The equation(s) of the bisectors(s) of that angles between the lines `x+2y-11=0,3x-6y-5=0` which contains the point (1,-3) is

Options

A. `3x=19`
B. `3y=7`
C. `3x=19` and `3y=7`
D. None of these

Answer

To find the equations of the angle bisectors of the lines given by the equations \( L_1: x + 2y - 11 = 0 \) and \( L_2: 3x - 6y - 5 = 0 \) that contain the point \( (1, -3) \), we can follow these steps:

### Step 1: Rewrite the equations in slope-intercept form

First, we need to express both lines in slope-intercept form (y = mx + b).

For \( L_1: x + 2y - 11 = 0 \):
\[ 2y = -x + 11 \implies y = -\frac{1}{2}x + \frac{11}{2} \]

For \( L_2: 3x - 6y - 5 = 0 \):
\[ -6y = -3x + 5 \implies y = \frac{1}{2}x - \frac{5}{6} \]

### Step 2: Find the slopes of the lines

From the equations, we can identify the slopes:
- Slope of \( L_1 \) (m1) = -1/2
- Slope of \( L_2 \) (m2) = 1/2

### Step 3: Find the angle bisectors

The angle bisectors can be found using the formula:
\[ \frac{y - y_1}{y - y_2} = \frac{m_1 - m_2}{1 + m_1 m_2} \]
where \( (x_1, y_1) \) and \( (x_2, y_2) \) are points on the lines.

However, a more straightforward approach is to use the formula for the angle bisectors directly:
\[ \frac{y - y_1}{y - y_2} = \frac{m_1 - m_2}{1 + m_1 m_2} \]

### Step 4: Substitute the point (1, -3)

We will use the point \( (1, -3) \) to find the specific angle bisector that contains this point.

### Step 5: Use the angle bisector formula

The angle bisector equations can be derived from the original line equations:
\[ \frac{x + 2y - 11}{\sqrt{1^2 + 2^2}} = \pm \frac{3x - 6y - 5}{\sqrt{3^2 + (-6)^2}} \]

Calculating the lengths:
\[ \sqrt{1^2 + 2^2} = \sqrt{5}, \quad \sqrt{3^2 + (-6)^2} = \sqrt{45} = 3\sqrt{5} \]

Thus, we have:
\[ \frac{x + 2y - 11}{\sqrt{5}} = \pm \frac{3x - 6y - 5}{3\sqrt{5}} \]

### Step 6: Simplifying the equation

This gives us two equations:
1. \( x + 2y - 11 = \frac{1}{3}(3x - 6y - 5) \)
2. \( x + 2y - 11 = -\frac{1}{3}(3x - 6y - 5) \)

### Step 7: Solve each equation

1. For the first equation:
\[ x + 2y - 11 = x - 2y + \frac{5}{3} \]
After simplifying, we find:
\[ 4y = 11 + \frac{5}{3} \implies y = \frac{38}{12} = \frac{19}{6} \]

2. For the second equation:
\[ x + 2y - 11 = -x + 2y + \frac{5}{3} \]
After simplifying, we find:
\[ 2x = 11 + \frac{5}{3} \implies x = \frac{38}{6} = \frac{19}{3} \]

### Step 8: Final equations

After substituting back, we can derive the final equations of the angle bisectors.