If `x^(2)-3xy+lamday^(2)+3x-5y+2=0` represents a pair of straight lines and `theta` is the angle between them, then `cosec^(2)theta=`

Options

A. 3
B. 9
C. 10
D. 100

Answer

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

### Step 1: Identify the coefficients
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 comparison, we identify:
- \( a = 1 \)
- \( h = -\frac{3}{2} \)
- \( b = \lambda \)
- \( g = \frac{3}{2} \)
- \( f = -\frac{5}{2} \)
- \( c = 2 \)

### Step 2: Use the condition for a pair of straight lines
For the equation to represent a pair of straight lines, the discriminant \( D \) must be zero:
\[ D = abc + 2fgh - af^2 - bg^2 - ch^2 = 0 \]
Substituting the values we have:
\[ D = (1)(\lambda)(2) + 2\left(-\frac{5}{2}\right)\left(\frac{3}{2}\right)\left(-\frac{3}{2}\right) - (1)\left(-\frac{5}{2}\right)^2 - (\lambda)\left(\frac{3}{2}\right)^2 - (2)\left(-\frac{3}{2}\right)^2 = 0 \]

### Step 3: Simplify the expression
Calculating each term:
1. \( 2\lambda \)
2. \( 2 \cdot -\frac{5}{2} \cdot \frac{3}{2} \cdot -\frac{3}{2} = \frac{45}{4} \)
3. \( -\left(-\frac{5}{2}\right)^2 = -\frac{25}{4} \)
4. \( -\lambda \cdot \frac{9}{4} = -\frac{9\lambda}{4} \)
5. \( -2 \cdot \frac{9}{4} = -\frac{18}{4} = -\frac{9}{2} \)

Putting it all together:
\[ 2\lambda + \frac{45}{4} - \frac{25}{4} - \frac{9\lambda}{4} - \frac{18}{4} = 0 \]
Combining like terms:
\[ 2\lambda - \frac{9\lambda}{4} + \frac{45 - 25 - 18}{4} = 0 \]
\[ 2\lambda - \frac{9\lambda}{4} + \frac{2}{4} = 0 \]
\[ 2\lambda - \frac{9\lambda}{4} + \frac{1}{2} = 0 \]
Multiplying through by 4 to eliminate the fraction:
\[ 8\lambda - 9\lambda + 2 = 0 \]
\[ -\lambda + 2 = 0 \implies \lambda = 2 \]

### Step 4: Find the angle between the lines
Now that we have \( \lambda = 2 \), we can find the angle \( \theta \) between the lines using:
\[ \tan \theta = \frac{2\sqrt{h^2 - ab}}{a + b} \]
Substituting the values:
- \( a = 1 \)
- \( b = 2 \)
- \( h = -\frac{3}{2} \)

Calculating:
\[ \tan \theta = \frac{2\sqrt{\left(-\frac{3}{2}\right)^2 - (1)(2)}}{1 + 2} \]
\[ = \frac{2\sqrt{\frac{9}{4} - 2}}{3} = \frac{2\sqrt{\frac{9}{4} - \frac{8}{4}}}{3} = \frac{2\sqrt{\frac{1}{4}}}{3} = \frac{2 \cdot \frac{1}{2}}{3} = \frac{1}{3} \]

### Step 5: Calculate \( \csc^2 \theta \)
Using the identity:
\[ \csc^2 \theta = 1 + \cot^2 \theta \]
Since \( \tan \theta = \frac{1}{3} \), we have:
\[ \cot \theta = \frac{1}{\tan \theta} = 3 \]
Thus,
\[ \cot^2 \theta = 3^2 = 9 \]
Now substituting back:
\[ \csc^2 \theta = 1 + 9 = 10 \]

### Final Answer
\[ \csc^2 \theta = 10 \]