For points `P -= (x_(1) ,y_(1)) and Q = (x_(2),y_(2))` of the coordinate 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) ,A -= (1,2), B -= (2,3) and C-= (4,3)` are four fixed points on x-y plane
Let `S(x,y)` such that `S` is equidistant from points `O` and `B` with respect to new distance and if `x ge 2 and 0 le y lt 3` then locus of `S` is

Options

A. (a) a line segment of infinite length
B. (b) a line of infinite length
C. (c) a ray of finite length
D. (d) a ray of infinite length

Answer

To find the locus of the point \( S(x, y) \) which is equidistant from the points \( O(0, 0) \) and \( B(2, 3) \) using the new distance defined as \( d(P, Q) = |x_1 - x_2| + |y_1 - y_2| \), we can follow these steps:

### Step 1: Write the distance equations
The distance from \( S \) to \( O \) is given by:
\[ d(S, O) = |x - 0| + |y - 0| = |x| + |y| \]
Since \( x \geq 2 \) and \( y \geq 0 \), we can simplify this to:
\[ d(S, O) = x + y \]

The distance from \( S \) to \( B \) is given by:
\[ d(S, B) = |x - 2| + |y - 3| \]
Since \( x \geq 2 \) and \( y < 3 \), we can simplify this to:
\[ d(S, B) = (x - 2) + (3 - y) = x - 2 + 3 - y = x - y + 1 \]

### Step 2: Set the distances equal
Since \( S \) is equidistant from \( O \) and \( B \), we set the two distance equations equal to each other:
\[ x + y = x - y + 1 \]

### Step 3: Solve for \( y \)
Now, we can simplify the equation:
\[ x + y = x - y + 1 \implies y + y = 1 \implies 2y = 1 \implies y = \frac{1}{2} \]

### Step 4: Determine the locus
The equation \( y = \frac{1}{2} \) represents a horizontal line. However, we need to consider the constraints given in the problem: \( x \geq 2 \) and \( 0 \leq y < 3 \).

### Step 5: Identify the valid region
Since \( y \) must equal \( \frac{1}{2} \) and \( x \) can take any value greater than or equal to \( 2 \), the locus of \( S \) is a ray starting from the point \( (2, \frac{1}{2}) \) and extending infinitely to the right along the line \( y = \frac{1}{2} \).

### Final Answer
Thus, the locus of the point \( S \) is a ray of infinite length starting from \( (2, \frac{1}{2}) \) and extending to the right.

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