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In the triangle OAB, M is the midpoint o...

In the `triangle OAB`, M is the midpoint of AB, C is a point on OM, such that `2 OC = CM`. X is a point on the side OB such that OX = 2XB. The line XC is produced to meet OA in Y. Then `(OY)/(YA)=`

A

`(1)/(3)`

B

`2/7`

C

`3/2`

D

`2/5`

Text Solution

Verified by Experts

The correct Answer is:
B

`OA=a,BC=b`
`thereforeOM=(a+b)/(2)`
`therefore OC=(a+b)/(6)`
`OX=(2)/(3)b`
Let `(OY)/(YA)=lamda`
`thereforeOY=(lamda)/(lamda+1)a`
Now points, Y,C and X are collinear.
`therefore YC=mCX`
`therefore(a+b)/(6)-(lamda)/(lamda+1)a=m(2b)/(3)-m(a+b)/(6)`
Comparing coefficients of a and b
`therefore(1)/(6)-(1)/(lamda+1)=-(m)/(6)`
and `(1)/(6)=(2m)/(3)-(m)/(6)`
`thereforem=(1)/(3) and lamda=(2)/(7)`.
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