(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:

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

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

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

(a) To find \(k\), substitute \(x = 2\) into the curve equation:

\(y = (2 - 3)\sqrt{2 + 1} + 3\)

\(y = (-1)\sqrt{3} + 3\)

\(y = 3 - \sqrt{3}\)

\(k = 1.2679\)

(b) The gradient of \(AE\) is given by:

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

\(\text{Gradient} = \frac{3 - 1.2679}{3 - 2}\)

\(\text{Gradient} = 1.7321\)

(c) The gradients of \(BE, CE, DE\) are approaching 2 as \(x\) approaches 3. This suggests that the gradient of the curve at \(E\) is the limit of these gradients, which is approximately 2.

Part (i):

The slope of line \(AB\) is \(m = \frac{3 - 7}{8 - 0} = -\frac{1}{2}\).

The equation of line \(AB\) is \(y = -\frac{1}{2}x + 7\).

Substitute \(C(3k, k)\) into the line equation:

\(k = -\frac{1}{2}(3k) + 7\)

\(k = -\frac{3}{2}k + 7\)

\(2k = -3k + 14\)

\(5k = 14\)

\(k = 2.8\)

Part (ii):

The midpoint of \(AB\) is \(M(4, 5)\).

The perpendicular slope is \(2\), so the equation of the perpendicular bisector is:

\(y - 5 = 2(x - 4)\)

\(y = 2x - 3\)

Substitute \(C(3k, k)\) into the bisector equation:

\(k = 2(3k) - 3\)

\(k = 6k - 3\)

\(5k = 3\)

\(k = 0.6\)

(i) To find the equation of the perpendicular bisector of \(AB\):

1. Calculate the midpoint of \(AB\):

\(\left( \frac{5+9}{2}, \frac{7+(-1)}{2} \right) = (7, 3)\)

2. Find the slope of \(AB\):

\(m = \frac{-1 - 7}{9 - 5} = -2\)

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

\(m_{\text{perp}} = \frac{1}{2}\)

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

\(y - 3 = \frac{1}{2}(x - 7)\)

(ii) To find the distance \(BX\):

1. Find the equation of the line through \(C(1, 2)\) parallel to \(AB\):

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

2. Solve the system of equations:

\(\begin{align*} y - 3 &= \frac{1}{2}(x - 7) \\ y - 2 &= -2(x - 1) \end{align*}\)

3. Substitute and solve:

\(\frac{1}{2}x - \frac{1}{2} = -2x + 4\)

\(x = \frac{9}{5}, \quad y = \frac{2}{5}\)

4. Calculate the distance \(BX\):

\(BX = \sqrt{(9 - \frac{9}{5})^2 + (-1 - \frac{2}{5})^2}\)

\(BX = \sqrt{7.2^2 + 1.4^2}\)

\(BX \approx 7.33\)

(i) Calculate the distance AB:

\(AB = \sqrt{(5 - (-3))^2 + (1 - 7)^2} = \sqrt{8^2 + (-6)^2} = \sqrt{64 + 36} = 10\)

Calculate the distance BC:

\(BC = \sqrt{(5 - (-1))^2 + (1 - k)^2} = \sqrt{6^2 + (k - 1)^2}\)

Given AB = BC, set the equations equal:

\(10 = \sqrt{36 + (k - 1)^2}\)

Square both sides:

\(100 = 36 + (k - 1)^2\)

Simplify:

\(64 = (k - 1)^2\)

Take the square root:

\(k - 1 = \pm 8\)

So, \(k = 9\) or \(k = -7\).

(ii) Find the midpoint M of AB:

\(M = \left( \frac{-3 + 5}{2}, \frac{7 + 1}{2} \right) = (1, 4)\)

Calculate the slope of AB:

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

The slope of the perpendicular bisector is the negative reciprocal:

\(m_{\text{perp}} = \frac{4}{3}\)

Equation of the perpendicular bisector:

\(y - 4 = \frac{4}{3}(x - 1)\)

Set \(y = 0\) to find the x-intercept:

\(0 - 4 = \frac{4}{3}(x - 1)\) \(-4 = \frac{4}{3}(x - 1)\) \(-12 = 4(x - 1)\) \(-12 = 4x - 4\) \(-8 = 4x\) \(x = -2\)

Thus, the coordinates of D are \((-2, 0)\).