To solve the problem, we need to find the equation of the perpendicular bisector of side AB of triangle ABC, given the midpoints of the sides.
### Step 1: Identify the Midpoints
We are given the midpoints of the sides of triangle ABC:
- Midpoint of BC: \( M_{BC} = (2, 1) \)
- Midpoint of CA: \( M_{CA} = (-1, -2) \)
- Midpoint of AB: \( M_{AB} = (3, -3) \)
### Step 2: Determine the Coordinates of Points A and B
Using the midpoint formula, we can express the coordinates of points A and B in terms of the midpoints.
Let:
- \( A(x_1, y_1) \)
- \( B(x_2, y_2) \)
From the midpoint of AB:
\[
M_{AB} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = (3, -3)
\]
This gives us two equations:
1. \( \frac{x_1 + x_2}{2} = 3 \) → \( x_1 + x_2 = 6 \) (Equation 1)
2. \( \frac{y_1 + y_2}{2} = -3 \) → \( y_1 + y_2 = -6 \) (Equation 2)
### Step 3: Determine the Coordinates of Points C
Using the midpoints of CA and BC, we can express the coordinates of point C as well.
From the midpoint of CA:
\[
M_{CA} = \left( \frac{x_1 + x_3}{2}, \frac{y_1 + y_3}{2} \right) = (-1, -2)
\]
This gives us:
1. \( \frac{x_1 + x_3}{2} = -1 \) → \( x_1 + x_3 = -2 \) (Equation 3)
2. \( \frac{y_1 + y_3}{2} = -2 \) → \( y_1 + y_3 = -4 \) (Equation 4)
From the midpoint of BC:
\[
M_{BC} = \left( \frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2} \right) = (2, 1)
\]
This gives us:
1. \( \frac{x_2 + x_3}{2} = 2 \) → \( x_2 + x_3 = 4 \) (Equation 5)
2. \( \frac{y_2 + y_3}{2} = 1 \) → \( y_2 + y_3 = 2 \) (Equation 6)
### Step 4: Solve the System of Equations
Now we have six equations (1 to 6) to solve for \( x_1, y_1, x_2, y_2, x_3, y_3 \).
From Equations 1 and 2:
- \( x_2 = 6 - x_1 \)
- \( y_2 = -6 - y_1 \)
Substituting \( x_2 \) and \( y_2 \) into Equations 5 and 6:
1. \( (6 - x_1) + x_3 = 4 \) → \( x_3 = -2 + x_1 \) (Equation 7)
2. \( (-6 - y_1) + y_3 = 2 \) → \( y_3 = 8 + y_1 \) (Equation 8)
Now substitute \( x_3 \) and \( y_3 \) from Equations 7 and 8 into Equations 3 and 4:
1. \( x_1 + (-2 + x_1) = -2 \) → \( 2x_1 - 2 = -2 \) → \( 2x_1 = 0 \) → \( x_1 = 0 \)
2. \( y_1 + (8 + y_1) = -4 \) → \( 2y_1 + 8 = -4 \) → \( 2y_1 = -12 \) → \( y_1 = -6 \)
Now substitute \( x_1 \) and \( y_1 \) back to find \( x_2, y_2, x_3, y_3 \):
- From Equation 1: \( x_2 = 6 - 0 = 6 \)
- From Equation 2: \( y_2 = -6 - (-6) = 0 \)
- From Equation 7: \( x_3 = -2 + 0 = -2 \)
- From Equation 8: \( y_3 = 8 + (-6) = 2 \)
Thus, we have:
- \( A(0, -6) \)
- \( B(6, 0) \)
- \( C(-2, 2) \)
### Step 5: Find the Slope of Line AB
The slope \( m_{AB} \) of line AB can be calculated as:
\[
m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - (-6)}{6 - 0} = \frac{6}{6} = 1
\]
### Step 6: Find the Slope of the Perpendicular Bisector
The slope of the perpendicular bisector \( m_{perpendicular} \) is the negative reciprocal of \( m_{AB} \):
\[
m_{perpendicular} = -\frac{1}{m_{AB}} = -1
\]
### Step 7: Find the Midpoint of AB
The midpoint \( M_{AB} \) is already given as \( (3, -3) \).
### Step 8: Write the Equation of the Perpendicular Bisector
Using the point-slope form of the equation of a line:
\[
y - y_1 = m(x - x_1)
\]
Substituting \( (x_1, y_1) = (3, -3) \) and \( m = -1 \):
\[
y + 3 = -1(x - 3)
\]
Simplifying:
\[
y + 3 = -x + 3
\]
\[
x + y = 0
\]
### Step 9: Convert to Standard Form
The equation can be written in the form \( ax + by + c = 0 \):
\[
1x + 1y + 0 = 0
\]
Thus, \( a = 1, b = 1, c = 0 \).
### Step 10: Calculate \( a + b + c \)
\[
a + b + c = 1 + 1 + 0 = 2
\]
### Final Answer
The value of \( a + b + c \) is **2**.
