1. The line \(\frac{x}{a} + \frac{y}{b} = 1\) intersects the x-axis at \(A(a, 0)\) and the y-axis at \(B(0, b)\).

2. The distance \(AB\) is given by the Pythagorean theorem:

\(AB = \sqrt{a^2 + b^2} = 10\)

\(a^2 + b^2 = 100\)

3. The mid-point \(M\) of \(AB\) is \(\left( \frac{a}{2}, \frac{b}{2} \right)\).

4. \(M\) lies on the line \(2x + y = 10\):

\(2 \left( \frac{a}{2} \right) + \frac{b}{2} = 10\)

\(a + \frac{b}{2} = 10\)

5. Solve the system of equations:

From \(a + \frac{b}{2} = 10\):

\(b = 20 - 2a\)

Substitute into \(a^2 + b^2 = 100\):

\(a^2 + (20 - 2a)^2 = 100\)

\(a^2 + 400 - 80a + 4a^2 = 100\)

\(5a^2 - 80a + 300 = 0\)

\(a^2 - 16a + 60 = 0\)

6. Solve the quadratic equation:

\((a - 6)(a - 10) = 0\)

\(a = 6 \text{ or } a = 10\)

7. If \(a = 6\), then \(b = 20 - 2 \times 6 = 8\).

8. If \(a = 10\), then \(b = 0\), which is not possible as \(b\) must be positive.

9. Therefore, \(a = 6\) and \(b = 8\).