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Let's solve the given problem step by step:

**Given:**
- Point \( A \) has coordinates \( (p, 1) \).
- Point \( B \) has coordinates \( (9, 3p + 1) \).

### Part (i)
We are asked to find the possible values of \( p \) for the case where the distance \( AB \) is 13 units.

1. **Calculate the distance between two points:**
   The formula for the distance between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is:
   \[
   \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
   \]
   
   Applying this to points \( A \) and \( B \):
   \[
   \sqrt{(9 - p)^2 + ((3p + 1) - 1)^2} = 13
   \]
   Simplifying the equation:
   \[
   \sqrt{(9 - p)^2 + (3p)^2} = 13
   \]
   
2. **Square both sides to eliminate the square root:**
   \[
   (9 - p)^2 + (3p)^2 = 169
   \]
   Expand both squares:
   \[
   (9 - p)^2 = 81 - 18p + p^2
   \]
   \[
   (3p)^2 = 9p^2
   \]
   Combining these:
   \[
   81 - 18p + p^2 + 9p^2 = 169
   \]
   \[
   10p^2 - 18p + 81 = 169
   \]
   \[
   10p^2 - 18p - 88 = 0
   \]
   
3. **Solve the quadratic equation:**
   Use the quadratic formula \( p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 10 \), \( b = -18 \), and \( c = -88 \):
   \[
   p = \frac{-(-18) \pm \sqrt{(-18)^2 - 4(10)(-88)}}{2(10)}
   \]
   \[
   p = \frac{18 \pm \sqrt{324 + 3520}}{20}
   \]
   \[
   p = \frac{18 \pm \sqrt{3844}}{20}
   \]
   \[
   p = \frac{18 \pm 62}{20}
   \]
   Thus, the two solutions for \( p \) are:
   \[
   p = \frac{18 + 62}{20} = 4 \quad \text{and} \quad p = \frac{18 - 62}{20} = -2.2
   \]
   
### Part (ii)
We need to find the value of \( p \) for which the line \( 2x + 3y = 9 \) is perpendicular to the line segment \( AB \).

1. **Find the slope of the line \( 2x + 3y = 9 \):**
   Rewrite the equation in the slope-intercept form \( y = mx + c \):
   \[
   3y = -2x + 9
   \]
   \[
   y = -\frac{2}{3}x + 3
   \]
   The slope of this line is \( -\frac{2}{3} \).

2. **Find the slope of the line segment \( AB \):**
   The slope of a line between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:
   \[
   \text{Slope of } AB = \frac{(3p + 1) - 1}{9 - p} = \frac{3p}{9 - p}
   \]

3. **Set up the perpendicularity condition:**
   Two lines are perpendicular if the product of their slopes is \( -1 \):
   \[
   \left(\frac{3p}{9 - p}\right) \left(-\frac{2}{3}\right) = -1
   \]
   Simplify:
   \[
   -\frac{6p}{9 - p} = -1
   \]
   Cancel the negatives and solve:
   \[
   \frac{6p}{9 - p} = 1
   \]
   \[
   6p = 9 - p
   \]
   \[
   7p = 9
   \]
   \[
   p = \frac{9}{7}
   \]

### Summary:
1. For part (i), the possible values of \( p \) are \( 4 \) and \( -2.2 \).
2. For part (ii), the value of \( p \) is \( \frac{9}{7} \).

Would you like further details on any part of the solution? Here are some related questions to expand upon this topic:

1. How would the solution change if the distance between points \( A \) and \( B \) was different?
2. What is the geometric interpretation of having multiple solutions for \( p \)?
3. How can we find the equation of the line passing through points \( A \) and \( B \)?
4. What if the line equation in part (ii) was parallel instead of perpendicular to \( AB \)?
5. Can you derive a general formula for finding a point on a line given its distance from another point?

**Tip:** Always double-check your algebraic manipulations when dealing with quadratic equations to avoid errors.

The point A has coordinates (p, 1) and the point B has coordinates (9, 3p + 1), where p is a constant.
(i) For the case where the distance AB is 13 units, find the possible values of p.
(ii) For the case in which the line with equation 2x + 3y = 9 is perpendicular to AB, find the value of p.

Solve for the values of p in a geometry problem involving distance between two points and perpendicularity conditions. This problem uses the distance formula and the slope condition for perpendicular lines. Suitable for high school students in Grades 10-12.

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