The line AC has the equation \(2y = x + 4\), which simplifies to \(y = \frac{1}{2}x + 2\). The gradient of AC is \(\frac{1}{2}\).

The line BD is perpendicular to AC, so its gradient is the negative reciprocal of \(\frac{1}{2}\), which is \(-2\).

The equation of line BD passing through \(D(10, -3)\) with gradient \(-2\) is:

\(y + 3 = -2(x - 10)\)

Simplifying gives:

\(y + 3 = -2x + 20\)

\(y = -2x + 17\)

Equating the equations of lines AC and BD:

\(\frac{1}{2}x + 2 = -2x + 17\)

Solving for \(x\):

\(\frac{1}{2}x + 2x = 17 - 2\)

\(\frac{5}{2}x = 15\)

\(x = 6\)

Substitute \(x = 6\) back into \(y = \frac{1}{2}x + 2\):

\(y = \frac{1}{2}(6) + 2 = 3 + 2 = 5\)

Thus, B is \((6, 5)\).

Since \(AB = BC\), the vector from A to B is the same as from B to C. If A is \((0, 2)\), then the vector AB is \((6 - 0, 5 - 2) = (6, 3)\).

Adding this vector to B gives C:

C = \((6 + 6, 5 + 3) = (12, 8)\).

(i) The gradient of AC is \(-\frac{1}{2}\). The perpendicular gradient is 2. The equation of line BX is \(y - 2 = 2(x - 2)\), which simplifies to \(y = 2x - 2\). Solving the equations \(2y + x = 16\) and \(y = 2x - 2\) simultaneously gives the coordinates of X as \((4, 6)\).

(ii) Since X is the midpoint of BD, and using the coordinates of X \((4, 6)\), we find D as \((6, 10)\).

(iii) Using the distance formula, \(AB = \sqrt{(14)^2 + (2)^2} = \sqrt{200}\) and \(BC = \sqrt{(2)^2 + (6)^2} = \sqrt{40}\). The perimeter is \(2\sqrt{200} + 2\sqrt{40} = 40.9\).

First, calculate the gradient of line AB:

Gradient of AB = \(\frac{2 - 8}{6 - 3} = -2\).

Since DC is parallel to AB, the gradient of DC is also -2. Using point C (10, 2), the equation of line DC is:

\(y - 2 = -2(x - 10)\)

Simplifying gives:

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

\(y = -2x + 22\)

For line DA, since it is perpendicular to AB, the gradient of DA is the negative reciprocal of -2, which is \(\frac{1}{2}\). Using point A (3, 8), the equation of line DA is:

\(y - 8 = \frac{1}{2}(x - 3)\)

Simplifying gives:

\(y = \frac{1}{2}x + \frac{1}{2} \times (-3) + 8\)

\(y = \frac{1}{2}x - \frac{3}{2} + 8\)

\(y = \frac{1}{2}x + \frac{13}{2}\)

Now, solve the simultaneous equations:

1. \(y = -2x + 22\)

2. \(y = \frac{1}{2}x + \frac{13}{2}\)

Equating the two expressions for \(y\):

\(-2x + 22 = \frac{1}{2}x + \frac{13}{2}\)

Multiply through by 2 to eliminate the fraction:

\(-4x + 44 = x + 13\)

\(44 - 13 = 4x + x\)

\(31 = 5x\)

\(x = \frac{31}{5} = 6.2\)

Substitute \(x = 6.2\) back into equation 1:

\(y = -2(6.2) + 22\)

\(y = -12.4 + 22\)

\(y = 9.6\)

Thus, the coordinates of D are (6.2, 9.6).

(i) To find the equation of BC, first calculate the slope of AB. The slope \(m\) of AB is given by:

\(m = \frac{14 - 8}{2 - (-2)} = \frac{6}{4} = \frac{3}{2}\)

Since BC is perpendicular to AB, the slope of BC is the negative reciprocal of \(\frac{3}{2}\), which is \(-\frac{2}{3}\).

Using point B (-2, 8) and the slope \(-\frac{2}{3}\), the equation of line BC is:

\(y - 8 = -\frac{2}{3}(x + 2)\)

Simplifying, we get:

\(3y + 2x = 20\)

(ii) Since C lies on the x-axis, its y-coordinate is 0. Substitute \(y = 0\) into the equation of BC:

\(3(0) + 2x = 20\)

\(2x = 20\)

\(x = 10\)

Thus, the coordinates of C are (10, 0).

To find D, use the vector move from B to C and apply it to A. The vector from B to C is:

\((10 - (-2), 0 - 8) = (12, -8)\)

Apply this vector to A (2, 14):

\((2 + 12, 14 - 8) = (14, 6)\)

Thus, the coordinates of D are (14, 6).

(i) To find the equation of CD, first calculate the slope of AB. The slope \(m\) of a line through points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:

\(m = \frac{y_2 - y_1}{x_2 - x_1}\)

For AB, \(m = \frac{11 - 3}{13 - 1} = \frac{8}{12} = \frac{2}{3}\).

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

Using point C (6, 15) and the slope \(-\frac{3}{2}\), the equation of CD is:

\(y - 15 = -\frac{3}{2}(x - 6)\)

(ii) To find the coordinates of D, first find the equation of AB using point A (1, 3):

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

Rearrange to get:

\(3y - 9 = 2x - 2\)

\(3y = 2x + 7\)

\(y = \frac{2}{3}x + \frac{7}{3}\)

Now solve the system of equations:

1. \(y - 15 = -\frac{3}{2}(x - 6)\)

2. \(y = \frac{2}{3}x + \frac{7}{3}\)

Substitute equation 2 into equation 1:

\(\frac{2}{3}x + \frac{7}{3} - 15 = -\frac{3}{2}(x - 6)\)

\(\frac{2}{3}x + \frac{7}{3} - 15 = -\frac{3}{2}x + 9\)

Clear fractions by multiplying through by 6:

\(4x + 14 - 90 = -9x + 54\)

\(4x - 76 = -9x + 54\)

\(13x = 130\)

\(x = 10\)

Substitute \(x = 10\) back into equation 2:

\(y = \frac{2}{3}(10) + \frac{7}{3} = \frac{20}{3} + \frac{7}{3} = 9\)

Thus, the coordinates of D are (10, 9).