The vertices of a triangle ABC are (1,1),(4,-2) and (5,5) respectively. Then equation of perpendiculart dropped from C to the internal bisector of angle A is

Options

A. `y-5=0`
B. `x-5=0`
C. `2x+3y-7=0`
D. none

Answer

To find the equation of the perpendicular dropped from point C to the internal bisector of angle A in triangle ABC with vertices A(1, 1), B(4, -2), and C(5, 5), we follow these steps:

### Step 1: Determine the slopes of sides AB and AC
- **Slope of AB (m1)**:
\[ m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-2 - 1}{4 - 1} = \frac{-3}{3} = -1 \]

- **Slope of AC (m2)**:
\[ m_{AC} = \frac{y_3 - y_1}{x_3 - x_1} = \frac{5 - 1}{5 - 1} = \frac{4}{4} = 1 \]

### Step 2: Find the angle bisector's slope
The angle bisector of angle A can be found using the formula for the slope of the angle bisector between two lines. The angle bisector's slope (m) can be computed using the formula:
\[ m = \frac{m_1 + m_2}{1 - m_1 m_2} \]
Substituting the values:
\[ m = \frac{-1 + 1}{1 - (-1)(1)} = \frac{0}{2} = 0 \]

### Step 3: Determine the equation of the angle bisector
Since the slope of the angle bisector is 0, it is a horizontal line. The equation of the line passing through point A(1, 1) with slope 0 is:
\[ y = 1 \]

### Step 4: Find the slope of the perpendicular from C to the angle bisector
The slope of the perpendicular line to a line with slope m is given by:
\[ m_{perpendicular} = -\frac{1}{m} \]
Since the slope of the angle bisector is 0, the slope of the perpendicular line from C will be undefined, indicating a vertical line.

### Step 5: Write the equation of the perpendicular from C(5, 5)
The equation of a vertical line passing through C(5, 5) is:
\[ x = 5 \]

### Step 6: Write the final equations
The equations of the perpendicular dropped from C to the internal bisector of angle A are:
1. \( y = 5 \) (horizontal line)
2. \( x = 5 \) (vertical line)

### Summary of the solution:
The equations of the perpendicular dropped from C to the internal bisector of angle A are:
- \( y = 5 \)
- \( x = 5 \)