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: \(y - 12\frac{1}{2} = x - 2\frac{1}{2}\)