To solve the problem, we need to find the value of \( \lambda \) in triangle \( \Delta ABC \) where a line drawn through the centroid divides \( AB \) in the ratio \( 2:1 \) and \( AC \) in the ratio \( \lambda:1 \).
### Step-by-Step Solution:
1. **Understand the Geometry**:
- Let \( A \) be at the origin \( O(0, 0) \), \( B \) at \( B(b_x, b_y) \), and \( C \) at \( C(c_x, c_y) \).
- The centroid \( G \) of triangle \( ABC \) is given by the formula:
\[
G = \left( \frac{b_x + c_x}{3}, \frac{b_y + c_y}{3} \right)
\]
2. **Finding the Point on \( AB \)**:
- The point \( D \) that divides \( AB \) in the ratio \( 2:1 \) can be calculated using the section formula:
\[
D = \frac{2B + 1A}{2 + 1} = \frac{2(b_x, b_y) + 1(0, 0)}{3} = \left( \frac{2b_x}{3}, \frac{2b_y}{3} \right)
\]
3. **Finding the Point on \( AC \)**:
- The point \( E \) that divides \( AC \) in the ratio \( \lambda:1 \) is:
\[
E = \frac{\lambda C + 1A}{\lambda + 1} = \frac{\lambda(c_x, c_y) + 1(0, 0)}{\lambda + 1} = \left( \frac{\lambda c_x}{\lambda + 1}, \frac{\lambda c_y}{\lambda + 1} \right)
\]
4. **Finding the Coordinates of the Centroid**:
- The centroid \( G \) also divides the median \( AD \) in the ratio \( 2:1 \). The coordinates of \( G \) can also be expressed as:
\[
G = \frac{D + E}{2} = \frac{\left( \frac{2b_x}{3}, \frac{2b_y}{3} \right) + \left( \frac{\lambda c_x}{\lambda + 1}, \frac{\lambda c_y}{\lambda + 1} \right)}{2}
\]
5. **Equating the Centroid Coordinates**:
- Set the two expressions for \( G \) equal:
\[
\left( \frac{b_x + c_x}{3}, \frac{b_y + c_y}{3} \right) = \frac{1}{2} \left( \frac{2b_x}{3} + \frac{\lambda c_x}{\lambda + 1}, \frac{2b_y}{3} + \frac{\lambda c_y}{\lambda + 1} \right)
\]
6. **Solving the x-coordinate Equation**:
- From the x-coordinates:
\[
\frac{b_x + c_x}{3} = \frac{1}{2} \left( \frac{2b_x}{3} + \frac{\lambda c_x}{\lambda + 1} \right)
\]
- Cross-multiplying and simplifying gives:
\[
2(b_x + c_x) = 3 \left( \frac{2b_x}{3} + \frac{\lambda c_x}{\lambda + 1} \right)
\]
- This leads to:
\[
2b_x + 2c_x = 2b_x + \frac{3\lambda c_x}{\lambda + 1}
\]
- Thus:
\[
2c_x = \frac{3\lambda c_x}{\lambda + 1}
\]
7. **Solving the y-coordinate Equation**:
- Similarly, for the y-coordinates:
\[
\frac{b_y + c_y}{3} = \frac{1}{2} \left( \frac{2b_y}{3} + \frac{\lambda c_y}{\lambda + 1} \right)
\]
- This gives a similar equation leading to:
\[
2c_y = \frac{3\lambda c_y}{\lambda + 1}
\]
8. **Finding \( \lambda \)**:
- From either equation (x or y), we can simplify to find \( \lambda \):
\[
\lambda + 1 = \frac{3}{2} \implies \lambda = 2
\]
### Final Answer:
The value of \( \lambda \) is \( 2 \).
