(i) To find the value of h, we use the fact that ∠ABC = 90° and lines BC and AD are parallel. The gradient of AB is \(-\frac{1}{2}\) and the gradient of BC is \(\frac{3h - 2}{h}\). Since BC is perpendicular to AB, the product of their gradients is \(-1\):

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

Solving for h, we get:

\(\frac{3h - 2}{h} = 2\)

\(3h - 2 = 2h\)

\(h = 2\)

(ii) With h = 2, the coordinates of C are (2, 6). Since CD is parallel to the x-axis, the y-coordinate of D is the same as C, which is 6. The x-coordinate of D can be found using the fact that AD is parallel to BC, so the gradient of AD is also 2:

\(\frac{6 - 0}{x - 4} = 2\)

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

\(6 = 2x - 8\)

\(2x = 14\)

\(x = 7\)

Thus, the coordinates of D are (7, 6).

(i) To find the equation of the perpendicular bisector of AC:

The midpoint of AC is \(\left( \frac{-1 + 5}{2}, \frac{2 + 4}{2} \right) = (2, 3)\).

The gradient of AC is \(\frac{4 - 2}{5 + 1} = \frac{1}{3}\).

The gradient of the perpendicular bisector is the negative reciprocal, \(-3\).

Using the point-slope form, the equation is \(y - 3 = -3(x - 2)\).

(ii) To find the coordinates of B and D:

Since B lies on the y-axis, its x-coordinate is 0. Using the perpendicular bisector equation, \(y - 3 = -3(0 - 2)\), we find \(y = 9\). Thus, \(B(0, 9)\).

To find D, use the vector move from A to C, which is \((5 - (-1), 4 - 2) = (6, 2)\).

Since D is opposite B, apply the vector move to B: \((0 + 4, 9 - 12) = (4, -3)\). Thus, \(D(4, -3)\).

(iii) To find the area of the rhombus:

The length of AC is \(\sqrt{(5 - (-1))^2 + (4 - 2)^2} = \sqrt{40}\).

The length of BD is \(\sqrt{(4 - 0)^2 + (-3 - 9)^2} = \sqrt{160}\).

The area of the rhombus is \(\frac{1}{2} \times \sqrt{40} \times \sqrt{160} = 40\).

The gradient of OB is calculated as follows:

Gradient of OB = \(\frac{1 - 0}{3 - 0} = \frac{1}{3}\).

Since L₁ is parallel to OB, it has the same gradient: \(\frac{1}{3}\).

The equation of L₁ passing through A (-1, 3) is:

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

The gradient of AB is:

Gradient of AB = \(\frac{1 - 3}{3 + 1} = -\frac{1}{2}\).

The gradient of L₂, being perpendicular to AB, is the negative reciprocal:

\(Gradient of L₂ = 2.\)

The equation of L₂ passing through B (3, 1) is:

\(y - 1 = 2(x - 3)\).

Solving the simultaneous equations:

1. \(y = \frac{1}{3}x + \frac{10}{3}\)

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

Equating the two expressions for \(y\):

\(\frac{1}{3}x + \frac{10}{3} = 2x - 5\)

Multiply through by 3 to clear the fraction:

\(x + 10 = 6x - 15\)

Rearrange to solve for \(x\):

\(5x = 25\)

\(x = 5\)

Substitute \(x = 5\) back into one of the equations to find \(y\):

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

Thus, the coordinates of C are (5, 5).

(i) The gradient of AB is calculated as:

\(\frac{22 - (-2)}{15 - 3} = \frac{24}{12} = 2\)

Since the gradient of AB is given as 2m, equating gives:

\(2m = 2\)

Thus, \(m = 1\).

(ii) The equation of line AC is:

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

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

\(y = -2x + 4\)

The equation of line BC is:

\(y - 22 = 1(x - 15)\)

\(y = x + 7\)

Solving the simultaneous equations:

\(y = -2x + 4\)

\(y = x + 7\)

Equating gives:

\(-2x + 4 = x + 7\)

\(-3x = 3\)

\(x = -1\)

Substitute \(x = -1\) into \(y = x + 7\):

\(y = -1 + 7 = 6\)

Thus, \(C = (-1, 6)\).

(iii) The midpoint M of AB is:

\(M = \left( \frac{3 + 15}{2}, \frac{-2 + 22}{2} \right) = (9, 10)\)

The gradient of the perpendicular bisector is:

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

The equation of the perpendicular bisector is:

\(y - 10 = -\frac{1}{2}(x - 9)\)

\(2y - 20 = -x + 9\)

\(2y + x = 29\)

Solving the equations \(y = x + 7\) and \(2y + x = 29\):

Substitute \(y = x + 7\) into \(2y + x = 29\):

\(2(x + 7) + x = 29\)

\(2x + 14 + x = 29\)

\(3x = 15\)

\(x = 5\)

Substitute \(x = 5\) into \(y = x + 7\):

\(y = 5 + 7 = 12\)

Thus, \(D = (5, 12)\).

(i) The y-coordinate of D is the same as the y-coordinate of the midpoint of AC. The midpoint of AC is \(\left( \frac{0+12}{2}, \frac{-2+14}{2} \right) = (6, 6)\). Therefore, the y-coordinate of D is 6.

(ii) The gradient of AD is calculated as \(\frac{y_2 - y_1}{x_2 - x_1} = \frac{6 - (-2)}{h - 0} = \frac{8}{h}\) or \(\frac{h-12}{8}\). The gradient of CD is \(\frac{6 - 14}{h - 12} = \frac{-8}{h-12}\) or \(\frac{-h}{8}\).

(iii) Since BD is parallel to the x-axis, the product of the gradients of AD and CD is -1. Solving \(\frac{8}{h} \times \frac{-8}{h-12} = -1\) gives \(h^2 - 12h - 64 = 0\). Solving this quadratic equation, we find \(h = 16\) or \(h = -4\). Thus, \(x_D = 16\) and \(x_B = -4\).

(iv) The area of rectangle ABCD can be calculated using the lengths of the sides. The length of AC is \(\sqrt{(12-0)^2 + (14-(-2))^2} = 20\). The length of BD is \(\sqrt{(16-(-4))^2 + (6-6)^2} = 20\). The area is \(20 \times 8 = 160\).

The line AC has the equation \(2y = x + 4\), which simplifies to \(y = \frac{1}{2}x + 2\). The gradient of AC is \(\frac{1}{2}\).

The line BD is perpendicular to AC, so its gradient is the negative reciprocal of \(\frac{1}{2}\), which is \(-2\).

The equation of line BD passing through \(D(10, -3)\) with gradient \(-2\) is:

\(y + 3 = -2(x - 10)\)

Simplifying gives:

\(y + 3 = -2x + 20\)

\(y = -2x + 17\)

Equating the equations of lines AC and BD:

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

Solving for \(x\):

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

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

\(x = 6\)

Substitute \(x = 6\) back into \(y = \frac{1}{2}x + 2\):

\(y = \frac{1}{2}(6) + 2 = 3 + 2 = 5\)

Thus, B is \((6, 5)\).

Since \(AB = BC\), the vector from A to B is the same as from B to C. If A is \((0, 2)\), then the vector AB is \((6 - 0, 5 - 2) = (6, 3)\).

Adding this vector to B gives C:

C = \((6 + 6, 5 + 3) = (12, 8)\).

(i) The gradient of AC is \(-\frac{1}{2}\). The perpendicular gradient is 2. The equation of line BX is \(y - 2 = 2(x - 2)\), which simplifies to \(y = 2x - 2\). Solving the equations \(2y + x = 16\) and \(y = 2x - 2\) simultaneously gives the coordinates of X as \((4, 6)\).

(ii) Since X is the midpoint of BD, and using the coordinates of X \((4, 6)\), we find D as \((6, 10)\).

(iii) Using the distance formula, \(AB = \sqrt{(14)^2 + (2)^2} = \sqrt{200}\) and \(BC = \sqrt{(2)^2 + (6)^2} = \sqrt{40}\). The perimeter is \(2\sqrt{200} + 2\sqrt{40} = 40.9\).

First, calculate the gradient of line AB:

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

Since DC is parallel to AB, the gradient of DC is also -2. Using point C (10, 2), the equation of line DC is:

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

Simplifying gives:

\(y = -2x + 20 + 2\)

\(y = -2x + 22\)

For line DA, since it is perpendicular to AB, the gradient of DA is the negative reciprocal of -2, which is \(\frac{1}{2}\). Using point A (3, 8), the equation of line DA is:

\(y - 8 = \frac{1}{2}(x - 3)\)

Simplifying gives:

\(y = \frac{1}{2}x + \frac{1}{2} \times (-3) + 8\)

\(y = \frac{1}{2}x - \frac{3}{2} + 8\)

\(y = \frac{1}{2}x + \frac{13}{2}\)

Now, solve the simultaneous equations:

1. \(y = -2x + 22\)

2. \(y = \frac{1}{2}x + \frac{13}{2}\)

Equating the two expressions for \(y\):

\(-2x + 22 = \frac{1}{2}x + \frac{13}{2}\)

Multiply through by 2 to eliminate the fraction:

\(-4x + 44 = x + 13\)

\(44 - 13 = 4x + x\)

\(31 = 5x\)

\(x = \frac{31}{5} = 6.2\)

Substitute \(x = 6.2\) back into equation 1:

\(y = -2(6.2) + 22\)

\(y = -12.4 + 22\)

\(y = 9.6\)

Thus, the coordinates of D are (6.2, 9.6).

(i) To find the equation of BC, first calculate the slope of AB. The slope \(m\) of AB is given by:

\(m = \frac{14 - 8}{2 - (-2)} = \frac{6}{4} = \frac{3}{2}\)

Since BC is perpendicular to AB, the slope of BC is the negative reciprocal of \(\frac{3}{2}\), which is \(-\frac{2}{3}\).

Using point B (-2, 8) and the slope \(-\frac{2}{3}\), the equation of line BC is:

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

Simplifying, we get:

\(3y + 2x = 20\)

(ii) Since C lies on the x-axis, its y-coordinate is 0. Substitute \(y = 0\) into the equation of BC:

\(3(0) + 2x = 20\)

\(2x = 20\)

\(x = 10\)

Thus, the coordinates of C are (10, 0).

To find D, use the vector move from B to C and apply it to A. The vector from B to C is:

\((10 - (-2), 0 - 8) = (12, -8)\)

Apply this vector to A (2, 14):

\((2 + 12, 14 - 8) = (14, 6)\)

Thus, the coordinates of D are (14, 6).

(i) To find the equation of CD, first calculate the slope of AB. The slope \(m\) of a line through points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:

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

For AB, \(m = \frac{11 - 3}{13 - 1} = \frac{8}{12} = \frac{2}{3}\).

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

Using point C (6, 15) and the slope \(-\frac{3}{2}\), the equation of CD is:

\(y - 15 = -\frac{3}{2}(x - 6)\)

(ii) To find the coordinates of D, first find the equation of AB using point A (1, 3):

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

Rearrange to get:

\(3y - 9 = 2x - 2\)

\(3y = 2x + 7\)

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

Now solve the system of equations:

1. \(y - 15 = -\frac{3}{2}(x - 6)\)

2. \(y = \frac{2}{3}x + \frac{7}{3}\)

Substitute equation 2 into equation 1:

\(\frac{2}{3}x + \frac{7}{3} - 15 = -\frac{3}{2}(x - 6)\)

\(\frac{2}{3}x + \frac{7}{3} - 15 = -\frac{3}{2}x + 9\)

Clear fractions by multiplying through by 6:

\(4x + 14 - 90 = -9x + 54\)

\(4x - 76 = -9x + 54\)

\(13x = 130\)

\(x = 10\)

Substitute \(x = 10\) back into equation 2:

\(y = \frac{2}{3}(10) + \frac{7}{3} = \frac{20}{3} + \frac{7}{3} = 9\)

Thus, the coordinates of D are (10, 9).

First, find the midpoint M of BD. The formula for the midpoint is:

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

Substitute the coordinates of B (2, 10) and D (6, 2):

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

Next, find the gradient of BD:

\(m_{BD} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 10}{6 - 2} = -2\)

Since ABCD is a rhombus, the diagonals are perpendicular. Therefore, the gradient of AC is the negative reciprocal of \(m_{BD}\):

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

The equation of line AC passing through M (4, 6) is:

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

Since A lies on the x-axis, \(y = 0\):

\(0 - 6 = \frac{1}{2}(x - 4)\)

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

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

\(x = -8\)

Thus, A is (-8, 0).

To find C, use the vector method. The vector from A to M is:

\(\overrightarrow{AM} = (4 - (-8), 6 - 0) = (12, 6)\)

Since M is the midpoint, C is:

\(C = M + \overrightarrow{AM} = (4, 6) + (12, 6) = (16, 12)\)

(i) To find the equation of AC, use the points A(0, 4) and C(8, 0). The gradient \(m\) is given by \(m = \frac{0 - 4}{8 - 0} = -\frac{1}{2}\). The equation of the line is \(y = mx + c\). Substituting the point A(0, 4), we get \(4 = -\frac{1}{2}(0) + c\), so \(c = 4\). Therefore, the equation of AC is \(y = -\frac{1}{2}x + 4\).

For OB, since AC is the line of symmetry, OB is perpendicular to AC. The gradient of OB is the negative reciprocal of \(-\frac{1}{2}\), which is 2. Therefore, the equation of OB is \(y = 2x\).

(ii) To find the coordinates of B, use the midpoint formula for OB. Since O is the origin (0, 0), the midpoint of OB is \(\left( \frac{0 + x}{2}, \frac{0 + y}{2} \right)\). Given that the midpoint is on AC, substitute into the equation \(y = -\frac{1}{2}x + 4\). Solving the simultaneous equations \(y = 2x\) and \(y = -\frac{1}{2}x + 4\), we find \(x = 3.2\) and \(y = 6.4\). Therefore, the coordinates of B are (3.2, 6.4).

(i) To find the equation of BC, calculate the slope using points B(4, 6) and C(x, y):

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

The equation of line BC is:

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

For CD, since BC is perpendicular to CD, \(m_{BC} \times m_{CD} = -1\).

\(m_{CD} = -2\)

Using point D(12, 5), the equation of line CD is:

\(y - 5 = -2(x - 12)\)

(ii) To find the coordinates of C, solve the simultaneous equations:

\(2y = x + 8\)

\(y + 2x = 29\)

Substitute \(x = 2y - 8\) into the second equation:

\(y + 2(2y - 8) = 29\)

\(y + 4y - 16 = 29\)

\(5y = 45\)

\(y = 9\)

Substitute \(y = 9\) back into \(x = 2y - 8\):

\(x = 2(9) - 8 = 10\)

Thus, the coordinates of C are (10, 9).

(i) The slope of AB is calculated as \(m = \frac{6 - 2}{1 - 3} = -2\). Since BC is perpendicular to AB, the slope of BC is \(\frac{1}{2}\). Using point B (1, 6), the equation of BC is \(y - 6 = \frac{1}{2}(x - 1)\) or \(2y = x + 11\).

(ii) To find the coordinates of C, solve the equations \(y = x - 1\) and \(y - 6 = \frac{1}{2}(x - 1)\). Substituting \(y = x - 1\) into the second equation gives \(x - 1 - 6 = \frac{1}{2}(x - 1)\). Solving this, we find \(x = 13\) and \(y = 12\), so C is (13, 12).

(iii) The length of AB is \(\sqrt{(3 - 1)^2 + (2 - 6)^2} = \sqrt{20}\). The length of BC is \(\sqrt{(13 - 1)^2 + (12 - 6)^2} = \sqrt{180}\). The perimeter is \(2 \times \sqrt{20} + 2 \times \sqrt{180} = 35.8\) or \(35.7\) or \(16\sqrt{5}\) or \(\sqrt{1280}\).

(i) To find the equation of AD, we first find the slope of AB which is \(m_{AB} = \frac{-3 - 6}{5 - 2} = -3\). Since AD is perpendicular to AB, the slope of AD is \(m_{AD} = \frac{1}{3}\). Using point A (2, 6), the equation of AD is \(y - 6 = \frac{1}{3}(x - 2)\) or \(3y = x + 16\).

(ii) To find the coordinates of D, we use the equation of CD: \(y - 3 = -3(x - 8)\) or \(y = -3x + 27\). Solving the equations of AD and CD simultaneously, we find \(D = (6\frac{1}{2}, 7\frac{1}{2})\).

(iii) To find the length of BE, we use the midpoint formula to find E such that ABCE is a parallelogram. The midpoint of AC is \((5, 4.5)\), so E is \((5, 12)\). The length of BE is \(\sqrt{(5 - 5)^2 + (12 + 3)^2} = 15\).

1. The midpoint of the diagonals is given as \(E \left( 6 \frac{1}{2}, 8 \frac{1}{2} \right)\).

2. Use the midpoint formula for diagonals: \(\frac{x_1 + x_3}{2} = 6.5\) and \(\frac{y_1 + y_3}{2} = 8.5\).

3. Solve the equations for \(A\) and \(C\):

\(y = 3x\) and \(4y = x + 11\).

4. Solving these simultaneously gives \(A(1, 3)\).

5. Use the midpoint formula to find \(C\):

\(x_3 = 2 \times 6.5 - 1 = 12\) and \(y_3 = 2 \times 8.5 - 3 = 14\).

6. Thus, \(C(12, 14)\).

7. Find the equation of \(BC\) using the midpoint and slope:

\(4y = x + 44\).

8. Solve for \(B\) using \(y = 3x\) and \(4y = x + 44\):

\(B(4, 12)\).

9. Use the midpoint formula to find \(D\):

\(x_4 = 2 \times 6.5 - 4 = 9\) and \(y_4 = 2 \times 8.5 - 12 = 5\).

10. Thus, \(D(9, 5)\).

Given the points A (0, 8) and B (4, 0), and the equation of diagonal AC as \(8y + x = 64\).

First, find the slope of AB: \(m_{AB} = \frac{8 - 0}{0 - 4} = -2\).

Since ABCD is a rectangle, BC is perpendicular to AB, so \(m_{BC} \times m_{AB} = -1\).

Thus, \(m_{BC} = \frac{1}{2}\).

The equation of line BC is \(y - 0 = \frac{1}{2}(x - 4)\), which simplifies to \(y = \frac{1}{2}x - 2\).

Now, solve the equations \(8y + x = 64\) and \(y = \frac{1}{2}x - 2\) simultaneously:

Substitute \(y = \frac{1}{2}x - 2\) into \(8y + x = 64\):

\(8(\frac{1}{2}x - 2) + x = 64\)

\(4x - 16 + x = 64\)

\(5x = 80\)

\(x = 16\)

Substitute \(x = 16\) back into \(y = \frac{1}{2}x - 2\):

\(y = \frac{1}{2}(16) - 2 = 8 - 2 = 6\)

Thus, the coordinates of C are (16, 6).

For D, use the vector step method or the fact that AD is parallel to BC:

Since AD is parallel to BC, \(m_{AD} = \frac{1}{2}\).

Using the midpoint M of AC as (8, 7), and knowing D is symmetric to C about M, calculate:

\(D = (2 \times 8 - 16, 2 \times 7 - 6) = (12, 14)\).

Thus, the coordinates of D are (12, 14).

(i) To find the coordinates of \(X\), we first calculate the slope of \(AB\):

\(m_{AB} = \frac{6 - 14}{14 - 2} = -\frac{2}{3}\).

The slope of \(CX\), being perpendicular to \(AB\), is the negative reciprocal:

\(m_{CX} = \frac{3}{2}\).

The equation of line \(AB\) is:

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

The equation of line \(CX\) is:

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

Solving these equations simultaneously gives the coordinates of \(X\):

\(X (11, 8)\).

(ii) To find the ratio \(AX : XB\), calculate the distances:

\(AX = \sqrt{(11 - 2)^2 + (8 - 14)^2} = \sqrt{9^2 + 6^2} = \sqrt{117}\).

\(XB = \sqrt{(14 - 11)^2 + (6 - 8)^2} = \sqrt{3^2 + 2^2} = \sqrt{13}\).

The ratio \(AX : XB = \sqrt{117} : \sqrt{13} = 3:1\).

(i) To find the equation of BC, we first calculate the gradient of AB. The gradient \(m\) is given by \(m = \frac{y_2 - y_1}{x_2 - x_1}\). For AB, \(m = \frac{11 - 3}{5 - 1} = 2\).

Since ABC is a right angle, the gradient of BC is the negative reciprocal of the gradient of AB, so \(m_{BC} = -\frac{1}{2}\).

The equation of a line is \(y - y_1 = m(x - x_1)\). Using point B (5, 11), the equation of BC is \(y - 11 = -\frac{1}{2}(x - 5)\).

(ii) To find the coordinates of C, we first find the equation of AC. The gradient of AC (or AX) is \(\frac{1}{3}\) since X (4, 4) lies on AC.

The equation of AC using point A (1, 3) is \(y - 3 = \frac{1}{3}(x - 1)\).

Alternatively, using point X (4, 4), the equation is \(y - 4 = \frac{1}{3}(x - 4)\).

Solving the simultaneous equations of BC and AC gives the coordinates of C as (13, 7).

(i) To find the coordinates of M, we first find the equations of lines AC and BD. The gradient of AC is \(\frac{4 - (-1)}{9 - (-1)} = \frac{1}{2}\). The equation of line AC is \(y + 1 = \frac{1}{2}(x + 1)\).

The gradient of BD is \(-2\) because it is perpendicular to AC (\(m_1 m_2 = -1\)). The equation of line BD is \(y - 6 = -2(x - 3)\).

Solving these equations simultaneously gives the intersection point M as (5, 2).

\(To find D, use the midpoint formula or vector method. Since BM = MD, D is at (7, -2).\)

(ii) To find the ratio AM : MC, calculate the distances using the distance formula. \(AM = \sqrt{(5 - (-1))^2 + (2 - (-1))^2} = \sqrt{45}\) and \(MC = \sqrt{(9 - 5)^2 + (4 - 2)^2} = \sqrt{20}\).

The ratio is \(\sqrt{45} : \sqrt{20} = 3 : 2\).

(i) To find the coordinates of \(A\) and \(B\), set the equations equal: \(\frac{2}{1-x} = 3x + 4\).

Cross-multiply to get: \(2 = (1-x)(3x + 4)\).

Expand and simplify: \(2 = 3x + 4 - 3x^2 - 4x\).

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

Factorize: \((x+1)(3x-2) = 0\).

Solutions are \(x = -1\) and \(x = \frac{2}{3}\).

Substitute back to find \(y\):

For \(x = -1\), \(y = 3(-1) + 4 = 1\), so \(A = (-1, 1)\).

For \(x = \frac{2}{3}\), \(y = 3\left(\frac{2}{3}\right) + 4 = 6\), so \(B = \left(\frac{2}{3}, 6\right)\).

(ii) To find the length of \(AB\), use the distance formula:

\(AB = \sqrt{\left(\frac{2}{3} + 1\right)^2 + (6 - 1)^2}\).

\(AB = \sqrt{\left(\frac{5}{3}\right)^2 + 5^2}\).

\(AB = \sqrt{\frac{25}{9} + 25} = \sqrt{\frac{250}{9}}\).

\(AB \approx 5.27\).

The mid-point of \(AB\) is:

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