Coordinate Plane Problem

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C: Junior | Next Rank: 30 Posts; Posts: 23; Joined: Mon Feb 08, 2010 8:19 am; Location: Norway; Thanked: 1 times

Coordinate Plane Problem

by C » Sun Mar 21, 2010 9:18 pm

The line represented by equation y=x is the perpendicular bisector of line segment AB. If A has the coordinates (-3,3), what are the coordinates of B?

Can someone please explain?

Source is MGMAT.
No answer choices, OA is [spoiler](3, -3)[/spoiler]

Thanks.

Source: Beat The GMAT — Problem Solving | Sun Mar 21, 2010 9:18 pm

neoreaves: Master | Next Rank: 500 Posts; Posts: 208; Joined: Sun Sep 28, 2008 12:30 pm; Thanked: 22 times

by neoreaves » Sun Mar 21, 2010 9:42 pm

The equation of the perpendicular line will be y = -x . For me the answer comes by visually looking at such a bisector. Here is how i arrived at the solution:

C: Junior | Next Rank: 30 Posts; Posts: 23; Joined: Mon Feb 08, 2010 8:19 am; Location: Norway; Thanked: 1 times

by C » Sun Mar 21, 2010 10:19 pm

Thanks.

In most cases it is probably easiest to solve it visually.

Do you also know how to solve algebraically?

neoreaves: Master | Next Rank: 500 Posts; Posts: 208; Joined: Sun Sep 28, 2008 12:30 pm; Thanked: 22 times

by neoreaves » Sun Mar 21, 2010 10:38 pm

I wouldn't recommend algebra in this case. But if I had to use algebra this is how I will solve this.

We know a few things about the line that bisects y = mx

1) m = -1 thus y = -mx
2) One point is (-3,3)

Because the line passes through the center as y = -mx. We can calculate the distance from the origin

Distance = d^2 = 3^2 + (-3)^2

This distance will be same when extending the line to quadrant IV.

Distance = d^2 = 3^2 + 3^2

Thus the only way this is possible in the IV quadran is when x is positive and y is negative

Thus (x,y) = (3, -3)

In general I think it is always a good idea to draw the diagram out first and then use a mix of algebra and extrapolation to arrive at coordinate geometry question types.

C: Junior | Next Rank: 30 Posts; Posts: 23; Joined: Mon Feb 08, 2010 8:19 am; Location: Norway; Thanked: 1 times

by C » Sun Mar 21, 2010 10:52 pm

Thank you for the thorough explanation!

Helped me see the big picture.

C

kstv: Legendary Member; Posts: 610; Joined: Fri Jan 15, 2010 12:33 amFollowed by: 2 members; Thanked: 47 times

by kstv » Sun Mar 21, 2010 11:37 pm

If the answer choice are given then visually this can be done easily.
If we try to solve the problem using the eq of st line then
Let the perpendicular bisector of y=x (line I , this passes trough the origin) be
y=mx+c (line II) slope m will be -1 as it is pernedicular to y=x
plug in the value (-3,3) in I 3= -1(-3) +c so c= 0
so the eq of the bisector is y = -x and it also passes through the origin (0,0)
Distance of A (-3,3) from (0,0) is âˆš (3-0)Â²+(-3+0)Â²
So Distance of B should be also âˆš 3Â²+3Â² possible if B is (3,-3)
it can be (-3,-3) or (3,3) but they are is in the wrong Quadrant.

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