In a triangle, ABC, the equation of the perpendicular bisector of AC is `3x - 2y + 8 = 0`. If the coordinates of the points A and B are `(1, -1) & (3, 1)` respectively, then the equation of the line BC & the centre of the circum-circle of the triangle ABC will be

Answer

To solve the problem, we need to find the equation of line BC and the center of the circumcircle of triangle ABC, given the coordinates of points A and B, and the equation of the perpendicular bisector of AC.

### Step 1: Identify the coordinates of points A and B
We are given:
- Point A = (1, -1)
- Point B = (3, 1)

### Step 2: Find the equation of line AC
We have the equation of the perpendicular bisector of AC as \(3x - 2y + 8 = 0\).

To find the coordinates of point C, we need to find the midpoint of AC. Let the coordinates of point C be (x, y).

### Step 3: Find the midpoint of AC
The midpoint M of segment AC can be calculated using the midpoint formula:
\[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]
For points A(1, -1) and C(x, y), the midpoint M is:
\[ M = \left( \frac{1 + x}{2}, \frac{-1 + y}{2} \right) \]

### Step 4: Substitute the midpoint into the perpendicular bisector equation
Since M lies on the perpendicular bisector, we substitute the coordinates of M into the equation of the perpendicular bisector:
\[ 3\left(\frac{1 + x}{2}\right) - 2\left(\frac{-1 + y}{2}\right) + 8 = 0 \]
Multiplying through by 2 to eliminate the fractions:
\[ 3(1 + x) + 2(1 - y) + 16 = 0 \]
This simplifies to:
\[ 3 + 3x + 2 - 2y + 16 = 0 \implies 3x - 2y + 21 = 0 \]

### Step 5: Solve for y in terms of x
Rearranging gives:
\[ 2y = 3x + 21 \implies y = \frac{3}{2}x + \frac{21}{2} \]

### Step 6: Find the coordinates of point C
To find point C, we need to use the fact that point C lies on the line AC. We can find the equation of line AC using points A and C.

### Step 7: Find the slope of line AC
The slope \(m_{AC}\) can be calculated as:
\[ m_{AC} = \frac{y_C - y_A}{x_C - x_A} = \frac{y - (-1)}{x - 1} \]
Substituting \(y\) from the previous equation:
\[ m_{AC} = \frac{\frac{3}{2}x + \frac{21}{2} + 1}{x - 1} = \frac{\frac{3}{2}x + \frac{23}{2}}{x - 1} \]

### Step 8: Find the equation of line BC
Using points B(3, 1) and C(x, y), the slope \(m_{BC}\) is:
\[ m_{BC} = \frac{y - 1}{x - 3} \]
We can find the equation of line BC using the point-slope form:
\[ y - 1 = m_{BC}(x - 3) \]

### Step 9: Find the circumcenter of triangle ABC
The circumcenter can be found by solving the equations of the perpendicular bisectors of two sides of the triangle. We already have the perpendicular bisector of AC. We need to find the perpendicular bisector of BC.

### Step 10: Solve the equations to find the circumcenter
To find the circumcenter, we solve the equations of the two perpendicular bisectors simultaneously.

### Final Answer
1. The equation of line BC is \(x + 4y = 7\).
2. The coordinates of the circumcenter of triangle ABC are \(\left(\frac{4}{5}, \frac{26}{5}\right)\).