It is currently 23 Jun 2017, 09:03

### GMAT Club Daily Prep

#### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# In a xy-plane, the line y=x is the perpendicular bisector of

Author Message
Senior Manager
Joined: 07 Nov 2004
Posts: 453
In a xy-plane, the line y=x is the perpendicular bisector of [#permalink]

### Show Tags

21 Jan 2005, 11:35
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 2 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

In a xy-plane, the line y=x is the perpendicular bisector of segment AB. If the coordinates of A is (1,-2), what are the coordinates of B?
VP
Joined: 18 Nov 2004
Posts: 1433

### Show Tags

21 Jan 2005, 11:58
DLMD wrote:
banerjeea_98 wrote:
B = (-4, 3).....will explain if correct.

Baner

answer should be (-2, 1) ^_^

yep...calculation mistake...i edited my ans....realized it right after posting.
Senior Manager
Joined: 07 Nov 2004
Posts: 453

### Show Tags

21 Jan 2005, 12:02
baner, would you break it down for me? I dont' know where I did wrong since I got a number with squre root >_<
Director
Joined: 07 Jun 2004
Posts: 612
Location: PA

### Show Tags

21 Jan 2005, 12:04
( -2 , 1)

based on 2 slopes perpendicular then theor product = -1 concept
Manager
Joined: 06 Sep 2004
Posts: 69

### Show Tags

21 Jan 2005, 12:07

I don't think GMAT tests co-ordinate geometry to this level.
VP
Joined: 18 Nov 2004
Posts: 1433

### Show Tags

21 Jan 2005, 12:25
DLMD wrote:
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.
Senior Manager
Joined: 07 Nov 2004
Posts: 453

### Show Tags

21 Jan 2005, 12:30
ssumitsh wrote:

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
VP
Joined: 18 Nov 2004
Posts: 1433

### Show Tags

21 Jan 2005, 12:33
DLMD wrote:
ssumitsh wrote:

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.
SVP
Joined: 03 Jan 2005
Posts: 2233

### Show Tags

21 Jan 2005, 12:41
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.
Senior Manager
Joined: 07 Nov 2004
Posts: 453

### Show Tags

21 Jan 2005, 12:43
damn, I forgot the product of slopes of 2 lines perpendicular to each other is -1. I was trying to slove it using geometry from the beginning >_<

thanks, especially to baner, great explanation
SVP
Joined: 05 Apr 2005
Posts: 1710

### Show Tags

16 May 2005, 06:43
banerjeea_98 wrote:
DLMD wrote:
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.
Senior Manager
Joined: 02 Feb 2004
Posts: 344

### Show Tags

16 May 2005, 07:04
HongHu wrote:
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).
16 May 2005, 07:04
Display posts from previous: Sort by