The equations of perpendicular bisectors of the sides AB and AC of a triangle ABC are `x-y+5=0` and `x+2y=0` respectively
If the point A is (1,-2) the equation of the line BC is

Options

A. `23x+4y-40=0`
B. `23x+14y+40=0`
C. `14x+23y-40=0`
D. `14x+23y+40=0`

Answer

To find the equation of the line BC given the perpendicular bisectors of the sides AB and AC of triangle ABC, we can follow these steps:

### Step 1: Identify the Given Information
We have the following:
- The coordinates of point A: \( A(1, -2) \)
- The equations of the perpendicular bisectors:
- Perpendicular bisector of AB: \( x - y + 5 = 0 \)
- Perpendicular bisector of AC: \( x + 2y = 0 \)

### Step 2: Find the Slope of the Perpendicular Bisectors
1. For the first equation \( x - y + 5 = 0 \):
- Rearranging gives \( y = x + 5 \).
- The slope \( m_1 = 1 \).

2. For the second equation \( x + 2y = 0 \):
- Rearranging gives \( y = -\frac{1}{2}x \).
- The slope \( m_2 = -\frac{1}{2} \).

### Step 3: Find the Slopes of Lines AB and AC
Since the perpendicular bisectors are perpendicular to the sides of the triangle:
- The slope of line AB, \( m_{AB} = -\frac{1}{m_1} = -1 \).
- The slope of line AC, \( m_{AC} = -\frac{1}{m_2} = 2 \).

### Step 4: Write the Equations of Lines AB and AC
1. For line AB using point A(1, -2) and slope \( m_{AB} = -1 \):
\[ y - (-2) = -1(x - 1) \implies y + 2 = -x + 1 \implies y + x + 1 = 0 \]

2. For line AC using point A(1, -2) and slope \( m_{AC} = 2 \):
\[ y - (-2) = 2(x - 1) \implies y + 2 = 2x - 2 \implies y - 2x + 4 = 0 \]

### Step 5: Find the Coordinates of Points B and C
To find the coordinates of points B and C, we need to find the intersection points of the lines AB and AC with their respective perpendicular bisectors.

1. **Finding Point D (intersection of AB and the perpendicular bisector of AB)**:
- Solve the equations:
\[ y + x + 1 = 0 \quad \text{(1)} \]
\[ x - y + 5 = 0 \quad \text{(2)} \]
- From (2), \( y = x + 5 \).
- Substitute into (1):
\[ x + (x + 5) + 1 = 0 \implies 2x + 6 = 0 \implies x = -3 \]
\[ y = -3 + 5 = 2 \]
- Thus, \( D(-3, 2) \).

2. **Finding Point E (intersection of AC and the perpendicular bisector of AC)**:
- Solve the equations:
\[ y - 2x + 4 = 0 \quad \text{(3)} \]
\[ x + 2y = 0 \quad \text{(4)} \]
- From (4), \( y = -\frac{1}{2}x \).
- Substitute into (3):
\[ -\frac{1}{2}x - 2x + 4 = 0 \implies -\frac{5}{2}x + 4 = 0 \implies x = \frac{8}{5} \]
\[ y = -\frac{1}{2} \cdot \frac{8}{5} = -\frac{4}{5} \]
- Thus, \( E\left(\frac{8}{5}, -\frac{4}{5}\right) \).

### Step 6: Find the Equation of Line BC
Now we can find the equation of line BC using points B and C (which are points D and E).

1. **Finding the slope of line BC**:
- Coordinates of D: \( (-3, 2) \)
- Coordinates of E: \( \left(\frac{8}{5}, -\frac{4}{5}\right) \)
- Slope \( m_{BC} = \frac{-\frac{4}{5} - 2}{\frac{8}{5} + 3} = \frac{-\frac{4}{5} - \frac{10}{5}}{\frac{8}{5} + \frac{15}{5}} = \frac{-\frac{14}{5}}{\frac{23}{5}} = -\frac{14}{23} \).

2. **Using point-slope form to write the equation of line BC**:
\[ y - 2 = -\frac{14}{23}(x + 3) \]
Rearranging gives:
\[ 23y - 46 = -14x - 42 \implies 14x + 23y - 4 = 0 \]

### Final Answer
The equation of line BC is:
\[ 14x + 23y - 4 = 0 \]