The equation `x^2 - 3xy+ lambday^2 + 3x - 5y + 2 = 0` where `lambda` is a real number, represents a pair of straight lines. If `theta` is the angle between the lines, then `cosec^2theta =`

Options

A. 9
B. 10
C. 15
D. 26

Explanation

To solve the problem, we need to find the value of \( \csc^2 \theta \) for the given equation representing a pair of straight lines. The equation is:

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

### Step 1: Identify the coefficients
The general form of the equation of a pair of straight lines is given by:

\[ Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0 \]

From the given equation, we can identify the coefficients:
- \( A = 1 \)
- \( 2H = -3 \) → \( H = -\frac{3}{2} \)
- \( B = \lambda \)
- \( 2G = 3 \) → \( G = \frac{3}{2} \)
- \( 2F = -5 \) → \( F = -\frac{5}{2} \)
- \( C = 2 \)

### Step 2: Condition for a pair of straight lines
For the equation to represent a pair of straight lines, the determinant \( D \) must be zero:

\[ D = ABC + 2FGH - AF^2 - BG^2 - CH^2 = 0 \]

Substituting the values we have:

\[ D = (1)(\lambda)(2) + 2\left(-\frac{5}{2}\right)\left(-\frac{3}{2}\right)(\frac{3}{2}) - (1)\left(-\frac{5}{2}\right)^2 - (\lambda)\left(\frac{3}{2}\right)^2 - (2)\left(-\frac{3}{2}\right)^2 \]

Calculating each term:

1. \( 2\lambda \)
2. \( 2 \cdot \frac{15}{8} = \frac{15}{4} \)
3. \( -\frac{25}{4} \)
4. \( -\frac{9}{4}\lambda \)
5. \( -\frac{18}{4} = -\frac{9}{2} \)

Putting it all together:

\[ 2\lambda + \frac{15}{4} - \frac{25}{4} - \frac{9}{4}\lambda - \frac{18}{4} = 0 \]

Combining like terms:

\[ (2 - \frac{9}{4})\lambda + \frac{15 - 25 - 18}{4} = 0 \]

This simplifies to:

\[ \left(\frac{8}{4} - \frac{9}{4}\right)\lambda - \frac{28}{4} = 0 \]

\[ -\frac{1}{4}\lambda - 7 = 0 \]

Solving for \( \lambda \):

\[ \lambda = -28 \]

### Step 3: Finding \( \tan \theta \)
Using the formula for \( \tan \theta \):

\[ \tan \theta = \frac{2\sqrt{H^2 - AB}}{A + B} \]

Substituting \( A = 1 \), \( B = -28 \), and \( H = -\frac{3}{2} \):

\[ \tan \theta = \frac{2\sqrt{\left(-\frac{3}{2}\right)^2 - (1)(-28)}}{1 - 28} \]

Calculating:

1. \( H^2 = \frac{9}{4} \)
2. \( AB = -28 \)

Thus,

\[ \tan \theta = \frac{2\sqrt{\frac{9}{4} + 28}}{-27} \]

Calculating \( \frac{9}{4} + 28 = \frac{9 + 112}{4} = \frac{121}{4} \):

\[ \tan \theta = \frac{2\sqrt{\frac{121}{4}}}{-27} = \frac{2 \cdot \frac{11}{2}}{-27} = \frac{11}{-27} \]

### Step 4: Finding \( \csc^2 \theta \)
Using the identity \( \csc^2 \theta = 1 + \cot^2 \theta \):

\[ \cot \theta = \frac{1}{\tan \theta} = \frac{-27}{11} \]

Calculating \( \cot^2 \theta \):

\[ \cot^2 \theta = \left(\frac{-27}{11}\right)^2 = \frac{729}{121} \]

Thus,

\[ \csc^2 \theta = 1 + \frac{729}{121} = \frac{121 + 729}{121} = \frac{850}{121} \]

### Final Answer
The value of \( \csc^2 \theta \) is:

\[ \csc^2 \theta = \frac{850}{121} \]