The combined equation of the two lines `ax+by+c=0` and `a'x+b'y+c'=0` can be written as `(ax+by+c)(a'x+b'y+c')=0`.The equation of the angle bisectors of the lines represented by the equation `2x^(2)+xy-3y^(2)=0` is

Options

A. `3x^(2)+5xy+2y^(2)=0`
B. `x^(2)-y^(2)-10xy=0`
C. `3x^(2)+xy-2y^(2)=0`
D. `x^(2)-y^(2)+10xy=0`

Answer

To find the equation of the angle bisectors of the lines represented by the equation \(2x^2 + xy - 3y^2 = 0\), we will follow these steps:

### Step 1: Factor the given quadratic equation
The given equation \(2x^2 + xy - 3y^2 = 0\) can be factored into two linear factors. We can rewrite it in the form of \((L_1)(L_2) = 0\), where \(L_1\) and \(L_2\) are the equations of the lines.

To factor, we can look for two numbers that multiply to \(2 \times -3 = -6\) and add to \(1\) (the coefficient of \(xy\)). The numbers \(3\) and \(-2\) satisfy this condition.

Thus, we can rewrite the equation as:
\[ 2x^2 + 3xy - 2xy - 3y^2 = 0 \]
Grouping the terms:
\[ (2x^2 + 3xy) + (-2xy - 3y^2) = 0 \]
Factoring by grouping:
\[ x(2x + 3y) - y(2x + 3y) = 0 \]
This gives:
\[ (2x + 3y)(x - y) = 0 \]

### Step 2: Identify the lines
From the factored form, we have two lines:
1. \(2x + 3y = 0\) (Line 1)
2. \(x - y = 0\) (Line 2)

### Step 3: Find the angle bisectors
The angle bisectors of two lines given by \(L_1: A_1x + B_1y + C_1 = 0\) and \(L_2: A_2x + B_2y + C_2 = 0\) can be found using the formula:
\[ \frac{A_1x + B_1y + C_1}{\sqrt{A_1^2 + B_1^2}} = \pm \frac{A_2x + B_2y + C_2}{\sqrt{A_2^2 + B_2^2}} \]

For our lines:
- For Line 1: \(A_1 = 2\), \(B_1 = 3\), \(C_1 = 0\)
- For Line 2: \(A_2 = 1\), \(B_2 = -1\), \(C_2 = 0\)

Substituting into the angle bisector formula:
\[ \frac{2x + 3y}{\sqrt{2^2 + 3^2}} = \pm \frac{x - y}{\sqrt{1^2 + (-1)^2}} \]

Calculating the magnitudes:
\[ \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \]
\[ \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \]

Thus, we have:
\[ \frac{2x + 3y}{\sqrt{13}} = \pm \frac{x - y}{\sqrt{2}} \]

### Step 4: Cross-multiply and simplify
Cross-multiplying gives us two equations:
1. \(2x + 3y = \frac{\sqrt{13}}{\sqrt{2}}(x - y)\)
2. \(2x + 3y = -\frac{\sqrt{13}}{\sqrt{2}}(x - y)\)

### Step 5: Solve for the angle bisector equations
For the first equation:
\[ 2x + 3y = \frac{\sqrt{13}}{\sqrt{2}}x - \frac{\sqrt{13}}{\sqrt{2}}y \]
Rearranging gives:
\[ (2 - \frac{\sqrt{13}}{\sqrt{2}})x + (3 + \frac{\sqrt{13}}{\sqrt{2}})y = 0 \]

For the second equation:
\[ 2x + 3y = -\frac{\sqrt{13}}{\sqrt{2}}x + \frac{\sqrt{13}}{\sqrt{2}}y \]
Rearranging gives:
\[ (2 + \frac{\sqrt{13}}{\sqrt{2}})x + (3 - \frac{\sqrt{13}}{\sqrt{2}})y = 0 \]

### Final Answer
Thus, the equations of the angle bisectors are:
1. \((2 - \frac{\sqrt{13}}{\sqrt{2}})x + (3 + \frac{\sqrt{13}}{\sqrt{2}})y = 0\)
2. \((2 + \frac{\sqrt{13}}{\sqrt{2}})x + (3 - \frac{\sqrt{13}}{\sqrt{2}})y = 0\)