Two point P, Q are taken at random on a straight line OA of length a. The chance that `PQgtb`, where `blta` is

Options

A. `(|a-b|)/a`
B. `(|b-a|)/a`
C. `((a-b)^2)/(a^2)`
D. `((b-a)^2)/(b^2)`

Answer

To solve the problem step by step, we will analyze the situation of selecting two random points \( P \) and \( Q \) on a straight line segment \( OA \) of length \( a \) and find the probability that the distance \( PQ \) is greater than \( b \), where \( b < a \).

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a line segment \( OA \) of length \( a \).
- We need to find the probability that the distance between two randomly chosen points \( P \) and \( Q \) on this line segment is greater than \( b \).

2. **Defining the Points**:
- Let \( P \) be at position \( x \) and \( Q \) be at position \( y \) on the line segment \( OA \).
- The coordinates of \( P \) and \( Q \) can be represented as \( P(x) \) and \( Q(y) \), where \( 0 \leq x, y \leq a \).

3. **Distance Between Points**:
- The distance \( PQ \) can be expressed as \( |x - y| \).
- We want to find the condition where \( |x - y| > b \).

4. **Setting Up the Inequality**:
- The condition \( |x - y| > b \) means that either \( x - y > b \) or \( y - x > b \).
- This can be rewritten as two inequalities:
- \( x > y + b \)
- \( y > x + b \)

5. **Visualizing the Problem**:
- We can visualize the problem in a coordinate plane where the x-axis represents the position of point \( P \) and the y-axis represents the position of point \( Q \).
- The area of interest is the square defined by \( 0 \leq x \leq a \) and \( 0 \leq y \leq a \), which has an area of \( a^2 \).

6. **Finding the Area Where \( |x - y| > b \)**:
- The lines \( y = x + b \) and \( y = x - b \) divide the square into regions.
- The area where \( |x - y| \leq b \) is a band of width \( 2b \) around the line \( y = x \).
- The area of this band can be calculated as the area of the square minus the area of the two triangles formed outside this band.

7. **Calculating the Area of the Band**:
- The area of the band is given by:
\[ \text{Area of band} = a^2 - \text{Area of two triangles} \]
- Each triangle has a base and height of \( a - b \), so the area of one triangle is \( \frac{1}{2}(a-b)(a-b) = \frac{(a-b)^2}{2} \).
- Therefore, the total area of the two triangles is \( (a-b)^2 \).

8. **Final Area Calculation**:
- The area where \( |x - y| > b \) is:
\[ \text{Area} = a^2 - (a-b)^2 = a^2 - (a^2 - 2ab + b^2) = 2ab - b^2 \]

9. **Calculating the Probability**:
- The probability \( P \) that \( |x - y| > b \) is given by the ratio of the area where \( |x - y| > b \) to the total area of the square:
\[ P = \frac{2ab - b^2}{a^2} \]

### Final Result:
The probability that the distance \( PQ \) is greater than \( b \) is:
\[ P = \frac{(a-b)^2}{a^2} \]