Any second degree equation `ax ^(2) + by ^(2) + 2 hxy + 2gx + 2fy + c =0` will represent oertain curves under different conditions:
It will represent a pair of stragith lines if `Delta =0` where `Delta = 2fgh + abc -af ^(2) - bg ^(2)-ch^(2).` Similar to this rule, it will represent real circel if `Delta ne 0, a =b and h =0.`
It will represent parabola, ellipse and hyperbola if `(Delta ne 0)h^(2)=ab, h^(2) -ab lt 0 and h^(2) -ab gt 0` respectively for each of the given curves.
If `3x ^(2) -11 xy + 10 y ^(2) - 7x + 13 y + k =0` represent pair of straight lines then vlaue of k is

Options

A. 3
B. 4
C. 6
D. none of these

Answer

To find the value of \( k \) such that the equation \( 3x^2 - 11xy + 10y^2 - 7x + 13y + k = 0 \) represents a pair of straight lines, we will use the condition that \( \Delta = 0 \), where

\[ \Delta = 2fgh + abc - af^2 - bg^2 - ch^2. \]

### Step 1: Identify the coefficients
From the equation \( 3x^2 - 11xy + 10y^2 - 7x + 13y + k = 0 \), we can identify the coefficients:
- \( a = 3 \) (coefficient of \( x^2 \))
- \( b = 10 \) (coefficient of \( y^2 \))
- \( h = -\frac{11}{2} \) (since \( 2h = -11 \))
- \( g = -\frac{7}{2} \) (since \( 2g = -7 \))
- \( f = \frac{13}{2} \) (since \( 2f = 13 \))
- \( c = k \) (constant term)

### Step 2: Substitute the coefficients into the formula for \( \Delta \)
Now, we will substitute these values into the formula for \( \Delta \):

\[ \Delta = 2 \left(\frac{13}{2}\right) \left(-\frac{7}{2}\right) \left(-\frac{11}{2}\right) + 3 \cdot 10 \cdot k - 3 \left(\frac{13}{2}\right)^2 - 10 \left(-\frac{7}{2}\right)^2 - k \left(-\frac{11}{2}\right)^2. \]

### Step 3: Calculate each term
1. **Calculate \( 2fgh \)**:
\[ 2 \cdot \frac{13}{2} \cdot -\frac{7}{2} \cdot -\frac{11}{2} = \frac{13 \cdot 7 \cdot 11}{4} = \frac{1001}{4}. \]

2. **Calculate \( abc \)**:
\[ 3 \cdot 10 \cdot k = 30k. \]

3. **Calculate \( af^2 \)**:
\[ 3 \cdot \left(\frac{13}{2}\right)^2 = 3 \cdot \frac{169}{4} = \frac{507}{4}. \]

4. **Calculate \( bg^2 \)**:
\[ 10 \cdot \left(-\frac{7}{2}\right)^2 = 10 \cdot \frac{49}{4} = \frac{490}{4}. \]

5. **Calculate \( ch^2 \)**:
\[ k \cdot \left(-\frac{11}{2}\right)^2 = k \cdot \frac{121}{4}. \]

### Step 4: Substitute back into \( \Delta \)
Now, substituting these values back into the equation for \( \Delta \):

\[ \Delta = \frac{1001}{4} + 30k - \frac{507}{4} - \frac{490}{4} - \frac{121k}{4}. \]

### Step 5: Combine like terms
Combine the constant terms and the terms involving \( k \):

\[ \Delta = \frac{1001 - 507 - 490}{4} + \left(30 - \frac{121}{4}\right)k = \frac{4}{4} + \left(30 - \frac{121}{4}\right)k = 1 + \left(30 - 30.25\right)k = 1 - \frac{1}{4}k. \]

### Step 6: Set \( \Delta = 0 \)
Setting \( \Delta = 0 \):

\[ 1 - \frac{1}{4}k = 0. \]

### Step 7: Solve for \( k \)
Solving for \( k \):

\[ \frac{1}{4}k = 1 \implies k = 4. \]

Thus, the value of \( k \) is \( \boxed{4} \).