To solve the problem step by step, we will follow the reasoning outlined in the video transcript while providing detailed explanations for each step.
### Step 1: Identify the Given Information
We have:
- Point A = (1, 2)
- Perpendicular bisector of AB is given by the line \( y = x \).
- Perpendicular bisector of angle C is given by the line \( x - 2y + 1 = 0 \).
- The equation of line BC is given by \( ax + by - 5 = 0 \).
### Step 2: Find the Equation of Line AB
Since the line \( y = x \) is the perpendicular bisector of AB, the slope of line AB must be the negative reciprocal of the slope of \( y = x \).
- The slope of \( y = x \) is 1, so the slope of AB is \( -1 \).
Using point A (1, 2) and the slope (-1), we can find the equation of line AB using the point-slope form:
\[
y - y_1 = m(x - x_1)
\]
Substituting \( y_1 = 2 \), \( m = -1 \), and \( x_1 = 1 \):
\[
y - 2 = -1(x - 1)
\]
Simplifying:
\[
y - 2 = -x + 1 \implies x + y - 3 = 0
\]
### Step 3: Find the Midpoint E of AB
Since E is the midpoint of AB, we can denote the coordinates of point B as \( (x_B, y_B) \). The coordinates of E can be expressed as:
\[
E = \left( \frac{1 + x_B}{2}, \frac{2 + y_B}{2} \right)
\]
Since E lies on the line \( y = x \), we have:
\[
\frac{2 + y_B}{2} = \frac{1 + x_B}{2}
\]
This simplifies to:
\[
2 + y_B = 1 + x_B \implies y_B = x_B - 1
\]
### Step 4: Find the Coordinates of Point B
Now substituting \( y_B = x_B - 1 \) into the equation of line AB:
\[
x + (x_B - 1) - 3 = 0 \implies 2x_B - 4 = 0 \implies x_B = 2
\]
Thus, substituting \( x_B = 2 \) back, we find:
\[
y_B = 2 - 1 = 1
\]
So, point B is \( (2, 1) \).
### Step 5: Find the Coordinates of Point C
Next, we need to find point C using the perpendicular bisector of angle C, given by \( x - 2y + 1 = 0 \).
### Step 6: Find the Slope of Line AC
The slope of line AC can be determined since BF is perpendicular to AC. The slope of line BF can be derived from the equation \( x - 2y + 1 = 0 \):
\[
2y = x + 1 \implies y = \frac{1}{2}x + \frac{1}{2}
\]
Thus, the slope of BF is \( \frac{1}{2} \), making the slope of AC:
\[
\text{slope of AC} = -\frac{1}{\frac{1}{2}} = -2
\]
### Step 7: Find the Equation of Line AC
Using point A (1, 2) and the slope (-2):
\[
y - 2 = -2(x - 1) \implies y - 2 = -2x + 2 \implies 2x + y - 4 = 0
\]
### Step 8: Find the Coordinates of Point F
Now we can find the intersection of the lines \( 2x + y - 4 = 0 \) and \( x - 2y + 1 = 0 \):
Substituting \( y = 2x - 4 \) into \( x - 2y + 1 = 0 \):
\[
x - 2(2x - 4) + 1 = 0 \implies x - 4x + 8 + 1 = 0 \implies -3x + 9 = 0 \implies x = 3
\]
Substituting \( x = 3 \) back into \( y = 2x - 4 \):
\[
y = 2(3) - 4 = 2
\]
Thus, point F is \( (3, 2) \).
### Step 9: Find the Coordinates of Point C
Using the midpoint formula for F:
\[
F = \left( \frac{1 + x_C}{2}, \frac{2 + y_C}{2} \right) = (3, 2)
\]
This gives us:
\[
\frac{1 + x_C}{2} = 3 \implies 1 + x_C = 6 \implies x_C = 5
\]
\[
\frac{2 + y_C}{2} = 2 \implies 2 + y_C = 4 \implies y_C = 2
\]
Thus, point C is \( (5, 2) \).
### Step 10: Find the Equation of Line BC
Now we can find the slope of line BC:
\[
\text{slope of BC} = \frac{y_C - y_B}{x_C - x_B} = \frac{2 - 1}{5 - 2} = \frac{1}{3}
\]
Using point B (2, 1) to find the equation:
\[
y - 1 = \frac{1}{3}(x - 2) \implies 3y - 3 = x - 2 \implies x - 3y + 1 = 0
\]
### Step 11: Compare with the Given Equation
The equation of line BC is \( x - 3y + 1 = 0 \), which can be rewritten as:
\[
1x - 3y - (-1) = 0
\]
Thus, \( a = 1 \) and \( b = -3 \).
### Step 12: Calculate \( a - 2b \)
Now we compute:
\[
a - 2b = 1 - 2(-3) = 1 + 6 = 7
\]
### Final Answer
The value of \( a - 2b \) is \( 7 \).
