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

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Senior Manager
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In a xy-plane, the line y=x is the perpendicular bisector of [#permalink]  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?
VP
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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
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baner, would you break it down for me? I dont' know where I did wrong since I got a number with squre root >_<
Director
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( -2 , 1)

based on 2 slopes perpendicular then theor product = -1 concept
Manager
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I don't think GMAT tests co-ordinate geometry to this level.
VP
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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
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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GMAT math is much harder than OG, OG is kind outdated, especially for Q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
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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
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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
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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
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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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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).
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