The locus represented by `xy + yz = 0` is

Options

A. Pair of perpendicular lines
B. pair of parallel lines
C. Pair of parallel planes
D. Pair of perpendicular planes

Answer

To find the locus represented by the equation \( xy + yz = 0 \), we can follow these steps:

### Step 1: Factor the equation
We start with the equation:
\[ xy + yz = 0 \]
We can factor out \( y \):
\[ y(x + z) = 0 \]

### Step 2: Identify the conditions
From the factored equation, we can set each factor to zero:
1. \( y = 0 \)
2. \( x + z = 0 \)

### Step 3: Analyze the first condition
The first condition \( y = 0 \) represents a plane in three-dimensional space. This is the \( xz \)-plane, where the value of \( y \) is always zero.

### Step 4: Analyze the second condition
The second condition \( x + z = 0 \) can be rewritten as:
\[ z = -x \]
This represents another plane in three-dimensional space, which is inclined at 45 degrees to both the x-axis and z-axis.

### Step 5: Determine the relationship between the planes
We have two planes:
- Plane 1: \( y = 0 \) (the \( xz \)-plane)
- Plane 2: \( x + z = 0 \) (a plane that passes through the origin and is inclined)

### Step 6: Check the perpendicularity of the planes
To check if these two planes are perpendicular, we can find their normal vectors:
- The normal to the plane \( y = 0 \) is given by \( \mathbf{n_1} = \mathbf{j} \).
- The normal to the plane \( x + z = 0 \) is given by \( \mathbf{n_2} = \mathbf{i} + \mathbf{k} \).

Now, we calculate the dot product:
\[ \mathbf{n_1} \cdot \mathbf{n_2} = \mathbf{j} \cdot (\mathbf{i} + \mathbf{k}) = 0 + 0 = 0 \]
Since the dot product is zero, the planes are perpendicular to each other.

### Conclusion
Thus, the locus represented by the equation \( xy + yz = 0 \) consists of two planes: one is the \( xz \)-plane (where \( y = 0 \)), and the other is the plane defined by \( x + z = 0 \), which are perpendicular to each other.

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