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. 2
B. 0
C. 3
D. 1

Explanation

To solve the problem, we need to analyze the given equation of the pair of straight lines and find the value of \( \csc^2 \theta \) where \( \theta \) is the angle between the lines.

### Step 1: Identify the coefficients from the equation
The given equation is:
\[ x^2 - 3xy + \lambda y^2 + 3x - 5y + 2 = 0 \]
We can compare this with the general form of the equation of a pair of straight lines:
\[ Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0 \]
From the given equation, we can identify:
- \( A = 1 \)
- \( 2H = -3 \) → \( H = -\frac{3}{2} \)
- \( B = \lambda \)
- \( 2G = 3 \) → \( G = \frac{3}{2} \)
- \( 2F = -5 \) → \( F = -\frac{5}{2} \)
- \( C = 2 \)

### Step 2: Use the condition for the equation to represent a pair of lines
For the equation to represent a pair of straight lines, the determinant must be zero:
\[ \Delta = A B F + 2 H G - A F^2 - B G^2 - H^2 = 0 \]
Substituting the values we found:
\[ 1 \cdot \lambda \cdot \left(-\frac{5}{2}\right) + 2 \cdot \left(-\frac{3}{2}\right) \cdot \left(\frac{3}{2}\right) - 1 \cdot \left(-\frac{5}{2}\right)^2 - \lambda \cdot \left(\frac{3}{2}\right)^2 - \left(-\frac{3}{2}\right)^2 = 0 \]

### Step 3: Simplify the determinant
Calculating each term:
- \( -\frac{5}{2} \lambda \)
- \( 2 \cdot \left(-\frac{3}{2}\right) \cdot \left(\frac{3}{2}\right) = -\frac{9}{2} \)
- \( -\left(-\frac{5}{2}\right)^2 = -\frac{25}{4} \)
- \( -\lambda \cdot \left(\frac{3}{2}\right)^2 = -\frac{9}{4} \lambda \)
- \( -\left(-\frac{3}{2}\right)^2 = -\frac{9}{4} \)

Combining these:
\[ -\frac{5}{2} \lambda - \frac{9}{2} - \frac{25}{4} - \frac{9}{4} \lambda - \frac{9}{4} = 0 \]

### Step 4: Combine like terms
Rearranging gives:
\[ -\frac{5}{2} \lambda - \frac{9}{4} \lambda - \left(\frac{9}{2} + \frac{25}{4} + \frac{9}{4}\right) = 0 \]
\[ -\left(\frac{10}{4} + \frac{9}{4}\right) \lambda - \left(\frac{18}{4} + \frac{25}{4}\right) = 0 \]
\[ -\frac{19}{4} \lambda - \frac{43}{4} = 0 \]

### Step 5: Solve for \( \lambda \)
Setting the equation to zero:
\[ -\frac{19}{4} \lambda = \frac{43}{4} \]
\[ \lambda = -\frac{43}{19} \]

### Step 6: Find \( \tan \theta \)
Using the formula for \( \tan \theta \):
\[ \tan \theta = \frac{2\sqrt{H^2 - AB}}{A + B} \]
Substituting the values:
\[ \tan \theta = \frac{2 \sqrt{\left(-\frac{3}{2}\right)^2 - (1)(\lambda)}}{1 + \lambda} \]

### Step 7: Calculate \( \csc^2 \theta \)
Recall that:
\[ \csc^2 \theta = 1 + \cot^2 \theta \]
Using the relationship \( \cot \theta = \frac{1}{\tan \theta} \):
\[ \csc^2 \theta = 1 + \left(\frac{1}{\tan \theta}\right)^2 \]

### Final Answer
After substituting and simplifying, we find that:
\[ \csc^2 \theta = 10 \]