The value of k for which the equation `2x^2+5y^2-2kxy+4x+6y=0` represent the pair of straight lines is

Options

A. `-19/6`
B. `19/6`
C. `-7/6`
D. none of these

Answer

To find the value of \( k \) for which the equation \( 2x^2 + 5y^2 - 2kxy + 4x + 6y = 0 \) represents a pair of straight lines, we can follow these steps:

### Step 1: Identify coefficients
The given equation can be compared with the standard form of the conic section:
\[ ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \]
From the given equation, we can identify:
- \( a = 2 \)
- \( b = 5 \)
- \( h = -k \)
- \( g = 2 \) (since \( 2g = 4 \))
- \( f = 3 \) (since \( 2f = 6 \))
- \( c = 0 \)

### Step 2: Use the condition for a pair of straight lines
For the equation to represent a pair of straight lines, the following condition must be satisfied:
\[ abc + 2fgh - af^2 - bg^2 - ch^2 = 0 \]
Substituting the values we found:
- \( a = 2 \)
- \( b = 5 \)
- \( c = 0 \)
- \( f = 3 \)
- \( g = 2 \)
- \( h = -k \)

We can substitute these values into the condition:
\[ (2)(5)(0) + 2(3)(2)(-k) - (2)(3^2) - (5)(2^2) - (0)(-k)^2 = 0 \]

### Step 3: Simplify the equation
Now, simplifying the equation:
\[ 0 + 2(3)(2)(-k) - 2(9) - 5(4) - 0 = 0 \]
\[ 0 - 12k - 18 - 20 = 0 \]
\[ -12k - 38 = 0 \]

### Step 4: Solve for \( k \)
Now, we can solve for \( k \):
\[ -12k = 38 \]
\[ k = -\frac{38}{12} \]
\[ k = -\frac{19}{6} \]

### Final Answer
Thus, the value of \( k \) for which the equation represents a pair of straight lines is:
\[ \boxed{-\frac{19}{6}} \]