For points `P-=(x_1,y_1)` and `Q-=(x_2,y_2)` of the coordinates plane, a new distance d (P,Q) is defined by `d(P,Q) =|x_1-x_2|+|y_1-y_2|`. Let `O-=(0,0)` and `A-=(3,2)`. Consider the set of points P in the first quadrant which are equidistant (with respect to the new distance) from O and A.
The set of poitns P consists of

Options

A. one straight line only
B. union of two line segments
C. union of two infinite rays
D. union of a line segment of finite length and an infinite ray

Explanation

To solve the problem, we need to find the set of points \( P(x, y) \) in the first quadrant that are equidistant from the points \( O(0, 0) \) and \( A(3, 2) \) using the defined distance \( d(P, Q) = |x_1 - x_2| + |y_1 - y_2| \).

### Step-by-Step Solution

1. **Define the distances**:
- The distance from \( O(0, 0) \) to \( P(x, y) \) is:
\[ d(O, P) = |x - 0| + |y - 0| = x + y \]
- The distance from \( A(3, 2) \) to \( P(x, y) \) is:
\[ d(A, P) = |x - 3| + |y - 2| \]

2. **Set the distances equal**:
Since \( P \) is equidistant from \( O \) and \( A \), we have:
\[ x + y = |x - 3| + |y - 2| \]

3. **Consider cases based on the absolute values**:
We need to analyze the equation based on the values of \( x \) and \( y \).

**Case 1**: \( x \geq 3 \)

Here, \( |x - 3| = x - 3 \) and we need to consider two sub-cases for \( y \):

- **Sub-case 1.1**: \( y \geq 2 \)
\[ x + y = (x - 3) + (y - 2) \]
Simplifying gives:
\[ x + y = x + y - 5 \implies 5 = 0 \quad \text{(no solution)} \]

- **Sub-case 1.2**: \( y < 2 \)
\[ x + y = (x - 3) + (2 - y) \]
Simplifying gives:
\[ x + y = x - 3 + 2 - y \implies 2y = -1 \implies y = -\frac{1}{2} \quad \text{(not valid in the first quadrant)} \]

Thus, there are no solutions for \( x \geq 3 \).

**Case 2**: \( x < 3 \)

Here, \( |x - 3| = 3 - x \). Again, we consider two sub-cases for \( y \):

- **Sub-case 2.1**: \( y \geq 2 \)
\[ x + y = (3 - x) + (y - 2) \]
Simplifying gives:
\[ x + y = 3 - x + y - 2 \implies 2x = 1 \implies x = \frac{1}{2} \]
Since \( y \geq 2 \), this gives us a point \( P\left(\frac{1}{2}, y\right) \) for \( y \geq 2 \), which represents an infinite ray starting from \( \left(\frac{1}{2}, 2\right) \).

- **Sub-case 2.2**: \( y < 2 \)
\[ x + y = (3 - x) + (2 - y) \]
Simplifying gives:
\[ x + y = 5 - x - y \implies 2x + 2y = 5 \implies x + y = \frac{5}{2} \]
This represents a line segment where \( x < 3 \) and \( y < 2 \).

4. **Conclusion**:
The set of points \( P \) consists of:
- An infinite ray starting from \( \left(\frac{1}{2}, 2\right) \) extending upwards.
- A line segment defined by \( x + y = \frac{5}{2} \) for \( x < 3 \) and \( y < 2 \).

Thus, the final answer is that the set of points \( P \) consists of a union of a line segment of finite length and an infinite ray.