The lines y = mx , `y + 2x = 0 , y = 2x + k and y + mx = k ` form a rhombus if m equals

Options

A. `-1`
B. `1/2`
C. 1
D. 2

Answer

To determine the value of \( m \) for which the lines \( y = mx \), \( y + 2x = 0 \), \( y = 2x + k \), and \( y + mx = k \) form a rhombus, we will follow these steps:

### Step 1: Rewrite the equations in slope-intercept form
1. The first line is already in slope-intercept form: \( y = mx \).
2. The second line \( y + 2x = 0 \) can be rewritten as \( y = -2x \).
3. The third line \( y = 2x + k \) is also in slope-intercept form.
4. The fourth line \( y + mx = k \) can be rewritten as \( y = -mx + k \).

### Step 2: Identify the slopes
From the equations, we have the following slopes:
- Slope of \( y = mx \): \( m \)
- Slope of \( y = -2x \): \( -2 \)
- Slope of \( y = 2x + k \): \( 2 \)
- Slope of \( y = -mx + k \): \( -m \)

### Step 3: Set conditions for rhombus formation
For the lines to form a rhombus, the opposite sides must be parallel. This means:
1. The slopes of \( y = mx \) and \( y = -mx + k \) must be equal.
2. The slopes of \( y = -2x \) and \( y = 2x + k \) must also be equal.

### Step 4: Set up the equations
From the first condition:
\[ m = -m \]
This implies:
\[ 2m = 0 \quad \Rightarrow \quad m = 0 \]

From the second condition:
\[ -2 = 2 \]
This condition does not hold true, indicating that the lines cannot be parallel.

### Step 5: Check for valid \( m \)
Since we found \( m = 0 \) from the first condition, we need to check if this value satisfies the rhombus condition. When \( m = 0 \), the lines become:
1. \( y = 0 \) (the x-axis)
2. \( y = -2x \)
3. \( y = 2x + k \)
4. \( y = k \)

### Step 6: Analyze the intersection points
To form a rhombus, we need to check the intersection points of these lines:
- The intersection of \( y = 0 \) and \( y = -2x \) gives the point \( (0, 0) \).
- The intersection of \( y = 0 \) and \( y = 2x + k \) gives the point \( (-\frac{k}{2}, 0) \).
- The intersection of \( y = -2x \) and \( y = k \) gives the point \( (-\frac{k}{2}, k) \).
- The intersection of \( y = 2x + k \) and \( y = k \) gives the point \( (0, k) \).

### Conclusion
Since the lines intersect at four points forming a rhombus when \( m = 0 \), we conclude that:
\[ \boxed{0} \]