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:

baner, would you break it down for me? I dont' know where I did wrong since I got a number with squre root >_<

Eqn of the line that passes thro 1,-2 and slope of -1 (as it is perpendicular to y = x) is: y = -x -1....(1)

Dist of 1,-2 from y = x ....is 3/sqrt 2.

We know that B is also same distance from the line only on the other side....if u draw these points and line in the xy plane u wud realize that B will approx land in y +ve and x -ve quadrant. We know that y = -x-1 intersect y axis at -1....

Now you have right triangle.....where we know the hypotenuous. Let x,y be the B coordinates. So we have....

9/2 = (x+1)^2 + y^2......(2)

Solve eq (1) and (2).....u wud get x = 4/8 or -2....4/8 is impossible...so
x = -2....and from eq 1....y = 1

B = (-2,1)......double check ur ans by calculating the dist from B to y=x...u will see it is also 3/sqrt 2. Draw it and u will see.

I don't think GMAT tests co-ordinate geometry to this level.

Ssumitsh:

The question is from REAL GMAT TEST

GMAT math is much harder than OG, OG is kind outdated, especially for Q

Agree tha this is of GMAT level.....but in GMAT u will have ans choices ...which u can quickly use and eliminate....once u know the eqn of a line.....here the ques didn't have any choices....so took longer.

The way you do this question is not to solve it algebrally. You draw a picture any you'll see all you need to do is to switch the x, y coordinations for the two points.

baner, would you break it down for me? I dont' know where I did wrong since I got a number with squre root >_<

Eqn of the line that passes thro 1,-2 and slope of -1 (as it is perpendicular to y = x) is: y = -x -1....(1)

Dist of 1,-2 from y = x ....is 3/sqrt 2.

We know that B is also same distance from the line only on the other side....if u draw these points and line in the xy plane u wud realize that B will approx land in y +ve and x -ve quadrant. We know that y = -x-1 intersect y axis at -1....

Now you have right triangle.....where we know the hypotenuous. Let x,y be the B coordinates. So we have....

9/2 = (x+1)^2 + y^2......(2)

Solve eq (1) and (2).....u wud get x = 4/8 or -2....4/8 is impossible...so x = -2....and from eq 1....y = 1

B = (-2,1)......double check ur ans by calculating the dist from B to y=x...u will see it is also 3/sqrt 2. Draw it and u will see.

Banerjee, could you please plainly expalain this again.thanks.

The way you do this question is not to solve it algebrally. You draw a picture any you'll see all you need to do is to switch the x, y coordinations for the two points.

Can you not use reflection point formula. if a segment is perpendicularly bisected by line x=y coordinations for the two points will be (a,b) & (b,a).