The co-ordinates of A, B, C are respectively (-4, 0), (0, 2) and (-3, 2). Find the co-ordinates of the point of intersection of the line which bisects the angle CAB internally and the line joining C to the middle point of AB is

Answer

To find the coordinates of the point of intersection of the angle bisector of angle CAB and the line joining point C to the midpoint of line segment AB, we will follow these steps:

### Step 1: Identify the coordinates of points A, B, and C
Given:
- A = (-4, 0)
- B = (0, 2)
- C = (-3, 2)

### Step 2: Calculate the midpoint D of line segment AB
The midpoint D of a line segment joining points (x1, y1) and (x2, y2) is given by:
\[ D = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]
Substituting the coordinates of A and B:
\[ D = \left( \frac{-4 + 0}{2}, \frac{0 + 2}{2} \right) = \left( -2, 1 \right) \]

### Step 3: Find the slope of line segment AB
The slope \( m \) of a line through points (x1, y1) and (x2, y2) is given by:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
For points A and B:
\[ m_{AB} = \frac{2 - 0}{0 - (-4)} = \frac{2}{4} = \frac{1}{2} \]

### Step 4: Find the equation of line AB
Using the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
Using point A (-4, 0):
\[ y - 0 = \frac{1}{2}(x + 4) \]
This simplifies to:
\[ y = \frac{1}{2}x + 2 \]
Rearranging gives:
\[ x - 2y + 4 = 0 \quad \text{(Equation 1)} \]

### Step 5: Find the slope of line segment AC
For points A and C:
\[ m_{AC} = \frac{2 - 0}{-3 - (-4)} = \frac{2}{1} = 2 \]

### Step 6: Find the equation of line AC
Using point A (-4, 0):
\[ y - 0 = 2(x + 4) \]
This simplifies to:
\[ y = 2x + 8 \]
Rearranging gives:
\[ 2x - y + 8 = 0 \quad \text{(Equation 2)} \]

### Step 7: Find the equation of the angle bisector of angle CAB
Using the angle bisector theorem, we can derive the equation:
\[ \frac{x - 2y + 4}{\sqrt{1^2 + (-2)^2}} = \pm \frac{2x - y + 8}{\sqrt{2^2 + (-1)^2}} \]
This simplifies to:
\[ \frac{x - 2y + 4}{\sqrt{5}} = \pm \frac{2x - y + 8}{\sqrt{5}} \]
Thus, we have two cases to solve:
1. \( x - 2y + 4 = 2x - y + 8 \)
2. \( x - 2y + 4 = - (2x - y + 8) \)

### Step 8: Solve the first case
From \( x - 2y + 4 = 2x - y + 8 \):
\[ -x + y - 4 = 0 \quad \Rightarrow \quad x - y + 4 = 0 \quad \text{(Equation 3)} \]

### Step 9: Solve the second case
From \( x - 2y + 4 = - (2x - y + 8) \):
\[ 3x - y + 12 = 0 \quad \Rightarrow \quad 3x - y + 12 = 0 \quad \text{(Equation 4)} \]

### Step 10: Find the equation of line CD
Using points C (-3, 2) and D (-2, 1):
\[ m_{CD} = \frac{1 - 2}{-2 + 3} = \frac{-1}{1} = -1 \]
Using point C (-3, 2):
\[ y - 2 = -1(x + 3) \]
This simplifies to:
\[ y = -x - 1 \]
Rearranging gives:
\[ x + y + 1 = 0 \quad \text{(Equation 5)} \]

### Step 11: Solve for intersection of angle bisector and line CD
Substituting Equation 5 into Equation 3:
\[ x - (-x - 1) + 4 = 0 \]
This simplifies to:
\[ 2x + 5 = 0 \quad \Rightarrow \quad x = -\frac{5}{2} \]
Substituting \( x \) back into Equation 5:
\[ -\frac{5}{2} + y + 1 = 0 \quad \Rightarrow \quad y = \frac{3}{2} \]

### Final Answer
The coordinates of the point of intersection are:
\[ \left( -\frac{5}{2}, \frac{3}{2} \right) \]