If `x^(2)-3xy+lamday^(2)+3x-5y+2=0` represents a pair of straight lines, then the value of `lamda` is

Options

A. 1
B. 4
C. 3
D. 2

Answer

To solve the problem, we need to find the value of \(\lambda\) such that the equation

\[ x^2 - 3xy + \lambda y^2 + 3x - 5y + 2 = 0 \]

represents a pair of straight lines. We will use the condition for a conic section to represent a pair of straight lines.

### Step 1: Identify the coefficients

The general form of a conic section is given by:

\[ ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 \]

From the given equation, we can identify the coefficients:

- \(a = 1\)
- \(h = -\frac{3}{2}\) (since the coefficient of \(xy\) is \(-3\))
- \(b = \lambda\)
- \(g = \frac{3}{2}\) (since the coefficient of \(x\) is \(3\))
- \(f = -\frac{5}{2}\) (since the coefficient of \(y\) is \(-5\))
- \(c = 2\)

### Step 2: Use the condition for a pair of straight lines

For the equation to represent a pair of straight lines, the determinant formed by these coefficients must be equal to zero:

\[ \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} = 0 \]

Substituting the values we identified:

\[ \begin{vmatrix} 1 & -\frac{3}{2} & \frac{3}{2} \\ -\frac{3}{2} & \lambda & -\frac{5}{2} \\ \frac{3}{2} & -\frac{5}{2} & 2 \end{vmatrix} = 0 \]

### Step 3: Calculate the determinant

Calculating the determinant:

\[ = 1 \cdot \begin{vmatrix} \lambda & -\frac{5}{2} \\ -\frac{5}{2} & 2 \end{vmatrix} - \left(-\frac{3}{2}\right) \cdot \begin{vmatrix} -\frac{3}{2} & -\frac{5}{2} \\ \frac{3}{2} & 2 \end{vmatrix} + \frac{3}{2} \cdot \begin{vmatrix} -\frac{3}{2} & \lambda \\ \frac{3}{2} & -\frac{5}{2} \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1.
\[ \begin{vmatrix} \lambda & -\frac{5}{2} \\ -\frac{5}{2} & 2 \end{vmatrix} = \lambda \cdot 2 - \left(-\frac{5}{2}\right) \left(-\frac{5}{2}\right) = 2\lambda - \frac{25}{4} \]

2.
\[ \begin{vmatrix} -\frac{3}{2} & -\frac{5}{2} \\ \frac{3}{2} & 2 \end{vmatrix} = -\frac{3}{2} \cdot 2 - \left(-\frac{5}{2}\right) \cdot \frac{3}{2} = -3 + \frac{15}{4} = -\frac{12}{4} + \frac{15}{4} = \frac{3}{4} \]

3.
\[ \begin{vmatrix} -\frac{3}{2} & \lambda \\ \frac{3}{2} & -\frac{5}{2} \end{vmatrix} = -\frac{3}{2} \cdot -\frac{5}{2} - \lambda \cdot \frac{3}{2} = \frac{15}{4} - \frac{3\lambda}{2} \]

### Step 4: Substitute back into the determinant equation

Now substituting back into the determinant equation:

\[ 1 \cdot (2\lambda - \frac{25}{4}) + \frac{3}{2} \cdot \frac{3}{4} + \frac{3}{2} \cdot \left(\frac{15}{4} - \frac{3\lambda}{2}\right) = 0 \]

This simplifies to:

\[ 2\lambda - \frac{25}{4} + \frac{9}{8} + \frac{45}{8} - \frac{9\lambda}{4} = 0 \]

Combining like terms:

\[ 2\lambda - \frac{9\lambda}{4} - \frac{25}{4} + \frac{54}{8} = 0 \]

Converting \(\frac{54}{8}\) to \(\frac{27}{4}\):

\[ 2\lambda - \frac{9\lambda}{4} - \frac{25}{4} + \frac{27}{4} = 0 \]

This simplifies to:

\[ 2\lambda - \frac{9\lambda}{4} + \frac{2}{4} = 0 \]

### Step 5: Solve for \(\lambda\)

Multiplying through by 4 to eliminate the fractions:

\[ 8\lambda - 9\lambda + 2 = 0 \]

This gives:

\[ -\lambda + 2 = 0 \implies \lambda = 2 \]

### Final Answer

Thus, the value of \(\lambda\) is:

\[ \lambda = 2 \]