In a triangle ABC , if the equation of sides AB,BC and CA are `2x- y + 4 = 0 , x - 2y - 1 = 0 and x + 3y - 3 = 0 ` respectively ,
The equation of external bisector of angle B is

Options

A. `x-y-1 =0`
B. `x-y+1=0`
C. `x+y-5=0`
D. `x+y+5=0`

Answer

To find the equation of the external bisector of angle B in triangle ABC, we will use the equations of the sides AB, BC, and CA given as follows:

1. **Equation of AB**: \(2x - y + 4 = 0\)
2. **Equation of BC**: \(x - 2y - 1 = 0\)
3. **Equation of CA**: \(x + 3y - 3 = 0\)

### Step 1: Identify coefficients from the equations of sides AB and BC

From the equation of side AB \(2x - y + 4 = 0\):
- \(a_1 = 2\)
- \(b_1 = -1\)
- \(c_1 = 4\)

From the equation of side BC \(x - 2y - 1 = 0\):
- \(a_2 = 1\)
- \(b_2 = -2\)
- \(c_2 = -1\)

### Step 2: Check the sign for the external angle bisector

To determine whether to use the positive or negative sign in the angle bisector formula, we calculate \(a_1a_2 + b_1b_2\):
\[ a_1a_2 + b_1b_2 = (2)(1) + (-1)(-2) = 2 + 2 = 4 \]
Since \(4 > 0\), we will use the positive sign in the angle bisector formula.

### Step 3: Apply the angle bisector formula

The formula for the angle bisector is given by:
\[ \frac{a_1x + b_1y + c_1}{\sqrt{a_1^2 + b_1^2}} = \frac{a_2x + b_2y + c_2}{\sqrt{a_2^2 + b_2^2}} \]

Substituting the values we have:
\[ \frac{2x - y + 4}{\sqrt{2^2 + (-1)^2}} = \frac{x - 2y - 1}{\sqrt{1^2 + (-2)^2}} \]

Calculating the denominators:
\[ \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \]
\[ \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5} \]

Thus, the equation simplifies to:
\[ \frac{2x - y + 4}{\sqrt{5}} = \frac{x - 2y - 1}{\sqrt{5}} \]

### Step 4: Clear the denominators

Multiplying both sides by \(\sqrt{5}\):
\[ 2x - y + 4 = x - 2y - 1 \]

### Step 5: Rearranging the equation

Rearranging the equation gives:
\[ 2x - y + 4 - x + 2y + 1 = 0 \]
\[ x + y + 5 = 0 \]

### Final Result

The equation of the external bisector of angle B is:
\[ x + y + 5 = 0 \]