(i) The gradient of line \(AP\) is given by:

\(\frac{b - 1}{a + 1} = 2\)

Solving for \(b\):

\(b - 1 = 2(a + 1)\) \(b - 1 = 2a + 2\) \(b = 2a + 3\)

Thus, \(b = 2a + 3\).

(ii) The distance \(AB\) is:

\(AB = \sqrt{(10 - (-1))^2 + (-1 - 1)^2} = \sqrt{11^2 + 2^2} = \sqrt{125}\)

Since \(AP = AB\), we have:

\(\sqrt{(a + 1)^2 + (b - 1)^2} = \sqrt{125}\)

Substitute \(b = 2a + 3\):

\((a + 1)^2 + (2a + 3 - 1)^2 = 125\) \((a + 1)^2 + (2a + 2)^2 = 125\) \((a + 1)^2 + 4(a + 1)^2 = 125\) \(5(a + 1)^2 = 125\) \((a + 1)^2 = 25\) \(a + 1 = 5 \quad \text{or} \quad a + 1 = -5\) \(a = 4 \quad \text{or} \quad a = -6\)

For \(a = 4\):

\(b = 2(4) + 3 = 11\)

For \(a = -6\):

\(b = 2(-6) + 3 = -9\)

Thus, the possible coordinates of \(P\) are \((4, 11)\) and \((-6, -9)\).

Step-by-step Solution:

(i) Find the equation of line AB:

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

Using the point-slope form of the equation of a line, \(y - y_1 = m(x - x_1)\), where \(m\) is the gradient and \((x_1, y_1)\) is a point on the line:

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

Simplifying:

\(y - 6 = -\frac{2}{3}x - \frac{4}{3}\)

\(y = -\frac{2}{3}x + \frac{14}{3}\)

Multiplying through by 3 to eliminate fractions:

\(3y = -2x + 14\)

Rearranging gives:

\(3y + 2x = 14\)

(ii) Find the coordinates of point B:

The midpoint of AB is the point where the perpendicular bisector intersects AB. Solving the equations \(2y = 3x + 5\) and \(3y + 2x = 14\) simultaneously:

From \(2y = 3x + 5\), express \(y\):

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

Substitute into \(3y + 2x = 14\):

\(3\left(\frac{3}{2}x + \frac{5}{2}\right) + 2x = 14\)

\(\frac{9}{2}x + \frac{15}{2} + 2x = 14\)

\(\frac{13}{2}x + \frac{15}{2} = 14\)

\(\frac{13}{2}x = \frac{13}{2}\)

\(x = 1\)

Substitute \(x = 1\) back into \(y = \frac{3}{2}x + \frac{5}{2}\):

\(y = \frac{3}{2}(1) + \frac{5}{2} = 4\)

The midpoint is \((1, 4)\). Since A is \((-2, 6)\), use the midpoint formula to find B:

\(\left(\frac{-2 + x}{2}, \frac{6 + y}{2}\right) = (1, 4)\)

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

\(\frac{6 + y}{2} = 4 \Rightarrow y = 2\)

Thus, the coordinates of B are \((4, 2)\).

(i) Since B is the midpoint of AC, we have:

\(\frac{2 + x}{2} = n \quad \Rightarrow \quad x = 2n - 2\)

\(\frac{m + y}{2} = -6 \quad \Rightarrow \quad y = -12 - m\)

Thus, the coordinates of C are \((2n - 2, -12 - m)\).

(ii) The line \(y = x + 1\) passes through C, so:

\(-12 - m = (2n - 2) + 1\)

\(-12 - m = 2n - 1\)

\(m + 2n = -11\)

The line is perpendicular to AB, so the slope of AB is the negative reciprocal of 1:

\(\frac{m + 6}{2 - n} = -1\)

\(m + 6 = -2 + n\)

\(m - n = -8\)

Solving the equations:

\(m + 2n = -11\)

\(m - n = -8\)

Subtract the second from the first:

\(3n = -3\)

\(n = -1\)

Substitute \(n = -1\) into \(m - n = -8\):

\(m + 1 = -8\)

\(m = -9\)

Thus, \(m = -9\) and \(n = -1\).

1. The line \(\frac{x}{a} + \frac{y}{b} = 1\) intersects the x-axis at \(A(a, 0)\) and the y-axis at \(B(0, b)\).

2. The distance \(AB\) is given by the Pythagorean theorem:

\(AB = \sqrt{a^2 + b^2} = 10\)

\(a^2 + b^2 = 100\)

3. The mid-point \(M\) of \(AB\) is \(\left( \frac{a}{2}, \frac{b}{2} \right)\).

4. \(M\) lies on the line \(2x + y = 10\):

\(2 \left( \frac{a}{2} \right) + \frac{b}{2} = 10\)

\(a + \frac{b}{2} = 10\)

5. Solve the system of equations:

From \(a + \frac{b}{2} = 10\):

\(b = 20 - 2a\)

Substitute into \(a^2 + b^2 = 100\):

\(a^2 + (20 - 2a)^2 = 100\)

\(a^2 + 400 - 80a + 4a^2 = 100\)

\(5a^2 - 80a + 300 = 0\)

\(a^2 - 16a + 60 = 0\)

6. Solve the quadratic equation:

\((a - 6)(a - 10) = 0\)

\(a = 6 \text{ or } a = 10\)

7. If \(a = 6\), then \(b = 20 - 2 \times 6 = 8\).

8. If \(a = 10\), then \(b = 0\), which is not possible as \(b\) must be positive.

9. Therefore, \(a = 6\) and \(b = 8\).

First, find the midpoint C of line AB:

\(C = \left( \frac{14 + (-6)}{2}, \frac{-7 + 3}{2} \right) = (4, -2)\)

Next, find the slope of AB:

\(m_{AB} = \frac{3 - (-7)}{-6 - 14} = \frac{10}{-20} = -\frac{1}{2}\)

The slope of line CD, being perpendicular to AB, is the negative reciprocal:

\(m_{CD} = 2\)

Using point C(4, -2) in the point-slope form:

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

Simplifying to slope-intercept form:

\(y = 2x - 8 - 2\)

\(y = 2x - 10\)

Since the line crosses the y-axis at D, set x = 0:

\(y = 2(0) - 10 = -10\)

So, the equation of CD is:

\(y = 2x - 10\)

For the distance AD, use the distance formula:

\(AD = \sqrt{(14 - 4)^2 + (-7 - (-2))^2}\)

\(AD = \sqrt{10^2 + (-5)^2}\)

\(AD = \sqrt{100 + 25}\)

\(AD = \sqrt{125} = 5\sqrt{5}\)

Therefore, the distance AD is:

\(AD = 5\sqrt{5}\)