OG11-Q199

givemeanid · Posted: Fri Jul 06, 2007 4:03 pm Post subject:

Since AB is perpendicular to line x=y (slope = 1), its slope is -1.
A = (2,3)
Equation of line: y = mx + b
In case of AB: 3 = -1*2 + b => b = 5
Hence, equation of AB: y = -x + 5 OR x + y = 5

The line x=y and AB intersect at (2.5, 2.5)
Since its the perpendicular bisector of AB, B is the same distance from x=y on the other side as A is.
So, B = (3, 2)

Now, X axis is perpendicular bisector of BC.
B is 2 units above X axis. So, C is 2 units below X axis.
C = (3, -2)

Answer is (D).

discreet · Posted: Fri Jul 06, 2007 11:29 pm Post subject:

Here's how I approached the problem

The line y = x will intersect AB at a point that has to be equidistant from A as well as B.Also,since y=x, its easy to observe that the co-ordinate of this mid-point wd be (3,3).Lets name it D.

Now, A is (2,3).So, B has to be (3,2) in order to be equidistant from D.
Now,X axis is perp. bisector of BC.

So,X co-ordinate will remain same.....Y co-ordinate obviously has to be -2

So,C (3,-2)