1. The gradient of line AB is \(\frac{q - 3}{p - 8}\).

2. The perpendicular bisector has a gradient of -2, so the gradient of AB is \(\frac{1}{2}\).

3. Set \(\frac{q - 3}{p - 8} = \frac{1}{2}\) and solve for q:

\(q - 3 = \frac{1}{2}(p - 8)\)

\(2q - p = -2\)

4. The midpoint of AB is \(\left( \frac{8 + p}{2}, \frac{3 + q}{2} \right)\).

5. Substitute the midpoint into the equation of the perpendicular bisector:

\(\frac{3 + q}{2} = -2 \left( \frac{8 + p}{2} \right) + 4\)

\(3 + q = -2(8 + p) + 8\)

\(3 + q = -16 - 2p + 8\)

\(q = -11 - 2p\)

6. Solve the system of equations:

\(2q - p = -2\)

\(q = -11 - 2p\)

Substitute \(q = -11 - 2p\) into \(2q - p = -2\):

\(2(-11 - 2p) - p = -2\)

\(-22 - 4p - p = -2\)

\(-5p = 20\)

\(p = -4\)

Substitute \(p = -4\) into \(q = -11 - 2p\):

\(q = -11 - 2(-4)\)

\(q = -11 + 8\)

\(q = -3\)

(i) To find the gradient of line \(AB\), use the formula:

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

Substitute the coordinates of \(A(-3k - 1, k + 3)\) and \(B(k + 3, 3k + 5)\):

\(m = \frac{(3k + 5) - (k + 3)}{(k + 3) - (-3k - 1)} = \frac{2k + 2}{4k + 4}\)

Simplify:

\(m = \frac{2(k + 1)}{4(k + 1)} = \frac{1}{2}\)

The gradient is \(\frac{1}{2}\), independent of \(k\).

(ii) To find the equation of the perpendicular bisector, first find the midpoint of \(AB\):

\(\text{Midpoint} = \left( \frac{-3k - 1 + k + 3}{2}, \frac{k + 3 + 3k + 5}{2} \right) = \left( \frac{-2k + 2}{2}, \frac{4k + 8}{2} \right)\)

Simplify:

\(\left( -k + 1, 2k + 4 \right)\)

The gradient of the perpendicular bisector is the negative reciprocal of \(\frac{1}{2}\):

\(-2\)

Using the point-slope form of a line equation:

\(y - (2k + 4) = -2(x - (-k + 1))\)

Simplify:

\(y - 2k - 4 = -2(x + k - 1)\) \(y - 2k - 4 = -2x - 2k + 2\) \(y + 2x = 6\)

1. Find the gradient of line \(AB\):

\(\text{Gradient of } AB = \frac{5h - h}{4h + 6 - h} = \frac{4h}{3h + 6}\)

2. The gradient of the perpendicular bisector is \(-\frac{3}{2}\), so:

\(\frac{4h}{3h + 6} \times \left(-\frac{3}{2}\right) = -1\)

3. Solve for \(h\):

\(\frac{4h}{3h + 6} = \frac{2}{3}\)

\(12h = 2(3h + 6)\)

\(12h = 6h + 12\)

\(6h = 12\)

\(h = 2\)

4. Find the midpoint of \(AB\):

\(\left(\frac{5h + 6}{2}, \frac{3h}{2}\right) = (8, 6)\)

5. Substitute the midpoint into the equation of the bisector:

\(3(8) + 2(6) = k\)

\(24 + 12 = k\)

\(k = 36\)

(i) To find the intersection points, set the equations equal:

\(\frac{1}{x} + c = cx - 3\)

Multiply through by \(x\) to clear the fraction:

\(1 + cx = cx^2 - 3x\)

Rearrange to form a quadratic equation:

\(cx^2 - 3x - cx + 1 = 0\)

\(cx^2 - cx - 3x + 1 = 0\)

Use the quadratic formula \(b^2 - 4ac\) to find the discriminant:

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

For real solutions, the discriminant must be non-negative:

\(c^2 + 2c + 9 \geq 0\)

Critical values are \(c = -1\) and \(c = -9\). Thus, \(c \leq -9\) or \(c \geq -1\).

(ii) For tangency, the discriminant must be zero. Substitute \(c = -1\) and \(c = -9\) into the quadratic equation:

For \(c = -1\):

\(-x^2 - 2x - 1 = 0\)

Solving gives \(x = -1\).

For \(c = -9\):

\(-9x^2 + 6x - 1 = 0\)

Solving gives \(x = \frac{1}{3}\).

Part (i):

1. Find the midpoint of AB:

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

2. Calculate the gradient of AB:

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

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

\(-\frac{1}{2}\)

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

\(y - 5 = -\frac{1}{2}(x - 3)\)

5. Simplify to get:

\(2y = 13 - x\)

Part (ii):

1. Substitute the equation of the perpendicular bisector into the curve equation:

\(6\left( \frac{13-x}{2} \right) = 5x^2 - 18x + 19\)

2. Simplify and solve the quadratic equation:

\(5x^2 - 3x - 4 = 0\)

3. Factorize to find x:

\((x-4)(x+1) = 0\)

4. Solutions are x = 4 and x = -1.

5. Find corresponding y values:

For x = 4, \(y = 4.5\); for x = -1, \(y = 7\).

6. Calculate the distance CD:

\(CD^2 = (4 - (-1))^2 + (4.5 - 7)^2 = 5^2 + (-4.5)^2 = 25 + 6.25 = 31.25\)

7. Therefore, \(CD = \sqrt{\frac{125}{4}}\).

(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:

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

(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 \((-2, 0)\).

(i) To find the distance \(AB\), use the distance formula:

\(\sqrt{(9 - p)^2 + (3p)^2} = 13\)

Squaring both sides, we get:

\((9 - p)^2 + (3p)^2 = 169\)

Expanding and simplifying:

\(81 - 18p + p^2 + 9p^2 = 169\)

\(10p^2 - 18p + 81 = 169\)

\(10p^2 - 18p - 88 = 0\)

Solving this quadratic equation using the quadratic formula:

\(p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

where \(a = 10\), \(b = -18\), \(c = -88\):

\(p = \frac{18 \pm \sqrt{(-18)^2 - 4 \times 10 \times (-88)}}{20}\)

\(p = \frac{18 \pm \sqrt{324 + 3520}}{20}\)

\(p = \frac{18 \pm \sqrt{3844}}{20}\)

\(p = \frac{18 \pm 62}{20}\)

\(p = 4 \quad \text{or} \quad p = -\frac{11}{5}\)

(ii) The gradient of the line \(2x + 3y = 9\) is \(-\frac{2}{3}\).

The gradient of \(AB\) is \(\frac{3p}{9 - p}\).

For the lines to be perpendicular, the product of their gradients must be \(-1\):

\(\left(-\frac{2}{3}\right) \left(\frac{3p}{9 - p}\right) = -1\)

\(\frac{2p}{9 - p} = 1\)

Solving for \(p\):

\(2p = 9 - p\)

\(3p = 9\)

\(p = 3\)

(i) Find the equation of the perpendicular bisector of AB.

1. Calculate the midpoint M of AB:

\(M = \left( \frac{4 + 10}{2}, \frac{6 + 2}{2} \right) = (7, 4)\)

2. Find the slope of AB:

\(m_{AB} = \frac{2 - 6}{10 - 4} = -\frac{2}{3}\)

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

\(m_{\text{perpendicular}} = \frac{3}{2}\)

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

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

(ii) Calculate the coordinates of C.

1. Equation of line parallel to AB through (3, 11):

\(y - 11 = -\frac{2}{3}(x - 3)\)

2. Solve the system of equations:

Equation 1: \(y - 4 = \frac{3}{2}(x - 7)\)

Equation 2: \(y - 11 = -\frac{2}{3}(x - 3)\)

3. Substitute and solve:

\(\frac{3}{2}(x - 7) + 4 = -\frac{2}{3}(x - 3) + 11\)

4. Solve for x:

\(\frac{3}{2}x - \frac{21}{2} + 4 = -\frac{2}{3}x + 2 + 11\)

5. Simplify and solve:

\(\frac{3}{2}x + \frac{2}{3}x = \frac{21}{2} - 4 + 13\)

6. Find x:

\(x = 9\)

7. Substitute x back to find y:

\(y - 4 = \frac{3}{2}(9 - 7)\)

\(y = 7\)

Coordinates of C: (9, 7)

(i) The equation of the line with gradient \(-2\) passing through \(P(3t, 2t)\) is:

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

Rearranging gives:

\(y = -2x + 6t + 2t\)

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

To find \(A\), set \(y = 0\):

\(0 = -2x + 8t\)

\(x = 4t\)

So, \(A(4t, 0)\).

To find \(B\), set \(x = 0\):

\(y = 8t\)

So, \(B(0, 8t)\).

The area of triangle \(AOB\) is:

\(\text{Area} = \frac{1}{2} \times 4t \times 8t = 16t^2\)

(ii) The line through \(P\) perpendicular to \(AB\) has gradient \(\frac{1}{2}\). Its equation is:

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

Rearranging gives:

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

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

To find \(C\), set \(y = 0\):

\(0 = \frac{1}{2}x + \frac{1}{2}t\)

\(x = -t\)

So, \(C(-t, 0)\).

The midpoint of \(PC\) is:

\(\left( \frac{3t - t}{2}, \frac{2t + 0}{2} \right) = (t, t)\)

This lies on the line \(y = x\).

(i) Find the gradient of line AB:

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

\(m = \frac{3a + 9 - (2a - 1)}{2a + 4 - a} = \frac{a + 10}{a + 4}\)

The gradient of a line perpendicular to \(AB\) is the negative reciprocal:

\(m_{\perp} = -\frac{a + 4}{a + 10}\)

(ii) Find the possible values of \(a\):

The distance \(AB\) is given by:

\(\sqrt{(2a + 4 - a)^2 + (3a + 9 - (2a - 1))^2} = \sqrt{260}\)

Simplify the expression:

\(\sqrt{(a + 4)^2 + (a + 10)^2} = \sqrt{260}\)

Square both sides:

\((a + 4)^2 + (a + 10)^2 = 260\)

Expand and simplify:

\(a^2 + 8a + 16 + a^2 + 20a + 100 = 260\) \(2a^2 + 28a + 116 = 260\) \(2a^2 + 28a - 144 = 0\)

Divide by 2:

\(a^2 + 14a - 72 = 0\)

Factorize:

\((a - 4)(a + 18) = 0\)

Thus, \(a = 4\) or \(a = -18\).

Step-by-step Solution:

1. Substitute point A (8, -4) into the line equation 4x + ky = 20:
   \(4(8) + k(-4) = 20\)
   \(32 - 4k = 20\)
   Solve for \(k\):
   \(32 - 20 = 4k\)
   \(12 = 4k\)
   \(k = 3\)
2. Substitute point B (b, 2b) into the line equation:
   \(4b + 3(2b) = 20\)
   \(4b + 6b = 20\)
   \(10b = 20\)
   Solve for \(b\):
   \(b = 2\)
3. Find the mid-point of AB:
   Mid-point formula: \(\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\)
   \(A (8, -4), B (2, 4)\)
   Mid-point: \(\left( \frac{8 + 2}{2}, \frac{-4 + 4}{2} \right) = (5, 0)\)

1. Find the midpoint of the line segment joining (2, 7) and (10, 3):

\(\left( \frac{2 + 10}{2}, \frac{7 + 3}{2} \right) = (6, 5)\)

2. Calculate the gradient (slope) of the line segment:

\(m = \frac{3 - 7}{10 - 2} = -\frac{1}{2}\)

3. Determine the gradient of the perpendicular bisector:

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

4. Write the equation of the perpendicular bisector using point-slope form:

\(y - 5 = 2(x - 6)\)

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

\(0 - 5 = 2(x - 6)\)

\(-5 = 2x - 12\)

\(2x = 7\)

\(x = \frac{7}{2}\)

6. The coordinates where the perpendicular bisector meets the x-axis are:

\(\left( \frac{7}{2}, 0 \right)\)

1. Use the distance formula between points \((a, 2)\) and \((3, b)\):

\(\sqrt{(a - 3)^2 + (2 - b)^2} = \sqrt{125}\)

Squaring both sides:

\((a - 3)^2 + (2 - b)^2 = 125\)

2. Use the gradient formula for the line \(AB\):

\(\frac{2 - b}{a - 3} = 2\)

Rearrange to find \(b\):

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

\(b = 2 - 2a + 6\)

\(b = 8 - 2a\)

3. Substitute \(b = 8 - 2a\) into the distance equation:

\((a - 3)^2 + (2 - (8 - 2a))^2 = 125\)

\((a - 3)^2 + (2a - 6)^2 = 125\)

4. Expand and simplify:

\((a - 3)^2 + (2a - 6)^2 = 125\)

\((a^2 - 6a + 9) + (4a^2 - 24a + 36) = 125\)

\(5a^2 - 30a + 45 = 125\)

\(5a^2 - 30a - 80 = 0\)

5. Factorize:

\(5(a^2 - 6a - 16) = 0\)

\(5(a + 2)(a - 8) = 0\)

6. Solve for \(a\):

\(a = -2\) or \(a = 8\)

7. Substitute back to find \(b\):

If \(a = -2\), \(b = 8 - 2(-2) = 12\)

If \(a = 8\), \(b = 8 - 2(8) = -8\)

Thus, the possible values are \(a = -2\) or \(8\), \(b = 12\) or \(-8\).

1. Find the midpoint M of the points (3, 7) and (-1, 1):

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

2. Find the slope of the line \(\frac{x}{3} + \frac{y}{2} = 1\):

Rewriting in slope-intercept form:

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

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

The slope is \(-\frac{2}{3}\).

3. Use the point-slope form to find the equation of the line through M (1, 4) with the same slope:

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

(i) The slope of the line y = 2x - 7 is 2. The slope of a line perpendicular to this is the negative reciprocal, which is -\frac{1}{2}. Using the point-slope form of a line equation, y - y_1 = m(x - x_1), where m is the slope and (x_1, y_1) is a point on the line, we have:

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

(ii) The midpoint C of AB is calculated as:

C = \left(\frac{3 + (-21)}{2}, \frac{1 + 11}{2}\right) = (-9, 6)

The distance AC is calculated using the distance formula:

AC = \sqrt{(3 - (-9))^2 + (1 - 6)^2} = \sqrt{12^2 + (-5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13

(i) To find the perpendicular bisector of AB:

1. Calculate the midpoint of AB:

Midpoint = \(\left( \frac{-1 + 7}{2}, \frac{6 + 2}{2} \right) = (3, 4)\)

2. Find the slope of AB:

Slope of AB = \(\frac{2 - 6}{7 + 1} = -\frac{1}{2}\)

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

Slope = 2

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

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

\(y = 2x - 6 + 4\)

\(y = 2x - 2\)

(ii) For point C on the perpendicular bisector:

1. The equation of the bisector is \(q = 2p - 2\)

2. The distance OC is 2 units:

\(p^2 + q^2 = 4\)

3. Substitute \(q = 2p - 2\) into \(p^2 + q^2 = 4\):

\(p^2 + (2p - 2)^2 = 4\)

\(p^2 + 4p^2 - 8p + 4 = 4\)

\(5p^2 - 8p = 0\)

\(p(5p - 8) = 0\)

\(p = 0\) or \(p = \frac{8}{5}\)

4. Find q for each p:

If \(p = 0\), \(q = 2(0) - 2 = -2\)

If \(p = \frac{8}{5}\), \(q = 2\left(\frac{8}{5}\right) - 2 = \frac{16}{5} - \frac{10}{5} = \frac{6}{5}\)

5. Possible positions of C are (0, -2) and \(\left( \frac{8}{5}, \frac{6}{5} \right)\).

1. The equation of the line is given as \(3y + 2x = 33\). Rearranging gives \(y = -\frac{2}{3}x + 11\).

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

3. The equation of the line perpendicular to 3y + 2x = 33 and passing through (-1, 3) is \(y - 3 = \frac{3}{2}(x + 1)\).

4. Simplifying gives the equation \(y = \frac{3}{2}x + \frac{9}{2}\).

5. Solving the system of equations \(3y + 2x = 33\) and \(y = \frac{3}{2}x + \frac{9}{2}\) gives the intersection point (3, 9).

6. The reflection of (-1, 3) across the line is found by using the midpoint formula. The midpoint of (-1, 3) and the reflection point R is (3, 9). Solving gives R = (7, 15).

To find the points of intersection, set the equations equal: \(x^2 - 4x + 4 = x\).

Rearrange to form a quadratic equation: \(x^2 - 5x + 4 = 0\).

Factor the quadratic: \((x - 1)(x - 4) = 0\).

Thus, \(x = 1\) and \(x = 4\).

Substitute back to find \(y\) coordinates: \(y = 1\) when \(x = 1\) and \(y = 4\) when \(x = 4\).

Points \(A\) and \(B\) are \((1, 1)\) and \((4, 4)\).

The midpoint of \(AB\) is \(\left( \frac{1 + 4}{2}, \frac{1 + 4}{2} \right) = \left( \frac{5}{2}, \frac{5}{2} \right)\).

1. Substitute \(k = 8\) into the line equation: \(2y + x = 8\).

2. Solve for \(x\) in terms of \(y\): \(x = 8 - 2y\).

3. Substitute \(x = 8 - 2y\) into the curve equation \(xy = 6\):

\(y(8 - 2y) = 6\)

4. Simplify and solve the quadratic equation:

\(2y^2 - 8y + 6 = 0\)

5. Factor or use the quadratic formula to find \(y\):

\((y - 1)(y - 3) = 0\)

6. The solutions are \(y = 1\) and \(y = 3\).

7. Find corresponding \(x\) values:

For \(y = 1\), \(x = 8 - 2(1) = 6\).

For \(y = 3\), \(x = 8 - 2(3) = 2\).

8. Points \(A\) and \(B\) are \((6, 1)\) and \((2, 3)\).

9. Find the midpoint \(M\) of \(AB\):

\(M = \left( \frac{6 + 2}{2}, \frac{1 + 3}{2} \right) = (4, 2)\)

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

\(m = \frac{3 - 1}{2 - 6} = -\frac{1}{2}\)

11. The perpendicular slope is \(2\) (since \(m_1 \cdot m_2 = -1\)).

12. Use point-slope form for the perpendicular bisector:

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

1. Find the midpoint M of AB:

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

2. Calculate the gradient of AB:

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

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

\(\text{Perpendicular gradient} = -\frac{4}{3}\)

4. Equation of the perpendicular bisector through M:

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

5. Find where it meets the x-axis (y = 0):

\(0 + 2 = -\frac{4}{3}(x - 3)\)

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

\(\frac{4}{3}x = 2\)

\(x = \frac{3}{2}\)

\(C = \left( \frac{3}{2}, 0 \right)\)

6. Find where it meets the y-axis (x = 0):

\(y + 2 = -\frac{4}{3}(0 - 3)\)

\(y + 2 = 4\)

\(y = 2\)

\(D = (0, 2)\)

7. Calculate the length of CD using the distance formula:

\(CD = \sqrt{\left( \frac{3}{2} - 0 \right)^2 + (0 - 2)^2}\)

\(CD = \sqrt{\left( \frac{3}{2} \right)^2 + (-2)^2}\)

\(CD = \sqrt{\frac{9}{4} + 4}\)

\(CD = \sqrt{\frac{9}{4} + \frac{16}{4}}\)

\(CD = \sqrt{\frac{25}{4}}\)

\(CD = \frac{5}{2} = 2.5\)

(i) The midpoint M of AC is calculated as:

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

The gradient of AC is:

\(\text{Gradient of } AC = \frac{6 - 2}{5 + 3} = \frac{1}{2}\)

The perpendicular bisector has a gradient of:

\(-2\)

The equation of MB is:

\(y - 4 = -2(x - 1)\)

Simplifying gives:

\(y = -2x + 6\)

Setting \(y = 0\) to find B:

\(0 = -2x + 6\) \(x = 3\)

So, B is (3, 0).

(ii) The gradient of AB is:

\(\frac{2 - 0}{-3 - 3} = -\frac{2}{6} = -\frac{1}{3}\)

The gradient of BC is:

\(\frac{6 - 0}{5 - 3} = \frac{6}{2} = 3\)

Since:

\(-\frac{1}{3} \times 3 = -1\)

AB is perpendicular to BC.

(iii) Since ABCD is a square, D is calculated as:

\(D = (-1, 8)\)

The length of AD is:

\(AD = \sqrt{(-1 + 3)^2 + (8 - 2)^2} = \sqrt{4 + 36} = \sqrt{40}\)

or approximately 6.32.

First, substitute \(k = 2\) into the equations:

Line: \(y = 2x + 6\)

Curve: \(y = x^2 + 3x + 4\)

Set the equations equal to find intersection points:

\(x^2 + 3x + 4 = 2x + 6\)

Simplify to: \(x^2 + x - 2 = 0\)

Factorize: \((x - 1)(x + 2) = 0\)

Solutions: \(x = 1\) and \(x = -2\)

Find corresponding \(y\)-values:

For \(x = 1\), \(y = 2(1) + 6 = 8\)

For \(x = -2\), \(y = 2(-2) + 6 = 2\)

Points \(A(1, 8)\) and \(B(-2, 2)\)

Distance \(AB\):

\(AB = \sqrt{(1 - (-2))^2 + (8 - 2)^2} = \sqrt{3^2 + 6^2} = \sqrt{45} = 3\sqrt{5}\)

Midpoint of \(AB\):

\(\left(\frac{1 + (-2)}{2}, \frac{8 + 2}{2}\right) = \left(-\frac{1}{2}, 5\right)\)

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

The distance \(PQ\) is given by:

\(PQ = \sqrt{a^2 + b^2} = \sqrt{45}\)

Squaring both sides, we get:

\(a^2 + b^2 = 45 \quad (1)\)

The gradient of the line \(PQ\) is:

\(\text{Gradient} = \frac{0 - b}{a - 0} = -\frac{b}{a} = -\frac{1}{2}\)

Solving for \(b\), we have:

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

\(b = \frac{a}{2} \quad (2)\)

Substitute equation (2) into equation (1):

\(a^2 + \left(\frac{a}{2}\right)^2 = 45\)

\(a^2 + \frac{a^2}{4} = 45\)

\(\frac{4a^2 + a^2}{4} = 45\)

\(5a^2 = 180\)

\(a^2 = 36\)

\(a = 6\)

Substitute \(a = 6\) back into equation (2):

\(b = \frac{6}{2} = 3\)

Thus, \(a = 6\) and \(b = 3\).

To solve the problem, follow these steps:

1. Find the gradient of \(L_1\) using points \(A(2, 5)\) and \(B(10, 9)\):

\(\text{Gradient of } L_1 = \frac{9 - 5}{10 - 2} = \frac{1}{2}\)

2. Since \(L_2\) is parallel to \(L_1\), its equation is:

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

3. The line \(AC\) is perpendicular to \(L_2\), so its gradient is the negative reciprocal:

\(\text{Gradient of } AC = -2\)

4. Using point \(A(2, 5)\), the equation of line \(AC\) is:

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

5. Solve the equations \(y = \frac{1}{2}x\) and \(y - 5 = -2(x - 2)\) simultaneously to find \(C\):

Substitute \(y = \frac{1}{2}x\) into \(y - 5 = -2(x - 2)\):

\(\frac{1}{2}x - 5 = -2(x - 2)\)

\(\frac{1}{2}x - 5 = -2x + 4\)

\(\frac{1}{2}x + 2x = 9\)

\(\frac{5}{2}x = 9\)

\(x = \frac{18}{5} = 3.6\)

\(y = \frac{1}{2}(3.6) = 1.8\)

Thus, \(C(3.6, 1.8)\).

6. Calculate the distance \(AC\):

\(AC = \sqrt{(3.6 - 2)^2 + (1.8 - 5)^2}\)

\(AC = \sqrt{1.6^2 + 3.2^2}\)

\(AC = \sqrt{2.56 + 10.24}\)

\(AC = \sqrt{12.8}\)

\(AC \approx 3.58\)

Step 1: Find the coordinates of D.

The midpoint D of AB is given by:

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

Step 2: Write an equation for CD² = 20.

The distance squared from C to D is:

\((x - 5)^2 + (y - 1)^2 = 20\)

Step 3: Find an equation for AC = BC and simplify.

The distances squared are:

\((x - 1)^2 + (y - 3)^2 = (x - 9)^2 + (y + 1)^2\)

Expanding both sides:

\(x^2 - 2x + 1 + y^2 - 6y + 9 = x^2 - 18x + 81 + y^2 + 2y + 1\)

Simplifying gives:

\(-2x - 6y + 10 = -18x + 2y + 82\)

\(16x - 8y = 72\)

\(y = 2x - 9\)

Step 4: Find the possible coordinates of C.

Substitute \(y = 2x - 9\) into \((x - 5)^2 + (y - 1)^2 = 20\):

\((x - 5)^2 + (2x - 10)^2 = 20\)

\((x - 5)^2 + (2x - 10)^2 = 20\)

\(5x^2 - 50x + 105 = 0\)

Solving gives \(x = 3\) or \(x = 7\).

For \(x = 3\), \(y = 2(3) - 9 = -3\).

For \(x = 7\), \(y = 2(7) - 9 = 5\).

Thus, the possible coordinates of C are (3, -3) and (7, 5).

To find the coordinates of \(Q\), substitute \(y = x + 4\) into the curve equation:

\(x + 4 = 2x^2 - 4x + 1\)

Simplify to:

\(2x^2 - 5x - 3 = 0\)

Factorize:

\((2x + 1)(x - 3) = 0\)

Solutions are \(x = -\frac{1}{2}\) and \(x = 3\). Since \(P = (3, 7)\), \(Q = \left( -\frac{1}{2}, \frac{3}{2} \right)\).

For part (iii), find the mid-point of \(AP\):

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

Gradient of line \(Q\) to mid-point:

\(\frac{3/2 - 3}{-1/2 - 2} = \frac{-3/2}{-5/2} = \frac{1}{5}\)

Equation of line:

\(y - 3 = -\frac{1}{5}(x - 2)\)

(i) To find the midpoint of AB, use the midpoint formula:

\(\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\)

Substitute \(A(2, 5)\) and \(B(5, -1)\):

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

(ii) First, find the slope of AB:

\(m = \frac{-1 - 5}{5 - 2} = \frac{-6}{3} = -2\)

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

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

Using point C(8, 6) and the point-slope form:

\(y - 6 = \frac{1}{2}(x - 8)\)

Simplify to get the equation in the form \(ax + by + c = 0\):

\(y - 6 = \frac{1}{2}x - 4\)

\(2y - 12 = x - 8\)

\(x - 2y + 4 = 0\)

First, solve the system of equations:

1. \(y^2 = 12x\)

2. \(3y = 4x + 6\)

From equation (2), express \(x\) in terms of \(y\):

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

Substitute \(x\) in equation (1):

\(y^2 = 12 \left( \frac{3y - 6}{4} \right)\)

Simplify:

\(y^2 = 9y - 18\)

Rearrange to form a quadratic equation:

\(y^2 - 9y + 18 = 0\)

Factor the quadratic:

\((y - 3)(y - 6) = 0\)

Thus, \(y = 3\) or \(y = 6\).

Find corresponding \(x\) values:

For \(y = 3\):

\(x = \frac{3(3) - 6}{4} = \frac{9 - 6}{4} = \frac{3}{4}\)

For \(y = 6\):

\(x = \frac{3(6) - 6}{4} = \frac{18 - 6}{4} = 3\)

The points of intersection are \(\left( \frac{3}{4}, 3 \right)\) and \((3, 6)\).

Calculate the distance between the points:

\(\text{Distance} = \sqrt{\left( 3 - \frac{3}{4} \right)^2 + (6 - 3)^2}\)

\(= \sqrt{\left( \frac{9}{4} \right)^2 + 3^2}\)

\(= \sqrt{\frac{81}{16} + 9}\)

\(= \sqrt{\frac{81}{16} + \frac{144}{16}}\)

\(= \sqrt{\frac{225}{16}}\)

\(= \frac{15}{4}\)

\(= 3.75\)

To solve the problem, we follow these steps:

1. Find the midpoint \(M\) of \(AB\):

\(M = \left( \frac{2 + 8}{2}, \frac{6 + 10}{2} \right) = (5, 8)\)

2. Calculate the gradient of \(AB\):

\(\text{Gradient of } AB = \frac{10 - 6}{8 - 2} = \frac{2}{3}\)

3. Find the gradient of the perpendicular bisector:

\(\text{Perpendicular gradient} = -\frac{3}{2}\)

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

\(y - 8 = -\frac{3}{2}(x - 5)\)

Expanding and rearranging gives:

\(2y + 3x = 31\)

5. Find the equation of line \(BC\):

\(y = 5(x - 6) \Rightarrow y = 5x - 30\)

6. Solve the system of equations:

\(2y + 3x = 31\)

\(y = 5x - 30\)

Substitute \(y = 5x - 30\) into \(2y + 3x = 31\):

\(2(5x - 30) + 3x = 31\)

\(10x - 60 + 3x = 31\)

\(13x = 91\)

\(x = 7\)

Substitute \(x = 7\) back into \(y = 5x - 30\):

\(y = 5(7) - 30 = 5\)

Thus, the coordinates of \(D\) are \((7, 5)\).

(i) To find the intersection points \(A\) and \(B\), solve the equations simultaneously:

\(x^2 - 4x + 7 = 9 - 3x\)

\(x^2 - x - 2 = 0\)

Solving this quadratic equation gives \(x = 2\) or \(x = -1\).

For \(x = 2\), \(y = 3\). For \(x = -1\), \(y = 12\).

Thus, \(A = (2, 3)\) and \(B = (-1, 12)\).

The midpoint \(M\) is:

\(M = \left( \frac{2 + (-1)}{2}, \frac{3 + 12}{2} \right) = \left( \frac{1}{2}, \frac{15}{2} \right)\)

(ii) The derivative of the curve is:

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

Set \(\frac{dy}{dx} = -3\) (slope of the line):

\(2x - 4 = -3\)

\(2x = 1\)

\(x = \frac{1}{2}\)

Substitute \(x = \frac{1}{2}\) into the curve equation:

\(y = \left( \frac{1}{2} \right)^2 - 4 \left( \frac{1}{2} \right) + 7 = \frac{1}{4} - 2 + 7 = \frac{5}{4}\)

So, \(Q = \left( \frac{1}{2}, \frac{5}{4} \right)\).

(iii) The distance \(MQ\) is:

\(MQ = \sqrt{\left( \frac{1}{2} - \frac{1}{2} \right)^2 + \left( \frac{7}{2} - \frac{5}{4} \right)^2}\)

\(MQ = \sqrt{0 + \left( \frac{14}{4} - \frac{5}{4} \right)^2}\)

\(MQ = \sqrt{\left( \frac{9}{4} \right)^2} = \frac{9}{4}\)

(i) Finding the coordinates of the intersection points:

Substitute \(y = 9 - \frac{6}{x}\) into \(y + x = 8\):

\(9 - \frac{6}{x} + x = 8\)

Simplify to find:

\(x^2 + x - 6 = 0\)

Solving the quadratic equation:

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

Thus, \(x = 2\) or \(x = -3\).

For \(x = 2\), \(y = 9 - \frac{6}{2} = 6\).

For \(x = -3\), \(y = 9 - \frac{6}{-3} = 11\).

So, the points are \((2, 6)\) and \((-3, 11)\).

(ii) Finding the equation of the perpendicular bisector:

Midpoint of \((2, 6)\) and \((-3, 11)\):

\(\left( \frac{2 + (-3)}{2}, \frac{6 + 11}{2} \right) = \left( -\frac{1}{2}, \frac{17}{2} \right)\)

Gradient of the line joining the points:

\(m = \frac{11 - 6}{-3 - 2} = -1\)

Gradient of the perpendicular bisector:

\(m = 1\)

Equation of the perpendicular bisector:

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

Simplify to:

\(y = x + 9\)

(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: \(y - 12\frac{1}{2} = x - 2\frac{1}{2}\)

Part (i):

To find the gradient of line \(AB\), use the formula:

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

Substitute the coordinates of \(A(3a, -a)\) and \(B(-a, 2a)\):

\(m = \frac{2a - (-a)}{-a - 3a} = \frac{3a}{-4a} = -\frac{3}{4}\)

The equation of the line through the origin parallel to \(AB\) is:

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

Part (ii):

Use the distance formula to find the length of \(AB\):

\(AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)

Substitute the coordinates:

\(AB = \sqrt{(-a - 3a)^2 + (2a - (-a))^2} = \sqrt{(-4a)^2 + (3a)^2}\)

\(AB = \sqrt{16a^2 + 9a^2} = \sqrt{25a^2} = 5a\)

Given \(AB = 3\frac{1}{3} = \frac{10}{3}\), set up the equation:

\(5a = \frac{10}{3}\)

Solve for \(a\):

\(a = \frac{10}{3} \times \frac{1}{5} = \frac{2}{3}\)