The equation of the lines represented by `x^(2)+2xy-y^(2)=0` are

Options

A. `(1+sqrt(2))x-y=0 and (1-sqrt(2))x+y=0`
B. `(1+sqrt(2))x-y=0 and (1-sqrt(2))x-y=0`
C. `(sqrt(2)+1)x-y=0 and (sqrt(2)-1)x-y=0`
D. `(1+sqrt(2))x-y=0 and (sqrt(2)-1)x-y=0`

Answer

To solve the equation of the lines represented by \( x^2 + 2xy - y^2 = 0 \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ x^2 + 2xy - y^2 = 0 \]

### Step 2: Divide by \( -x^2 \)
To simplify the equation, we divide the entire equation by \( -x^2 \):
\[ \frac{x^2}{-x^2} + \frac{2xy}{-x^2} - \frac{y^2}{-x^2} = 0 \]
This simplifies to:
\[ -1 - \frac{2y}{x} + \frac{y^2}{x^2} = 0 \]
Rearranging gives:
\[ \frac{y^2}{x^2} - \frac{2y}{x} - 1 = 0 \]

### Step 3: Substitute \( \frac{y}{x} = m \)
Let \( \frac{y}{x} = m \). Then, we can rewrite the equation as:
\[ m^2 - 2m - 1 = 0 \]

### Step 4: Solve the quadratic equation
Now we solve the quadratic equation \( m^2 - 2m - 1 = 0 \) using the quadratic formula:
\[ m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -2, c = -1 \). Plugging in these values:
\[ m = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]
\[ m = \frac{2 \pm \sqrt{4 + 4}}{2} \]
\[ m = \frac{2 \pm \sqrt{8}}{2} \]
\[ m = \frac{2 \pm 2\sqrt{2}}{2} \]
\[ m = 1 \pm \sqrt{2} \]

### Step 5: Write the equations of the lines
Now, we have two values for \( m \):
1. \( m_1 = 1 + \sqrt{2} \)
2. \( m_2 = 1 - \sqrt{2} \)

Thus, the equations of the lines can be written as:
1. \( y = (1 + \sqrt{2})x \)
2. \( y = (1 - \sqrt{2})x \)

### Final Answer
The equations of the lines represented by \( x^2 + 2xy - y^2 = 0 \) are:
1. \( y = (1 + \sqrt{2})x \)
2. \( y = (1 - \sqrt{2})x \)

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