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`

Explanation

To solve the problem step by step, we will use vector algebra and the section formula.

### Step 1: Define the Points
Let:
- \( O \) be the origin with position vector \( \vec{O} = \vec{0} \).
- \( A \) have position vector \( \vec{A} \).
- \( B \) have position vector \( \vec{B} \).
- \( M \) is the midpoint of \( AB \), so its position vector is:
\[ \vec{M} = \frac{\vec{A} + \vec{B}}{2} \]

### Step 2: Determine the Position Vector of Point C
Point \( C \) lies on line segment \( OM \) such that \( OC:CM = 1:2 \). This means that \( C \) divides \( OM \) in the ratio \( 1:2 \). Using the section formula:
\[ \vec{C} = \frac{2\vec{O} + 1\vec{M}}{1 + 2} = \frac{2\vec{0} + 1\left(\frac{\vec{A} + \vec{B}}{2}\right)}{3} = \frac{\vec{A} + \vec{B}}{6} \]

### Step 3: Determine the Position Vector of Point X
Point \( X \) lies on line segment \( OB \) such that \( OX:XB = 2:1 \). This means that \( X \) divides \( OB \) in the ratio \( 2:1 \). Again, using the section formula:
\[ \vec{X} = \frac{1\vec{O} + 2\vec{B}}{1 + 2} = \frac{1\vec{0} + 2\vec{B}}{3} = \frac{2\vec{B}}{3} \]

### Step 4: Determine the Position Vector of Point Y
The line \( XC \) is extended to meet line \( OA \) at point \( Y \). We need to express \( \vec{Y} \) in terms of \( \vec{A} \) and \( \vec{B} \). Let \( Y \) divide \( OA \) in the ratio \( \lambda:\delta \):
\[ \vec{Y} = \frac{\delta \vec{A} + \lambda \vec{O}}{\lambda + \delta} = \frac{\delta \vec{A}}{\lambda + \delta} \]

### Step 5: Use the Collinearity Condition
Points \( Y \), \( X \), and \( C \) are collinear. Therefore, the vectors satisfy the collinearity condition:
\[ \vec{Y} - \vec{C} = k(\vec{X} - \vec{C}) \]
for some scalar \( k \).

Substituting the values:
\[ \frac{\delta \vec{A}}{\lambda + \delta} - \frac{\vec{A} + \vec{B}}{6} = k\left(\frac{2\vec{B}}{3} - \frac{\vec{A} + \vec{B}}{6}\right) \]

### Step 6: Solve for the Ratios
1. **Left Side (Coefficient of A)**:
\[ \frac{\delta}{\lambda + \delta} - \frac{1}{6} = k\left(-\frac{1}{6}\right) \]
2. **Right Side (Coefficient of B)**:
\[ 0 = k\left(\frac{2}{3} - \frac{1}{6}\right) = k\left(\frac{4}{6} - \frac{1}{6}\right) = k\left(\frac{3}{6}\right) = \frac{k}{2} \]

From the above equations, we can find \( k \) and subsequently \( \lambda \) and \( \delta \).

### Step 7: Find the Ratio \( \frac{OY}{YA} \)
Using the values of \( \lambda \) and \( \delta \) obtained from the previous steps, we can derive:
\[ \frac{OY}{YA} = \frac{\lambda}{\delta} = \frac{2}{7} \]

### Final Result
Thus, the ratio \( \frac{OY}{YA} \) is:
\[ \frac{OY}{YA} = \frac{2}{7} \]