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 \(\left( \frac{8}{5}, \frac{6}{5} \right)\).