To solve the problem, we need to find the area of the region bounded by the locus of points \( P \) that are equidistant from the points \( O(0,0) \) and \( A(3,2) \) using the defined distance \( d(P,Q) = |x_1 - x_2| + |y_1 - y_2| \), and then find the area of the region bounded by this locus and the line \( y = 4 \) in the first quadrant.
### Step 1: Find the locus of points P equidistant from O and A
The distance from point \( P(x_1, y_1) \) to \( O(0,0) \) is:
\[
d(P,O) = |x_1 - 0| + |y_1 - 0| = |x_1| + |y_1| = x_1 + y_1 \quad \text{(since } P \text{ is in the first quadrant)}
\]
The distance from point \( P(x_1, y_1) \) to \( A(3,2) \) is:
\[
d(P,A) = |x_1 - 3| + |y_1 - 2| = |x_1 - 3| + |y_1 - 2|
\]
Setting these two distances equal gives us the equation:
\[
x_1 + y_1 = |x_1 - 3| + |y_1 - 2|
\]
### Step 2: Analyze the absolute value expressions
We need to consider different cases based on the values of \( x_1 \) and \( y_1 \):
**Case 1:** \( x_1 \geq 3 \) and \( y_1 \geq 2 \)
\[
x_1 + y_1 = (x_1 - 3) + (y_1 - 2)
\]
This simplifies to:
\[
x_1 + y_1 = x_1 + y_1 - 5 \implies 5 = 0 \quad \text{(not possible)}
\]
**Case 2:** \( x_1 < 3 \) and \( y_1 \geq 2 \)
\[
x_1 + y_1 = (3 - x_1) + (y_1 - 2)
\]
This simplifies to:
\[
x_1 + y_1 = 3 - x_1 + y_1 - 2 \implies 2x_1 = 1 \implies x_1 = \frac{1}{2}
\]
Thus, \( y_1 \geq 2 \).
**Case 3:** \( x_1 \geq 3 \) and \( y_1 < 2 \)
\[
x_1 + y_1 = (x_1 - 3) + (2 - y_1)
\]
This simplifies to:
\[
x_1 + y_1 = x_1 - 3 + 2 - y_1 \implies 2y_1 = -1 \quad \text{(not possible)}
\]
**Case 4:** \( x_1 < 3 \) and \( y_1 < 2 \)
\[
x_1 + y_1 = (3 - x_1) + (2 - y_1)
\]
This simplifies to:
\[
x_1 + y_1 = 5 - x_1 - y_1 \implies 2x_1 + 2y_1 = 5 \implies x_1 + y_1 = \frac{5}{2}
\]
### Step 3: Determine the locus
From the cases, we find that the locus of points \( P \) is given by:
1. \( x_1 + y_1 = \frac{5}{2} \) for \( x_1 < 3 \) and \( y_1 < 2 \)
2. \( x_1 = \frac{1}{2} \) for \( y_1 \geq 2 \)
### Step 4: Find the area bounded by the locus and the line \( y = 4 \)
The line \( x_1 + y_1 = \frac{5}{2} \) intersects the line \( y = 4 \) at \( x_1 + 4 = \frac{5}{2} \) which is not in the first quadrant.
The area bounded by the line \( y = 4 \) and the locus in the first quadrant consists of a rectangle and a trapezium.
1. **Rectangle**: The rectangle formed by the points \( (0,2) \), \( (0,4) \), \( (3,4) \), and \( (3,2) \).
- Length = 3 (from \( x = 0 \) to \( x = 3 \))
- Height = 2 (from \( y = 2 \) to \( y = 4 \))
- Area of rectangle = \( 3 \times 2 = 6 \)
2. **Trapezium**: The trapezium formed by the points \( (0,2) \), \( (0,4) \), \( (3,4) \), and \( (3,2) \).
- Area of trapezium = \( \frac{1}{2} \times (b_1 + b_2) \times h \)
- Base 1 \( b_1 = 3 \), Base 2 \( b_2 = 0 \), Height \( h = 2 \)
- Area of trapezium = \( \frac{1}{2} \times (3 + 0) \times 2 = 3 \)
### Step 5: Total Area
Total area = Area of rectangle + Area of trapezium = \( 6 + 3 = 9 \) square units.
### Final Answer
The area of the region bounded by the locus of \( P \) and the line \( y = 4 \) in the first quadrant is \( 9 \) square units.
