If the distance of any point (x,y) from the origin is defined as `d(x,y)="max"{|x|,|y|},d(x,y)=a` none zero constant, then the locus is

Options

A. a circle
B. parallel straight lines
C. a square
D. a triangle

Answer

To solve the problem, we need to find the locus of points (x, y) such that the distance from the origin, defined as \( d(x,y) = \max\{|x|, |y|\} \), is equal to a non-zero constant \( a \).

### Step-by-Step Solution:

1. **Understanding the Distance Definition**:
The distance \( d(x, y) \) is defined as \( d(x, y) = \max\{|x|, |y|\} \). This means that for any point (x, y), the distance is determined by the larger of the absolute values of x and y.

2. **Setting Up the Equation**:
We set the distance equal to a constant \( a \):
\[ \max\{|x|, |y|\} = a \]

3. **Analyzing Cases**:
We will analyze two cases based on the definition of the maximum function.

- **Case 1**: \( |x| \geq |y| \)
In this case, we have:
\[ |x| = a \]
This leads to two equations:
\[ x = a \quad \text{or} \quad x = -a \]
This means that for these values of x, y can take any value, leading to the vertical lines \( x = a \) and \( x = -a \).

- **Case 2**: \( |y| > |x| \)
Here, we have:
\[ |y| = a \]
This leads to two equations:
\[ y = a \quad \text{or} \quad y = -a \]
This means that for these values of y, x can take any value, leading to the horizontal lines \( y = a \) and \( y = -a \).

4. **Combining the Results**:
From both cases, we can conclude that the locus of points (x, y) such that \( d(x, y) = a \) consists of:
- The vertical lines \( x = a \) and \( x = -a \)
- The horizontal lines \( y = a \) and \( y = -a \)

5. **Final Locus Representation**:
Therefore, the locus of the points is represented by the four lines:
\[ x = a, \quad x = -a, \quad y = a, \quad y = -a \]

### Conclusion:
The locus of the points (x, y) such that \( d(x, y) = a \) is a pair of vertical lines and a pair of horizontal lines, forming a square-like shape centered at the origin with vertices at \( (a, a), (a, -a), (-a, a), (-a, -a) \).