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

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In a xy-plane, the line y=x is the perpendicular bisector of [#permalink] New post 21 Jan 2005, 10:35
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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?
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 [#permalink] New post 21 Jan 2005, 10:58
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. :?
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 [#permalink] New post 21 Jan 2005, 11:02
baner, would you break it down for me? I dont' know where I did wrong since I got a number with squre root >_<
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 [#permalink] New post 21 Jan 2005, 11:04
( -2 , 1)

based on 2 slopes perpendicular then theor product = -1 concept
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 [#permalink] New post 21 Jan 2005, 11:07
(-2,1) should be answer.

I don't think GMAT tests co-ordinate geometry to this level.
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 [#permalink] New post 21 Jan 2005, 11:25
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.
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 [#permalink] New post 21 Jan 2005, 11:30
ssumitsh wrote:
(-2,1) should be answer.

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
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 [#permalink] New post 21 Jan 2005, 11:33
DLMD wrote:
ssumitsh wrote:
(-2,1) should be answer.

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.
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 [#permalink] New post 21 Jan 2005, 11:41
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.
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 [#permalink] New post 21 Jan 2005, 11:43
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
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 [#permalink] New post 16 May 2005, 05:43
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.
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 [#permalink] New post 16 May 2005, 06:04
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).
  [#permalink] 16 May 2005, 06:04
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