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In other words AB is divided by four equal segments and AC is one of them.
Let F be a midpoint of AB. F's X=3+(6-3)/2=4.5, and F's Y=2+(4-2)/2=3.
C is a midpoint of F.

Solution: coordinate geometry [#permalink]
17 Oct 2003, 06:01

good try by all. amar's answer is correct.
Stolyar the only mistake that u did was to find C such that AC = 1/3*CB.
Question asks for AC = 3*CB.
Find the midpoint of AB (say M) M = (9/2,3) [avg. of coordinates of A & B]

Now, find the midpoint of MB (say C) C = (21/4,7/2) [avg. of coordinates of M & B]. Point C divides AB in the ratio 3:1.

This was easy. What if the given ratio is 3:2.
Then above method will be more involved and lengthy.
In such problems i wud directly use the following formula:

If point C divides line segemnt joining A(x1, y1) and B(x2,y2) in the ratio
m:n , then cordinates of C is given by:

(m*x2 + n*x1)/(m+n), (m*y2 + n*y1)/(m+n)

For above example: cordinates of C are:
(3*6+ 1*3)/(3+1), (3*4+ 1*2)/(3+1)
= 21/4,7/2

Center of (3,2) and (6,4) is (3+6/2,4+2/2) = (9/2,3)--this is the center of the A and B. No find the center of (9/2,3) and (6,4) it will give (21/4,7/2)