To solve the problem, we need to find the range of values of the slope \( m \) for the line passing through the point \( (2, 4) \) such that the line segment \( AB \) subtends a right angle at points \( P \) and \( Q \) on that line, where \( A(0, 0) \) and \( B(6, 0) \).
### Step-by-Step Solution:
1. **Identify the Points and Circle**:
- The points \( A(0, 0) \) and \( B(6, 0) \) form a line segment \( AB \).
- The midpoint \( M \) of \( AB \) is \( (3, 0) \).
- The radius \( r \) of the circle with diameter \( AB \) is half the distance between \( A \) and \( B \), which is \( 3 \).
2. **Equation of the Circle**:
- The equation of the circle with center \( (3, 0) \) and radius \( 3 \) is:
\[
(x - 3)^2 + y^2 = 3^2
\]
Simplifying this gives:
\[
(x - 3)^2 + y^2 = 9
\]
Expanding it results in:
\[
x^2 - 6x + 9 + y^2 = 9 \implies x^2 + y^2 - 6x = 0
\]
3. **Equation of the Line through (2, 4)**:
- The line passing through point \( (2, 4) \) with slope \( m \) can be expressed as:
\[
y - 4 = m(x - 2)
\]
Rearranging gives:
\[
y = mx - 2m + 4
\]
4. **Finding Points of Intersection**:
- To find the points \( P \) and \( Q \), we substitute the equation of the line into the equation of the circle:
\[
x^2 + (mx - 2m + 4)^2 - 6x = 0
\]
- Expanding this leads to a quadratic equation in \( x \).
5. **Condition for Right Angle**:
- For \( P \) and \( Q \) to subtend a right angle at \( A \) and \( B \), the distance \( PQ \) must satisfy the condition:
\[
|PQ|^2 = |AB|^2
\]
- The distance \( |AB| = 6 \), hence \( |PQ|^2 = 36 \).
6. **Quadratic Equation**:
- The quadratic equation derived from the intersection will be of the form:
\[
Ax^2 + Bx + C = 0
\]
- The discriminant \( D \) of this equation must be non-negative for real intersections:
\[
D = B^2 - 4AC \geq 0
\]
7. **Finding the Range of \( m \)**:
- After simplifying the conditions, we arrive at a quadratic inequality in terms of \( m \):
\[
8m^2 - 8m - 7 > 0
\]
- Solving this quadratic inequality gives us the range of values for \( m \).
8. **Final Range**:
- The roots of the equation \( 8m^2 - 8m - 7 = 0 \) can be found using the quadratic formula:
\[
m = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \cdot 8 \cdot (-7)}}{2 \cdot 8}
\]
- This simplifies to:
\[
m = \frac{8 \pm \sqrt{64 + 224}}{16} = \frac{8 \pm 16}{16}
\]
- Thus, the roots are:
\[
m = \frac{24}{16} = \frac{3}{2} \quad \text{and} \quad m = \frac{-8}{16} = -\frac{1}{2}
\]
- The range of \( m \) is:
\[
m \in (-\infty, -\frac{1}{2}) \cup (\frac{3}{2}, \infty)
\]
