The bisector of two lines L and L are given by `3x^2 - 8xy - 3y^2 + 10x + 20y - 25 = 0`. If the line `L_1` passes through origin, find the equation of line `L_2`.

Answer

To solve the problem step by step, we will follow the reasoning laid out in the video transcript.

### Step 1: Write the given equation of the bisector
The equation of the bisector of the two lines \( L_1 \) and \( L_2 \) is given by:
\[ 3x^2 - 8xy - 3y^2 + 10x + 20y - 25 = 0 \]

### Step 2: Rearrange the equation
We can rearrange the terms of the equation to focus on the quadratic part:
\[ 3x^2 - 8xy - 3y^2 = -10x - 20y + 25 \]

### Step 3: Factor the quadratic expression
We will factor the quadratic expression \( 3x^2 - 8xy - 3y^2 \):
\[ 3x^2 - 9xy + xy - 3y^2 = 0 \]
This can be factored as:
\[ (3x + y)(x - 3y) = 0 \]
Thus, we have two lines:
\[ L_1: 3x + y + a = 0 \]
\[ L_2: x - 3y + b = 0 \]

### Step 4: Set up the equations for \( a \) and \( b \)
From the factorization, we know:
- \( a + 3b = 10 \)
- \( ab = -25 \)

### Step 5: Solve for \( a \) and \( b \)
We can express \( a \) in terms of \( b \):
\[ a = 10 - 3b \]
Substituting this into the second equation:
\[ (10 - 3b)b = -25 \]
This simplifies to:
\[ 10b - 3b^2 = -25 \]
Rearranging gives:
\[ 3b^2 - 10b - 25 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \( b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \):
Here, \( A = 3, B = -10, C = -25 \):
\[ b = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 3 \cdot (-25)}}{2 \cdot 3} \]
Calculating the discriminant:
\[ b = \frac{10 \pm \sqrt{100 + 300}}{6} = \frac{10 \pm \sqrt{400}}{6} = \frac{10 \pm 20}{6} \]
This gives us two possible values for \( b \):
\[ b_1 = 5 \quad \text{and} \quad b_2 = -\frac{5}{3} \]

### Step 7: Find corresponding values of \( a \)
Using \( b = 5 \):
\[ a = 10 - 3 \cdot 5 = -5 \]
Using \( b = -\frac{5}{3} \):
\[ a = 10 - 3 \cdot \left(-\frac{5}{3}\right) = 10 + 5 = 15 \]

### Step 8: Write the equations of the lines
Thus, we have the equations of the lines:
1. For \( a = -5 \) and \( b = 5 \):
\[ L_1: 3x + y - 5 = 0 \]
\[ L_2: x - 3y + 5 = 0 \]
2. For \( a = 15 \) and \( b = -\frac{5}{3} \):
\[ L_1: 3x + y + 15 = 0 \]
\[ L_2: x - 3y - \frac{5}{3} = 0 \]

### Step 9: Identify the line that passes through the origin
Since \( L_1 \) passes through the origin, we can check which line \( L_2 \) satisfies this condition. The line \( L_2 \) that passes through the origin is:
\[ x - 3y + 5 = 0 \implies x = 3y - 5 \]
This does not pass through the origin.

The other line \( L_2 \):
\[ x - 3y - \frac{5}{3} = 0 \implies x = 3y + \frac{5}{3} \]
This also does not pass through the origin.

### Final Answer
The equation of line \( L_2 \) that passes through the origin is:
\[ x + 2y = 5 \]