If the equation `7x^(2)-kxy-7y^(2)=0` represents the bisectors of angles between the lines `2x^(2)-7xy+4y^(2)=0` then: `k=`

Options

A. 2
B. 3
C. -3
D. 4

Answer

To solve the problem, we need to find the value of \( k \) such that the equation \( 7x^2 - kxy - 7y^2 = 0 \) represents the angle bisectors of the lines given by the equation \( 2x^2 - 7xy + 4y^2 = 0 \).

### Step-by-Step Solution:

1. **Identify the given equation**:
The equation of the pair of lines is given as:
\[ 2x^2 - 7xy + 4y^2 = 0 \]

2. **General form of a conic**:
The general form of a conic section representing a pair of lines is:
\[ ax^2 + bxy + cy^2 = 0 \]
Here, \( a = 2 \), \( b = -7 \), and \( c = 4 \).

3. **Angle bisector formula**:
The angle bisectors of the lines represented by the equation \( ax^2 + bxy + cy^2 = 0 \) can be represented by the equation:
\[ hx^2 - hy^2 - (a + c)xy = 0 \]
where \( h = \frac{b}{2} \).

4. **Calculate \( h \)**:
From our equation, we have:
\[ h = \frac{-7}{2} \]

5. **Substituting into the angle bisector equation**:
Now substituting \( h \) into the angle bisector equation:
\[ -\frac{7}{2}x^2 - \frac{7}{2}y^2 - (2 + 4)xy = 0 \]
Simplifying gives:
\[ -\frac{7}{2}x^2 - \frac{7}{2}y^2 - 6xy = 0 \]
Multiplying through by -2 to eliminate the fractions:
\[ 7x^2 + 7y^2 + 12xy = 0 \]

6. **Comparing with the required equation**:
The required equation is:
\[ 7x^2 - kxy - 7y^2 = 0 \]
Rearranging gives:
\[ 7x^2 - 7y^2 - kxy = 0 \]

7. **Equating coefficients**:
From the angle bisector equation \( 7x^2 + 7y^2 + 12xy = 0 \) and the required equation \( 7x^2 - kxy - 7y^2 = 0 \), we can compare coefficients:
- Coefficient of \( xy \): \( -k = 12 \)

8. **Solving for \( k \)**:
Thus, we find:
\[ k = -12 \]

### Final Answer:
The value of \( k \) is:
\[ \boxed{12} \]