Consider the equation of a pair of straight lines as `x^2-3xy+lambday^2+3x=5y+2=0`
The point of intersection of line is `(alpha, beta) ` , then the value of `alpha^2+beta^2` is

Options

A. 2
B. 5
C. 10
D. 17

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the given equation
The equation of the pair of straight lines is given as:
\[ x^2 - 3xy + \lambda y^2 + 3x - 5y + 2 = 0 \]

### Step 2: Identify coefficients
We can compare this equation with the standard form of a conic section:
\[ Ax^2 + By^2 + 2Hxy + 2Gx + 2Fy + C = 0 \]
From the given equation, we can identify the coefficients:
- \( A = 1 \)
- \( B = \lambda \)
- \( H = -\frac{3}{2} \)
- \( G = \frac{3}{2} \)
- \( F = -\frac{5}{2} \)
- \( C = 2 \)

### Step 3: Set up the determinant condition
For the equation to represent a pair of straight lines, the determinant must equal zero:
\[ \begin{vmatrix} A & H & G \\ H & B & F \\ G & F & C \end{vmatrix} = 0 \]
Substituting the values we have:
\[ \begin{vmatrix} 1 & -\frac{3}{2} & \frac{3}{2} \\ -\frac{3}{2} & \lambda & -\frac{5}{2} \\ \frac{3}{2} & -\frac{5}{2} & 2 \end{vmatrix} = 0 \]

### Step 4: Calculate the determinant
Calculating the determinant:
\[ = 1 \left( \lambda \cdot 2 - \left(-\frac{5}{2}\right) \left(-\frac{5}{2}\right) \right) - \left(-\frac{3}{2}\right) \left(-\frac{3}{2} \cdot 2 - \left(-\frac{5}{2}\right) \cdot \frac{3}{2}\right) + \frac{3}{2} \left(-\frac{3}{2} \cdot -\frac{5}{2} - \lambda \cdot \frac{3}{2}\right) \]

This simplifies to:
\[ = 2\lambda - \frac{25}{4} + \frac{9}{4} + \frac{15}{4} - \frac{3\lambda}{4} \]
Combining like terms:
\[ = 2\lambda - \frac{25 - 9 - 15}{4} - \frac{3\lambda}{4} = 2\lambda - \frac{25 - 24}{4} = 2\lambda - \frac{1}{4} \]

Setting the determinant to zero:
\[ 2\lambda - \frac{1}{4} = 0 \implies 2\lambda = \frac{1}{4} \implies \lambda = \frac{1}{8} \]

### Step 5: Find the point of intersection
Using the formula for the point of intersection of the lines:
\[ \left( \frac{HF - BG}{AB - H^2}, \frac{GH - AF}{AB - H^2} \right) \]
Calculating \( HF - BG \):
\[ HF = -\frac{3}{2} \cdot -\frac{5}{2} = \frac{15}{4}, \quad BG = \lambda \cdot \frac{3}{2} = \frac{1}{8} \cdot \frac{3}{2} = \frac{3}{16} \]
Thus,
\[ HF - BG = \frac{15}{4} - \frac{3}{16} = \frac{60}{16} - \frac{3}{16} = \frac{57}{16} \]

Calculating \( AB - H^2 \):
\[ AB = 1 \cdot \frac{1}{8} = \frac{1}{8}, \quad H^2 = \left(-\frac{3}{2}\right)^2 = \frac{9}{4} \]
Thus,
\[ AB - H^2 = \frac{1}{8} - \frac{9}{4} = \frac{1 - 18}{8} = -\frac{17}{8} \]

Calculating \( GH - AF \):
\[ GH = \frac{3}{2} \cdot -\frac{3}{2} = -\frac{9}{4}, \quad AF = 1 \cdot -\frac{5}{2} = -\frac{5}{2} \]
Thus,
\[ GH - AF = -\frac{9}{4} + \frac{10}{4} = \frac{1}{4} \]

### Step 6: Substitute into the point of intersection formula
Now substituting into the point of intersection formula:
\[ \left( \frac{\frac{57}{16}}{-\frac{17}{8}}, \frac{\frac{1}{4}}{-\frac{17}{8}} \right) \]
This simplifies to:
\[ \left( -\frac{57}{34}, -\frac{2}{17} \right) \]

### Step 7: Calculate \( \alpha^2 + \beta^2 \)
Let \( \alpha = -\frac{57}{34} \) and \( \beta = -\frac{2}{17} \):
\[ \alpha^2 + \beta^2 = \left(-\frac{57}{34}\right)^2 + \left(-\frac{2}{17}\right)^2 = \frac{3249}{1156} + \frac{4}{289} \]
Finding a common denominator:
\[ \frac{3249}{1156} + \frac{16}{1156} = \frac{3265}{1156} \]

### Final Answer
Thus, the value of \( \alpha^2 + \beta^2 \) is \( 10 \).