1. The equation of the line is given as \(3y + 2x = 33\). Rearranging gives \(y = -\frac{2}{3}x + 11\).

2. The gradient of the line is \(-\frac{2}{3}\). The gradient of the perpendicular line is the negative reciprocal, \(\frac{3}{2}\).

3. The equation of the line perpendicular to 3y + 2x = 33 and passing through (-1, 3) is \(y - 3 = \frac{3}{2}(x + 1)\).

4. Simplifying gives the equation \(y = \frac{3}{2}x + \frac{9}{2}\).

5. Solving the system of equations \(3y + 2x = 33\) and \(y = \frac{3}{2}x + \frac{9}{2}\) gives the intersection point (3, 9).

6. The reflection of (-1, 3) across the line is found by using the midpoint formula. The midpoint of (-1, 3) and the reflection point R is (3, 9). Solving gives R = (7, 15).

To find the points of intersection, set the equations equal: \(x^2 - 4x + 4 = x\).

Rearrange to form a quadratic equation: \(x^2 - 5x + 4 = 0\).

Factor the quadratic: \((x - 1)(x - 4) = 0\).

Thus, \(x = 1\) and \(x = 4\).

Substitute back to find \(y\) coordinates: \(y = 1\) when \(x = 1\) and \(y = 4\) when \(x = 4\).

Points \(A\) and \(B\) are \((1, 1)\) and \((4, 4)\).

The midpoint of \(AB\) is \(\left( \frac{1 + 4}{2}, \frac{1 + 4}{2} \right) = \left( \frac{5}{2}, \frac{5}{2} \right)\).

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:

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

1. Find the midpoint M of AB:

\(M = \left( \frac{-1 + 7}{2}, \frac{-5 + 1}{2} \right) = (3, -2)\)

2. Calculate the gradient of AB:

\(\text{Gradient of } AB = \frac{1 - (-5)}{7 - (-1)} = \frac{6}{8} = \frac{3}{4}\)

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

\(\text{Perpendicular gradient} = -\frac{4}{3}\)

4. Equation of the perpendicular bisector through M:

\(y + 2 = -\frac{4}{3}(x - 3)\)

5. Find where it meets the x-axis (y = 0):

\(0 + 2 = -\frac{4}{3}(x - 3)\)

\(2 = -\frac{4}{3}x + 4\)

\(\frac{4}{3}x = 2\)

\(x = \frac{3}{2}\)

\(C = \left( \frac{3}{2}, 0 \right)\)

6. Find where it meets the y-axis (x = 0):

\(y + 2 = -\frac{4}{3}(0 - 3)\)

\(y + 2 = 4\)

\(y = 2\)

\(D = (0, 2)\)

7. Calculate the length of CD using the distance formula:

\(CD = \sqrt{\left( \frac{3}{2} - 0 \right)^2 + (0 - 2)^2}\)

\(CD = \sqrt{\left( \frac{3}{2} \right)^2 + (-2)^2}\)

\(CD = \sqrt{\frac{9}{4} + 4}\)

\(CD = \sqrt{\frac{9}{4} + \frac{16}{4}}\)

\(CD = \sqrt{\frac{25}{4}}\)

\(CD = \frac{5}{2} = 2.5\)

(i) The midpoint M of AC is calculated as:

\(M = \left( \frac{-3 + 5}{2}, \frac{2 + 6}{2} \right) = (1, 4)\)

The gradient of AC is:

\(\text{Gradient of } AC = \frac{6 - 2}{5 + 3} = \frac{1}{2}\)

The perpendicular bisector has a gradient of:

\(-2\)

The equation of MB is:

\(y - 4 = -2(x - 1)\)

Simplifying gives:

\(y = -2x + 6\)

Setting \(y = 0\) to find B:

\(0 = -2x + 6\) \(x = 3\)

So, B is (3, 0).

(ii) The gradient of AB is:

\(\frac{2 - 0}{-3 - 3} = -\frac{2}{6} = -\frac{1}{3}\)

The gradient of BC is:

\(\frac{6 - 0}{5 - 3} = \frac{6}{2} = 3\)

Since:

\(-\frac{1}{3} \times 3 = -1\)

AB is perpendicular to BC.

(iii) Since ABCD is a square, D is calculated as:

\(D = (-1, 8)\)

The length of AD is:

\(AD = \sqrt{(-1 + 3)^2 + (8 - 2)^2} = \sqrt{4 + 36} = \sqrt{40}\)

or approximately 6.32.