If one of the lines of `my^(2)+(1-m^(2))xy-mx^(2)=0` is a bisector of the angle between the lines `xy=0`, then m is

Options

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

Explanation

To solve the problem, we need to find the value of \( m \) such that one of the lines represented by the equation \( my^2 + (1 - m^2)xy - mx^2 = 0 \) is a bisector of the angle between the lines represented by \( xy = 0 \).

### Step-by-Step Solution:

1. **Identify the Lines from \( xy = 0 \)**:
The equation \( xy = 0 \) represents two lines:
- \( x = 0 \) (the y-axis)
- \( y = 0 \) (the x-axis)

2. **Determine the Slopes of the Angle Bisectors**:
The angle bisectors between the x-axis and y-axis have slopes of:
- \( 1 \) (for the line \( y = x \))
- \( -1 \) (for the line \( y = -x \))

3. **Analyze the Given Equation**:
The given equation is:
\[ my^2 + (1 - m^2)xy - mx^2 = 0 \]
This represents a pair of straight lines. We can rearrange it to identify the slopes of the lines.

4. **Factor the Equation**:
We can factor the equation as follows:
\[ my^2 + (1 - m^2)xy - mx^2 = 0 \]
Rearranging gives:
\[ my^2 + (1 - m^2)xy - mx^2 = 0 \]
We can factor out \( y \):
\[ y(my + (1 - m^2)x) - mx^2 = 0 \]
This leads to:
\[ y(m y + (1 - m^2)x) = mx^2 \]

5. **Find the Slopes of the Lines**:
From the factored form, we can derive the slopes of the lines:
- The first line is \( y = mx \) (slope = \( m \))
- The second line can be derived from \( (1 - m^2)x = -my \), giving \( y = -\frac{1 - m^2}{m} x \) (slope = \( -\frac{1 - m^2}{m} \))

6. **Set the Slopes Equal to the Bisector Slopes**:
For one of the lines to be a bisector, its slope must equal \( 1 \) or \( -1 \):
- Case 1: Set \( m = 1 \)
- Case 2: Set \( -\frac{1 - m^2}{m} = 1 \)
- Case 3: Set \( m = -1 \)
- Case 4: Set \( -\frac{1 - m^2}{m} = -1 \)

7. **Solve for \( m \)**:
- From Case 1: \( m = 1 \)
- From Case 2: \( -1 + m^2 = m \) leads to \( m^2 - m - 1 = 0 \) (use quadratic formula)
- From Case 3: \( m = -1 \)
- From Case 4: \( 1 - m^2 = -m \) leads to \( m^2 - m - 1 = 0 \) (same as Case 2)

8. **Calculate Roots**:
Using the quadratic formula \( m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ m = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2} \]

9. **Final Values of \( m \)**:
Thus, the possible values of \( m \) are:
- \( m = 1 \)
- \( m = -1 \)
- \( m = \frac{1 + \sqrt{5}}{2} \)
- \( m = \frac{1 - \sqrt{5}}{2} \)

### Conclusion:
The values of \( m \) for which one of the lines is a bisector of the angle between the x-axis and y-axis are \( m = 1 \) and \( m = -1 \).