The lines represented by `x^(2)+2lambda xy+2y^(2)=0` and the lines represented by `(1+lambda)x^(2)-8xy+y^(2)=0` are equally inclined, then `lambda =`

Answer

To solve the problem, we need to find the value of \(\lambda\) such that the lines represented by the equations \(x^2 + 2\lambda xy + 2y^2 = 0\) and \((1 + \lambda)x^2 - 8xy + y^2 = 0\) are equally inclined.

### Step-by-Step Solution:

1. **Identify the coefficients**:
For the first equation \(x^2 + 2\lambda xy + 2y^2 = 0\), we can identify the coefficients:
- \(a_1 = 1\)
- \(h_1 = \lambda\)
- \(b_1 = 2\)

For the second equation \((1 + \lambda)x^2 - 8xy + y^2 = 0\), the coefficients are:
- \(a_2 = 1 + \lambda\)
- \(h_2 = -4\)
- \(b_2 = 1\)

2. **Set up the angle bisector equations**:
The angle bisector condition for two pairs of lines can be expressed as:
\[ \frac{x^2 - y^2}{a_1 - b_1} = \frac{xy}{h_1} \]
and
\[ \frac{x^2 - y^2}{a_2 - b_2} = \frac{xy}{h_2} \]

3. **Substituting the coefficients**:
For the first equation:
\[ \frac{x^2 - y^2}{1 - 2} = \frac{xy}{\lambda} \]
Simplifying gives:
\[ \frac{x^2 - y^2}{-1} = \frac{xy}{\lambda} \quad \Rightarrow \quad x^2 - y^2 = -\frac{xy}{\lambda} \]

For the second equation:
\[ \frac{x^2 - y^2}{(1 + \lambda) - 1} = \frac{xy}{-4} \]
Simplifying gives:
\[ \frac{x^2 - y^2}{\lambda} = -\frac{xy}{4} \quad \Rightarrow \quad x^2 - y^2 = -\frac{\lambda}{4}xy \]

4. **Equating the two expressions**:
Since the lines are equally inclined, we can set the two expressions for \(x^2 - y^2\) equal to each other:
\[ -\frac{xy}{\lambda} = -\frac{\lambda}{4}xy \]

5. **Canceling \(xy\)** (assuming \(xy \neq 0\)):
\[ \frac{1}{\lambda} = \frac{\lambda}{4} \]

6. **Cross-multiplying**:
\[ 4 = \lambda^2 \]

7. **Finding \(\lambda\)**:
Taking the square root of both sides gives:
\[ \lambda = \pm 2 \]

### Final Answer:
Thus, the values of \(\lambda\) are:
\[ \lambda = 2 \quad \text{or} \quad \lambda = -2 \]