Part (i):

1. Find the midpoint of AB:

\(\left( \frac{1+5}{2}, \frac{1+9}{2} \right) = (3, 5)\)

2. Calculate the gradient of AB:

\(\frac{9-1}{5-1} = 2\)

3. The gradient of the perpendicular bisector is the negative reciprocal:

\(-\frac{1}{2}\)

4. Use the point-slope form to find the equation:

\(y - 5 = -\frac{1}{2}(x - 3)\)

5. Simplify to get:

\(2y = 13 - x\)

Part (ii):

1. Substitute the equation of the perpendicular bisector into the curve equation:

\(6\left( \frac{13-x}{2} \right) = 5x^2 - 18x + 19\)

2. Simplify and solve the quadratic equation:

\(5x^2 - 3x - 4 = 0\)

3. Factorize to find x:

\((x-4)(x+1) = 0\)

4. Solutions are x = 4 and x = -1.

5. Find corresponding y values:

For x = 4, \(y = 4.5\); for x = -1, \(y = 7\).

6. Calculate the distance CD:

\(CD^2 = (4 - (-1))^2 + (4.5 - 7)^2 = 5^2 + (-4.5)^2 = 25 + 6.25 = 31.25\)

7. Therefore, \(CD = \sqrt{\frac{125}{4}}\).