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)=`

Options

A. `(1)/(3)`
B. `2/7`
C. `3/2`
D. `2/5`

Answer

B

Explanation

`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)`.