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The point C divides line segment AB such that AC=3*CB. If [#permalink]
 15 Oct 2003, 05:32
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The point C divides line segment AB such that AC=3*CB. If point A(3,2)
B(6,4). Find the cordinates of point C.
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CEO
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Re: PS: coordinate geom. [#permalink]
15 Oct 2003, 07:24
Vicky wrote: The point C divides line segment AB such that AC=3*CB. If point A(3,2)
B(6,4). Find the cordinates of point C.
i get (3/2,1)
i used distance formula,pythagorous theorem, and finally similarity.
thanks vicks
your problems are really good.
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I got (3┬╛, 2┬╜).
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wonder_gmat wrote: I got (3┬╛, 2┬╜).
no wonder, i got the ratio backwards..never mind...
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I am posting again with answer choices so u have more info to work with:
The point C divides line segment AB such that AC=3*CB. If point A(3,2)
B(6,4). Find the cordinates of point C.
A) (3/2, 1)
B) (3┬╛, 2┬╜)
C) (9/2, 3)
D) (21/4, 7/2)
E) (15/4, 5/2)
Hint: A & B are not correct. praet and wonder try again...
thanks
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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.
C's X=3+(4.5-3)/2=3.75, and C's Y=2+(3-2)/2=2.5
C (3.75; 2.5)
Looks like B.
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Manager
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Vicky wrote: I am posting again with answer choices so u have more info to work with:
The point C divides line segment AB such that AC=3*CB. If point A(3,2)
B(6,4). Find the cordinates of point C.
A) (3/2, 1)
B) (3┬╛, 2┬╜)
C) (9/2, 3)
D) (21/4, 7/2)
E) (15/4, 5/2)
Hint: A & B are not correct. praet and wonder try again...
thanks
Aren't (B) and (E) the same?
(D) can't be it, because too big. (C) has a chance but that's a little big. (E) seems like a good choice but it's the same as (B).
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I vote for D.
Let AB be divided into 4 parts AP, PQ, QC and CB such that AC = 3 * CB
Then AQ = QB So, Q = ((3 + 6)/2, (2+4)/2) = (4.5, 3)
C will be the mid point of Q and B
i.e ((4.5 + 6)/2, (3+4)/2) = (5.25, 3.5) = (21/4, 7/2)
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Solution: coordinate geometry [#permalink]
17 Oct 2003, 07:01
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
thanks.
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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)
Bhai
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Vicky wrote: I am posting again with answer choices so u have more info to work with:
The point C divides line segment AB such that AC=3*CB. If point A(3,2)
B(6,4). Find the cordinates of point C.
A) (3/2, 1)
B) (3¾, 2½)
C) (9/2, 3)
D) (21/4, 7/2)
E) (15/4, 5/2)
Hint: A & B are not correct. praet and wonder try again...
thanks
Find C, such that it cuts segment AB in m:n
A(x1, y1)
B(x2, y2)
P(x, y)
AC/CB = m/n=1/3
Then C(x) = (nx1+mx2)/(m+n)=(9+6)/4=3 3/4
C(y) = (ny1+my2)/(m+n)=(6+4)/4=2 1/2
What's wrong with this???
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edited by praet: virtual , i edited your post to make your answer more clear.  mirhaque, do you see the error in your solution?
mirhaque wrote :
Find C, such that it cuts segment AB in m:n
A(x1, y1)
B(x2, y2)
P(x, y)
AC/CB = m/n=1/3 .........
Virtual's comment : wrong here AC/CB is 3/1 an not 1/3 . 
Then C(x) = (nx1+mx2)/(m+n)=(9+6)/4=3 3/4
C(y) = (ny1+my2)/(m+n)=(6+4)/4=2 1/2
What's wrong with this???
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Virtual wrote: edited by praet: virtual , i edited your post to make your answer more clear. mirhaque, do you see the error in your solution?mirhaque wrote : Find C, such that it cuts segment AB in m:n A(x1, y1) B(x2, y2) P(x, y) AC/CB = m/n=1/3 ......... Virtual's comment : wrong here AC/CB is 3/1 an not 1/3 . Then C(x) = (nx1+mx2)/(m+n)=(9+6)/4=3 3/4 C(y) = (ny1+my2)/(m+n)=(6+4)/4=2 1/2 What's wrong with this???
thanks a lot! 
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