(i) To show that triangle ABC is isosceles, calculate the lengths of AB, BC, and AC:

\(AB^2 = (4 + 2)^2 + (6 + 1)^2 = 6^2 + 7^2 = 85\)

\(BC^2 = (6 - 4)^2 + (-3 - 6)^2 = 2^2 + 9^2 = 85\)

\(AC^2 = (6 + 2)^2 + (-3 + 1)^2 = 8^2 + 2^2 = 68\)

Since \(AB = BC\), triangle ABC is isosceles.

To find the area of triangle ABC, use the formula:

\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\)

Let M be the midpoint of AC, \(M = (2, -2)\).

Calculate \(BM\):

\(BM = \sqrt{(4 - 2)^2 + (6 + 2)^2} = \sqrt{2^2 + 8^2} = \sqrt{68}\)

Area \(\triangle ABC = \frac{1}{2} \times \sqrt{68} \times \sqrt{68} = 34\).

(ii) Find the gradient of AB:

\(\text{Gradient of } AB = \frac{6 - (-1)}{4 - (-2)} = \frac{7}{6}\)

Equation of AB:

\(y + 1 = \frac{7}{6}(x + 2)\)

Find the gradient of CD (perpendicular to AB):

\(\text{Gradient of } CD = -\frac{6}{7}\)

Equation of CD:

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

Solving the equations:

\(2 = -\frac{6}{7}x + \frac{36}{7}\)

\(\frac{7}{6}x - \frac{14}{6}\)

\(x = \frac{34}{85} = \frac{2}{5}\)