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
Le T(x,y) such that T is equisdistant from point O and C with respect to new distance and if T lie in first quadrant , then T consists of the union of a line segment of finite length and an infinite ray whose labelled diagram is

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Answer

To solve the problem, we need to find the points \( T(x, y) \) that are equidistant from points \( O(0, 0) \) and \( C(4, 3) \) using the defined distance \( d(P, Q) = |x_1 - x_2| + |y_1 - y_2| \).

### Step 1: Set Up the Equidistant Condition
We start with the condition that the distance from \( T \) to \( O \) is equal to the distance from \( T \) to \( C \):
\[ d(T, O) = d(T, C) \]
This translates to:
\[ |x - 0| + |y - 0| = |x - 4| + |y - 3| \]
Since \( T \) lies in the first quadrant, we can simplify the absolute values:
\[ x + y = |x - 4| + |y - 3| \]

### Step 2: Analyze Cases for the Absolute Values
We need to consider different cases based on the values of \( x \) and \( y \).

#### Case 1: \( 0 \leq x < 4 \) and \( 0 \leq y < 3 \)
In this case:
\[ x + y = (4 - x) + (3 - y) \]
Simplifying gives:
\[ x + y = 7 - x - y \]
\[ 2x + 2y = 7 \quad \Rightarrow \quad x + y = \frac{7}{2} \]

#### Case 2: \( 4 \leq x \) and \( 0 \leq y < 3 \)
Here:
\[ x + y = (x - 4) + (3 - y) \]
Simplifying gives:
\[ x + y = x - 4 + 3 - y \]
\[ 2y = -1 \quad \Rightarrow \quad y = -\frac{1}{2} \quad \text{(not valid since } y \geq 0\text{)} \]

#### Case 3: \( 0 \leq x < 4 \) and \( y \geq 3 \)
In this case:
\[ x + y = (4 - x) + (y - 3) \]
Simplifying gives:
\[ x + y = 4 - x + y - 3 \]
\[ 2x = 1 \quad \Rightarrow \quad x = \frac{1}{2} \]

#### Case 4: \( 4 \leq x \) and \( y \geq 3 \)
Here:
\[ x + y = (x - 4) + (y - 3) \]
Simplifying gives:
\[ x + y = x - 4 + y - 3 \]
\[ 0 = -7 \quad \text{(not valid)} \]

### Step 3: Summary of Results
From the cases analyzed, we have two valid results:
1. From Case 1: \( x + y = \frac{7}{2} \) for \( 0 \leq x < 4 \) and \( 0 \leq y < 3 \).
2. From Case 3: \( x = \frac{1}{2} \) for \( y \geq 3 \).

### Step 4: Graphical Representation
To visualize the results:
- The line \( x + y = \frac{7}{2} \) intersects the axes at \( (0, \frac{7}{2}) \) and \( (\frac{7}{2}, 0) \).
- The point \( (0.5, y) \) where \( y \geq 3 \) is a vertical line at \( x = 0.5 \).

### Final Answer
The set of points \( T \) consists of:
- A line segment from \( (0, \frac{7}{2}) \) to \( (4, 3) \) (finite length).
- An infinite ray starting from \( (0.5, 3) \) extending vertically upwards.