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 `R(x,y)` such that R is equidistant from the point `O` and `A` with respect to new distance and if ` 0 le x lt 1 and 0 le y lt 2`, then R lie on a line segment whose equation is

Options

A. (a) `x+y=3`
B. (b) `x+2y=3`
C. (c) `2x+y=3`
D. (d) `2x+2y=3`

Answer

To solve the problem step by step, we need to find the equation of the line segment on which the point \( R(x, y) \) lies, given that it is equidistant from points \( O(0, 0) \) and \( A(1, 2) \) with respect to the defined distance \( d(P, Q) = |x_1 - x_2| + |y_1 - y_2| \).

### Step 1: Write the distance equations
The distance from point \( R(x, y) \) to point \( O(0, 0) \) is:
\[ d(R, O) = |x - 0| + |y - 0| = |x| + |y| \]
Since \( 0 \leq x < 1 \) and \( 0 \leq y < 2 \), both \( x \) and \( y \) are non-negative, so:
\[ d(R, O) = x + y \]

The distance from point \( R(x, y) \) to point \( A(1, 2) \) is:
\[ d(R, A) = |x - 1| + |y - 2| \]

### Step 2: Analyze the distance from \( R \) to \( A \)
Given the constraints on \( x \) and \( y \):
- Since \( x < 1 \), we have \( |x - 1| = 1 - x \).
- Since \( y < 2 \), we have \( |y - 2| = 2 - y \).

Thus, we can write:
\[ d(R, A) = (1 - x) + (2 - y) = 3 - x - y \]

### Step 3: Set the distances equal
Since \( R \) is equidistant from \( O \) and \( A \), we set the two distance equations equal to each other:
\[ x + y = 3 - x - y \]

### Step 4: Solve for \( x \) and \( y \)
Rearranging the equation gives:
\[ x + y + x + y = 3 \]
\[ 2x + 2y = 3 \]

### Step 5: Simplify the equation
Dividing the entire equation by 2, we get:
\[ x + y = \frac{3}{2} \]

### Conclusion
The equation of the line segment on which point \( R \) lies is:
\[ 2x + 2y = 3 \]