To solve the problem, we follow these steps:

(ii) Finding the coordinates of \(A\) and \(B\):

1. Substitute \(k = 15\) into the line equation: \(y + x = 15\) gives \(y = 15 - x\).

2. Substitute \(y = 15 - x\) into the curve equation: \(15 - x = 2x + \frac{12}{x}\)

3. Rearrange and multiply through by \(x\) to clear the fraction: \(15x - x^2 = 2x^2 + 12\)

4. Rearrange to form a quadratic equation: \(3x^2 - 15x + 12 = 0\)

5. Solve the quadratic equation using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

6. Here, \(a = 3\), \(b = -15\), \(c = 12\). Calculate the discriminant: \(b^2 - 4ac = 225 - 144 = 81\).

7. Solve for \(x\): \(x = \frac{15 \pm 9}{6}\)

8. The solutions are \(x = 4\) and \(x = 1\).

9. Substitute back to find \(y\):

• For \(x = 4\), \(y = 15 - 4 = 11\).
• For \(x = 1\), \(y = 15 - 1 = 14\).

10. The coordinates are \((1, 14)\) and \((4, 11)\).

(iii) Finding the equation of the perpendicular bisector:

1. Calculate the midpoint of \(A\) and \(B\): \(\left( \frac{1+4}{2}, \frac{14+11}{2} \right) = \left( 2\frac{1}{2}, 12\frac{1}{2} \right)\)

2. The gradient of \(AB\) is \(\frac{11 - 14}{4 - 1} = -1\).

3. The perpendicular gradient is \(+1\).

4. Use the point-gradient form of the line equation: \(y - 12\frac{1}{2} = 1(x - 2\frac{1}{2})\)

5. Simplify to get the equation of the perpendicular bisector: \(y - 12\frac{1}{2} = x - 2\frac{1}{2}\)