean on...

This month I got selected by Stanford GSB to be included in “Best & Brightest, Class of 2017” by Poets & Quants. Besides feeling honored for being part of...

Joe Navarro is an ex FBI agent who was a founding member of the FBI’s Behavioural Analysis Program. He was a body language expert who he used his ability to successfully...