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Let's solve this problem step by step.

1. **Given Information:**
   - The equation of line \( m \) is \( y = 3x \), meaning the slope of line \( m \) is 3.
   - Point \( A(1,3) \) lies on line \( m \), and the line \( p \) is perpendicular to line \( m \) at point \( A \).
   - We need to determine which of the given points is on line \( p \).

2. **Slope of Line \( p \):**
   - Since line \( p \) is perpendicular to line \( m \), its slope is the negative reciprocal of the slope of line \( m \). The slope of line \( m \) is 3, so the slope of line \( p \) is:
   \[
   \text{slope of } p = -\frac{1}{3}
   \]

3. **Equation of Line \( p \):**
   - The equation of line \( p \) will have the form \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) = (1, 3) \) is point \( A \), and \( m = -\frac{1}{3} \).
   - Substituting into the point-slope form of a line equation:
   \[
   y - 3 = -\frac{1}{3}(x - 1)
   \]
   \[
   y - 3 = -\frac{1}{3}x + \frac{1}{3}
   \]
   \[
   y = -\frac{1}{3}x + \frac{1}{3} + 3
   \]
   \[
   y = -\frac{1}{3}x + \frac{10}{3}
   \]
   So the equation of line \( p \) is:
   \[
   y = -\frac{1}{3}x + \frac{10}{3}
   \]

4. **Testing the Given Points:**
   We will now check which of the points satisfies the equation \( y = -\frac{1}{3}x + \frac{10}{3} \).

   - **Point A: (3, 9)**:
     Substituting \( x = 3 \) into the equation of line \( p \):
     \[
     y = -\frac{1}{3}(3) + \frac{10}{3} = -1 + \frac{10}{3} = \frac{7}{3} \neq 9
     \]
     So, (3, 9) is not on line \( p \).

   - **Point B: \( \left(5, \frac{5}{3}\right) \)**:
     Substituting \( x = 5 \) into the equation of line \( p \):
     \[
     y = -\frac{1}{3}(5) + \frac{10}{3} = -\frac{5}{3} + \frac{10}{3} = \frac{5}{3}
     \]
     This matches the \( y \)-coordinate of point B. Therefore, point \( \left(5, \frac{5}{3}\right) \) is on line \( p \).

   - **Point C: (3, 0)**:
     Substituting \( x = 3 \) into the equation of line \( p \):
     \[
     y = -\frac{1}{3}(3) + \frac{10}{3} = \frac{7}{3} \neq 0
     \]
     So, (3, 0) is not on line \( p \).

   - **Point D: (0, 0)**:
     Substituting \( x = 0 \) into the equation of line \( p \):
     \[
     y = -\frac{1}{3}(0) + \frac{10}{3} = \frac{10}{3} \neq 0
     \]
     So, (0, 0) is not on line \( p \).

5. **Conclusion:**
   The correct answer is **B: \( \left(5, \frac{5}{3}\right) \)**.

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

1. How do you find the equation of a line perpendicular to another line?
2. What is the significance of the point-slope form of a linear equation?
3. Can we solve this problem graphically? How would you plot these lines?
4. How do you check if a point lies on a given line?
5. What are the different forms of the equation of a line?

**Tip:** Remember, perpendicular lines have slopes that are negative reciprocals of each other!

The straight line m has an equation y = 3x. The point A of coordinates (1,3) is on m. The line p is perpendicular to m at point A. Which of the following points is on p?

Explore how to find points on a line perpendicular to another line using the example of line m with equation y = 3x and line p passing through point (1,3). Understand the use of slopes, point-slope form, and coordinate geometry to solve this high school level problem.

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