The equation of the external bisector of `angleBAC" to "DeltaABC` with vertices `A(5, 2), B(2, 3) and C(6, 5)` is

Options

A. `2x+y+12=0`
B. `x+2y-12=0`
C. `2x+y-12=0`
D. `x-2y-1=0`

Answer

To find the equation of the external bisector of angle \( BAC \) in triangle \( ABC \) with vertices \( A(5, 2) \), \( B(2, 3) \), and \( C(6, 5) \), we can follow these steps:

### Step 1: Calculate the lengths of sides \( AB \) and \( AC \)

Using the distance formula:
\[ AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
For \( AB \):
\[ AB = \sqrt{(2 - 5)^2 + (3 - 2)^2} = \sqrt{(-3)^2 + (1)^2} = \sqrt{9 + 1} = \sqrt{10} \]

For \( AC \):
\[ AC = \sqrt{(6 - 5)^2 + (5 - 2)^2} = \sqrt{(1)^2 + (3)^2} = \sqrt{1 + 9} = \sqrt{10} \]

### Step 2: Identify that triangle \( ABC \) is isosceles

Since \( AB = AC \), triangle \( ABC \) is isosceles.

### Step 3: Find the midpoint \( D \) of segment \( BC \)

The coordinates of the midpoint \( D \) can be calculated as:
\[ D\left( \frac{x_B + x_C}{2}, \frac{y_B + y_C}{2} \right) = D\left( \frac{2 + 6}{2}, \frac{3 + 5}{2} \right) = D(4, 4) \]

### Step 4: Calculate the slope of line \( AD \)

Using the slope formula:
\[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} \]
For points \( A(5, 2) \) and \( D(4, 4) \):
\[ \text{slope of } AD = \frac{4 - 2}{4 - 5} = \frac{2}{-1} = -2 \]

### Step 5: Find the slope of the external bisector

The external bisector is perpendicular to the internal bisector. Therefore, if the slope of \( AD \) is \( -2 \), the slope of the external bisector is:
\[ \text{slope of external bisector} = \frac{1}{\text{slope of } AD} = \frac{1}{-2} = \frac{1}{2} \]

### Step 6: Write the equation of the external bisector

Using the point-slope form of the line equation:
\[ y - y_1 = m(x - x_1) \]
Substituting \( A(5, 2) \) and the slope \( \frac{1}{2} \):
\[ y - 2 = \frac{1}{2}(x - 5) \]

### Step 7: Rearranging the equation

Expanding and rearranging:
\[ y - 2 = \frac{1}{2}x - \frac{5}{2} \]
\[ 2y - 4 = x - 5 \]
\[ x - 2y - 1 = 0 \]

### Final Equation

Thus, the equation of the external bisector of angle \( BAC \) is:
\[ x - 2y - 1 = 0 \]