(i) The gradient of \(L_1\) is \(-2\) (since \(y = -2x + 8\)).

For \(L_2\) to be perpendicular to \(L_1\), its gradient is the negative reciprocal: \(\frac{1}{2}\).

Using the point-slope form \(y - y_1 = m(x - x_1)\) with point \(A(7, 4)\):

\(y - 4 = \frac{1}{2}(x - 7)\)

(ii) Solve \(2x + y = 8\) and \(y - 4 = \frac{1}{2}(x - 7)\) simultaneously:

Substitute \(y = \frac{1}{2}(x - 7) + 4\) into \(2x + y = 8\):

\(2x + \frac{1}{2}(x - 7) + 4 = 8\)

\(2x + \frac{1}{2}x - \frac{7}{2} + 4 = 8\)

\(\frac{5}{2}x - \frac{7}{2} + 4 = 8\)

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

\(x = 3\)

Substitute \(x = 3\) back to find \(y\):

\(y = \frac{1}{2}(3 - 7) + 4 = 2\)

Point \(B\) is \((3, 2)\).

Length of \(AB\):

\(AB = \sqrt{(7 - 3)^2 + (4 - 2)^2} = \sqrt{4^2 + 2^2} = \sqrt{20} \approx 4.47\)

1. Differentiate \(C_1: y = x^2 - 4x + 7\) to find the slope of the tangent:

\(\frac{dy}{dx} = 2x - 4\)

At \(x = 3\), the slope \(m = 2(3) - 4 = 2\).

2. The equation of the tangent at \(x = 3\) is:

\(y - 4 = 2(x - 3)\)

\(y = 2x - 2\)

3. For \(C_2: y^2 = 4x + k\), differentiate implicitly:

\(2y \frac{dy}{dx} = 4\)

\(\frac{dy}{dx} = \frac{2}{y}\)

4. Equate the slopes from \(C_1\) and \(C_2\):

\(2 = \frac{2}{y} \Rightarrow y = 1\)

5. Substitute \(y = 1\) into \(C_2\):

\(1^2 = 4x + k \Rightarrow 4x + k = 1\)

6. Solve for \(x\) using the tangent equation:

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

7. Substitute \(x = \frac{3}{2}\) into \(4x + k = 1\):

\(4 \left( \frac{3}{2} \right) + k = 1 \Rightarrow 6 + k = 1 \Rightarrow k = -5\)

8. The coordinates of \(P\) are \(\left( \frac{3}{2}, 1 \right)\) or \((-1, 1)\).

1. Find the midpoint of AB:

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

2. Calculate the slope of AB:

\(m_{AB} = \frac{-1 - 3}{9 - 1} = \frac{-4}{8} = -\frac{1}{2}\)

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

\(m = 2\)

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

\(y - 1 = 2(x - 5)\)

5. Simplify to find the equation:

\(y = 2x - 9\)

6. Set \(x = 0\) to find the intersection with the y-axis:

\(y = 2(0) - 9 = -9\)

7. Thus, the coordinates of C are \((0, -9)\).

Step 1: Find the gradient of line AB.

The gradient \(m\) of line AB is given by:

\(m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - 1}{3 - (-1)} = \frac{3}{4}\)

Step 2: Find the gradient of line BC.

Since line BC is perpendicular to AB, its gradient is the negative reciprocal of \(\frac{3}{4}\):

\(m_{BC} = -\frac{4}{3}\)

Step 3: Find the equation of line BC.

Using point B (3, 4) and the gradient \(-\frac{4}{3}\), the equation of line BC is:

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

Simplifying:

\(y = -\frac{4}{3}x + 8\)

Step 4: Find the x-coordinate of C.

Since C is on the x-axis, \(y = 0\):

\(0 = -\frac{4}{3}x + 8\) \(\frac{4}{3}x = 8\) \(x = 6\)

Step 5: Find the distance AC.

Coordinates of A are (-1, 1) and C are (6, 0). Using the distance formula:

\(AC = \sqrt{(6 - (-1))^2 + (0 - 1)^2} = \sqrt{7^2 + 1^2} = \sqrt{49 + 1} = \sqrt{50} = 7.071\)

Thus, the distance AC is 7.071.

To solve the problem, we follow these steps:

(ii) Finding the coordinates of \(A\) and \(B\):

1. Substitute \(k = 15\) into the line equation: \(y + x = 15\) gives \(y = 15 - x\).

2. Substitute \(y = 15 - x\) into the curve equation: \(15 - x = 2x + \frac{12}{x}\)

3. Rearrange and multiply through by \(x\) to clear the fraction: \(15x - x^2 = 2x^2 + 12\)

4. Rearrange to form a quadratic equation: \(3x^2 - 15x + 12 = 0\)

5. Solve the quadratic equation using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

6. Here, \(a = 3\), \(b = -15\), \(c = 12\). Calculate the discriminant: \(b^2 - 4ac = 225 - 144 = 81\).

7. Solve for \(x\): \(x = \frac{15 \pm 9}{6}\)

8. The solutions are \(x = 4\) and \(x = 1\).

9. Substitute back to find \(y\):

• For \(x = 4\), \(y = 15 - 4 = 11\).
• For \(x = 1\), \(y = 15 - 1 = 14\).

10. The coordinates are \((1, 14)\) and \((4, 11)\).

(iii) Finding the equation of the perpendicular bisector:

1. Calculate the midpoint of \(A\) and \(B\): \(\left( \frac{1+4}{2}, \frac{14+11}{2} \right) = \left( 2\frac{1}{2}, 12\frac{1}{2} \right)\)

2. The gradient of \(AB\) is \(\frac{11 - 14}{4 - 1} = -1\).

3. The perpendicular gradient is \(+1\).

4. Use the point-gradient form of the line equation: \(y - 12\frac{1}{2} = 1(x - 2\frac{1}{2})\)

5. Simplify to get the equation of the perpendicular bisector: \(y - 12\frac{1}{2} = x - 2\frac{1}{2}\)