If the sum of the distances of a moving point in a plane from the axes is `1,` then find the locus of the point.

Explanation

To find the locus of a moving point in a plane such that the sum of its distances from the x-axis and y-axis is equal to 1, we can follow these steps:

### Step 1: Define the point
Let the moving point be \( P(x, y) \).

### Step 2: Determine the distances
The distance of point \( P \) from the x-axis is given by the absolute value of the y-coordinate, which is \( |y| \). The distance from the y-axis is given by the absolute value of the x-coordinate, which is \( |x| \).

### Step 3: Set up the equation
According to the problem, the sum of these distances is equal to 1:
\[ |x| + |y| = 1 \]

### Step 4: Analyze the equation
The equation \( |x| + |y| = 1 \) represents a geometric figure in the coordinate plane. To understand this better, we can consider the different cases for \( x \) and \( y \):

1. **Case 1**: \( x \geq 0 \) and \( y \geq 0 \)
- The equation simplifies to \( x + y = 1 \).

2. **Case 2**: \( x \geq 0 \) and \( y < 0 \)
- The equation simplifies to \( x - y = 1 \).

3. **Case 3**: \( x < 0 \) and \( y \geq 0 \)
- The equation simplifies to \( -x + y = 1 \).

4. **Case 4**: \( x < 0 \) and \( y < 0 \)
- The equation simplifies to \( -x - y = 1 \).

### Step 5: Write the equations of the lines
From the cases above, we can derive the following lines:
1. \( x + y = 1 \) (First quadrant)
2. \( x - y = 1 \) (Fourth quadrant)
3. \( -x + y = 1 \) (Second quadrant)
4. \( -x - y = 1 \) (Third quadrant)

### Step 6: Identify the shape
These lines form a square in the coordinate plane with vertices at the points \( (1, 0) \), \( (0, 1) \), \( (-1, 0) \), and \( (0, -1) \).

### Final Answer
The locus of the point is the square defined by the lines:
\[ |x| + |y| = 1 \]

---