(i) To find the value of h, we use the fact that ∠ABC = 90° and lines BC and AD are parallel. The gradient of AB is \(-\frac{1}{2}\) and the gradient of BC is \(\frac{3h - 2}{h}\). Since BC is perpendicular to AB, the product of their gradients is \(-1\):

\(\left(-\frac{1}{2}\right) \times \frac{3h - 2}{h} = -1\)

Solving for h, we get:

\(\frac{3h - 2}{h} = 2\)

\(3h - 2 = 2h\)

\(h = 2\)

(ii) With h = 2, the coordinates of C are (2, 6). Since CD is parallel to the x-axis, the y-coordinate of D is the same as C, which is 6. The x-coordinate of D can be found using the fact that AD is parallel to BC, so the gradient of AD is also 2:

\(\frac{6 - 0}{x - 4} = 2\)

\(6 = 2(x - 4)\)

\(6 = 2x - 8\)

\(2x = 14\)

\(x = 7\)

Thus, the coordinates of D are (7, 6).

(i) To find the equation of the perpendicular bisector of AC:

The midpoint of AC is \(\left( \frac{-1 + 5}{2}, \frac{2 + 4}{2} \right) = (2, 3)\).

The gradient of AC is \(\frac{4 - 2}{5 + 1} = \frac{1}{3}\).

The gradient of the perpendicular bisector is the negative reciprocal, \(-3\).

Using the point-slope form, the equation is \(y - 3 = -3(x - 2)\).

(ii) To find the coordinates of B and D:

Since B lies on the y-axis, its x-coordinate is 0. Using the perpendicular bisector equation, \(y - 3 = -3(0 - 2)\), we find \(y = 9\). Thus, \(B(0, 9)\).

To find D, use the vector move from A to C, which is \((5 - (-1), 4 - 2) = (6, 2)\).

Since D is opposite B, apply the vector move to B: \((0 + 4, 9 - 12) = (4, -3)\). Thus, \(D(4, -3)\).

(iii) To find the area of the rhombus:

The length of AC is \(\sqrt{(5 - (-1))^2 + (4 - 2)^2} = \sqrt{40}\).

The length of BD is \(\sqrt{(4 - 0)^2 + (-3 - 9)^2} = \sqrt{160}\).

The area of the rhombus is \(\frac{1}{2} \times \sqrt{40} \times \sqrt{160} = 40\).

The gradient of OB is calculated as follows:

Gradient of OB = \(\frac{1 - 0}{3 - 0} = \frac{1}{3}\).

Since L₁ is parallel to OB, it has the same gradient: \(\frac{1}{3}\).

The equation of L₁ passing through A (-1, 3) is:

\(y - 3 = \frac{1}{3}(x + 1)\).

The gradient of AB is:

Gradient of AB = \(\frac{1 - 3}{3 + 1} = -\frac{1}{2}\).

The gradient of L₂, being perpendicular to AB, is the negative reciprocal:

\(Gradient of L₂ = 2.\)

The equation of L₂ passing through B (3, 1) is:

\(y - 1 = 2(x - 3)\).

Solving the simultaneous equations:

1. \(y = \frac{1}{3}x + \frac{10}{3}\)

2. \(y = 2x - 5\)

Equating the two expressions for \(y\):

\(\frac{1}{3}x + \frac{10}{3} = 2x - 5\)

Multiply through by 3 to clear the fraction:

\(x + 10 = 6x - 15\)

Rearrange to solve for \(x\):

\(5x = 25\)

\(x = 5\)

Substitute \(x = 5\) back into one of the equations to find \(y\):

\(y = 2(5) - 5 = 5\)

Thus, the coordinates of C are (5, 5).

(i) The gradient of AB is calculated as:

\(\frac{22 - (-2)}{15 - 3} = \frac{24}{12} = 2\)

Since the gradient of AB is given as 2m, equating gives:

\(2m = 2\)

Thus, \(m = 1\).

(ii) The equation of line AC is:

\(y + 2 = -2(x - 3)\)

\(y = -2x + 6 - 2\)

\(y = -2x + 4\)

The equation of line BC is:

\(y - 22 = 1(x - 15)\)

\(y = x + 7\)

Solving the simultaneous equations:

\(y = -2x + 4\)

\(y = x + 7\)

Equating gives:

\(-2x + 4 = x + 7\)

\(-3x = 3\)

\(x = -1\)

Substitute \(x = -1\) into \(y = x + 7\):

\(y = -1 + 7 = 6\)

Thus, \(C = (-1, 6)\).

(iii) The midpoint M of AB is:

\(M = \left( \frac{3 + 15}{2}, \frac{-2 + 22}{2} \right) = (9, 10)\)

The gradient of the perpendicular bisector is:

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

The equation of the perpendicular bisector is:

\(y - 10 = -\frac{1}{2}(x - 9)\)

\(2y - 20 = -x + 9\)

\(2y + x = 29\)

Solving the equations \(y = x + 7\) and \(2y + x = 29\):

Substitute \(y = x + 7\) into \(2y + x = 29\):

\(2(x + 7) + x = 29\)

\(2x + 14 + x = 29\)

\(3x = 15\)

\(x = 5\)

Substitute \(x = 5\) into \(y = x + 7\):

\(y = 5 + 7 = 12\)

Thus, \(D = (5, 12)\).

(i) The y-coordinate of D is the same as the y-coordinate of the midpoint of AC. The midpoint of AC is \(\left( \frac{0+12}{2}, \frac{-2+14}{2} \right) = (6, 6)\). Therefore, the y-coordinate of D is 6.

(ii) The gradient of AD is calculated as \(\frac{y_2 - y_1}{x_2 - x_1} = \frac{6 - (-2)}{h - 0} = \frac{8}{h}\) or \(\frac{h-12}{8}\). The gradient of CD is \(\frac{6 - 14}{h - 12} = \frac{-8}{h-12}\) or \(\frac{-h}{8}\).

(iii) Since BD is parallel to the x-axis, the product of the gradients of AD and CD is -1. Solving \(\frac{8}{h} \times \frac{-8}{h-12} = -1\) gives \(h^2 - 12h - 64 = 0\). Solving this quadratic equation, we find \(h = 16\) or \(h = -4\). Thus, \(x_D = 16\) and \(x_B = -4\).

(iv) The area of rectangle ABCD can be calculated using the lengths of the sides. The length of AC is \(\sqrt{(12-0)^2 + (14-(-2))^2} = 20\). The length of BD is \(\sqrt{(16-(-4))^2 + (6-6)^2} = 20\). The area is \(20 \times 8 = 160\).