A man strats from the point `P(-3,4)` and reaches the point `Q(0,1)` touching the x-axis at `R(alpha,0)` such that `PR+RQ` is minimum. Then `|alpha|=`.

Explanation

To solve the problem, we need to find the value of \( | \alpha | \) where a man travels from point \( P(-3, 4) \) to point \( Q(0, 1) \) touching the x-axis at point \( R(\alpha, 0) \) such that the total distance \( PR + RQ \) is minimized.

### Step-by-step Solution:

1. **Identify Points**:
- Point \( P \) is at \( (-3, 4) \).
- Point \( Q \) is at \( (0, 1) \).
- Point \( R \) is at \( (\alpha, 0) \) on the x-axis.

2. **Set Up Distances**:
- The distance \( PR \) from \( P \) to \( R \) can be calculated using the distance formula:
\[ PR = \sqrt{(\alpha - (-3))^2 + (0 - 4)^2} = \sqrt{(\alpha + 3)^2 + 16} \]
- The distance \( RQ \) from \( R \) to \( Q \) is:
\[ RQ = \sqrt{(0 - \alpha)^2 + (1 - 0)^2} = \sqrt{\alpha^2 + 1} \]

3. **Total Distance**:
- The total distance \( D \) is:
\[ D = PR + RQ = \sqrt{(\alpha + 3)^2 + 16} + \sqrt{\alpha^2 + 1} \]

4. **Minimizing the Distance**:
- To minimize \( D \), we can use the concept of reflection. Reflect point \( Q \) across the x-axis to get point \( Q'(0, -1) \).
- The straight line distance \( PQ' \) will be the shortest path from \( P \) to \( Q' \) that touches the x-axis.

5. **Finding the Intersection**:
- The slope of line \( PQ' \) is:
\[ \text{slope} = \frac{-1 - 4}{0 - (-3)} = \frac{-5}{3} \]
- The equation of line \( PQ' \) can be written as:
\[ y - 4 = \frac{-5}{3}(x + 3) \]
- Simplifying, we get:
\[ y = -\frac{5}{3}x - 5 + 4 = -\frac{5}{3}x - 1 \]

6. **Finding \( \alpha \)**:
- To find where this line intersects the x-axis (where \( y = 0 \)):
\[ 0 = -\frac{5}{3}x - 1 \]
\[ \frac{5}{3}x = -1 \implies x = -\frac{3}{5} \]
- Thus, \( \alpha = -\frac{3}{5} \).

7. **Finding \( | \alpha | \)**:
- The absolute value of \( \alpha \) is:
\[ | \alpha | = \left| -\frac{3}{5} \right| = \frac{3}{5} \]

### Final Answer:
\[ | \alpha | = \frac{3}{5} \]