1. Substitute \(k = 8\) into the line equation: \(2y + x = 8\).

2. Solve for \(x\) in terms of \(y\): \(x = 8 - 2y\).

3. Substitute \(x = 8 - 2y\) into the curve equation \(xy = 6\):

\(y(8 - 2y) = 6\)

4. Simplify and solve the quadratic equation:

\(2y^2 - 8y + 6 = 0\)

5. Factor or use the quadratic formula to find \(y\):

\((y - 1)(y - 3) = 0\)

6. The solutions are \(y = 1\) and \(y = 3\).

7. Find corresponding \(x\) values:

For \(y = 1\), \(x = 8 - 2(1) = 6\).

For \(y = 3\), \(x = 8 - 2(3) = 2\).

8. Points \(A\) and \(B\) are \((6, 1)\) and \((2, 3)\).

9. Find the midpoint \(M\) of \(AB\):

\(M = \left( \frac{6 + 2}{2}, \frac{1 + 3}{2} \right) = (4, 2)\)

10. Find the slope of \(AB\):

\(m = \frac{3 - 1}{2 - 6} = -\frac{1}{2}\)

11. The perpendicular slope is \(2\) (since \(m_1 \cdot m_2 = -1\)).

12. Use point-slope form for the perpendicular bisector:

\(y - 2 = 2(x - 4)\)