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 \(b = 8\).