Find the equations of the straight lines , bisectors of the angles formed by the following pairs of lines
`y-b=(2m)/(1-m^(2))(x-a)` and `y-b=(2m')/(1-m' ^(2))(x-a)`

Answer

To find the equations of the straight lines that bisect the angles formed by the given pairs of lines, we will follow these steps:

### Step 1: Write the equations of the given lines
The given lines are:
1. \( y - b = \frac{2m}{1 - m^2}(x - a) \)
2. \( y - b = \frac{2m'}{1 - m'^2}(x - a) \)

### Step 2: Rearrange the equations into standard form
For the first line:
\[ y - b = \frac{2m}{1 - m^2}(x - a) \]
Multiplying through by \(1 - m^2\):
\[ (1 - m^2)(y - b) = 2m(x - a) \]
Expanding:
\[ (1 - m^2)y - (1 - m^2)b = 2mx - 2ma \]
Rearranging gives:
\[ 2mx - (1 - m^2)y + (1 - m^2)b - 2ma = 0 \]

For the second line:
\[ y - b = \frac{2m'}{1 - m'^2}(x - a) \]
Multiplying through by \(1 - m'^2\):
\[ (1 - m'^2)(y - b) = 2m'(x - a) \]
Expanding:
\[ (1 - m'^2)y - (1 - m'^2)b = 2m'x - 2m'a \]
Rearranging gives:
\[ 2m'x - (1 - m'^2)y + (1 - m'^2)b - 2m'a = 0 \]

### Step 3: Identify coefficients for perpendicular distance
The equations of the lines can be expressed as:
1. \( A_1x + B_1y + C_1 = 0 \) where \( A_1 = 2m \), \( B_1 = -(1 - m^2) \), \( C_1 = (1 - m^2)b - 2ma \)
2. \( A_2x + B_2y + C_2 = 0 \) where \( A_2 = 2m' \), \( B_2 = -(1 - m'^2) \), \( C_2 = (1 - m'^2)b - 2m'a \)

### Step 4: Set up the condition for angle bisectors
The angle bisectors of the two lines can be found using the condition:
\[ \frac{A_1x + B_1y + C_1}{\sqrt{A_1^2 + B_1^2}} = \pm \frac{A_2x + B_2y + C_2}{\sqrt{A_2^2 + B_2^2}} \]

### Step 5: Cross-multiply and simplify
Cross-multiplying gives:
\[ (A_1x + B_1y + C_1)\sqrt{A_2^2 + B_2^2} = \pm (A_2x + B_2y + C_2)\sqrt{A_1^2 + B_1^2} \]

### Step 6: Solve for the equations of the bisectors
By simplifying the equation obtained from cross-multiplication, we can derive the equations of the angle bisectors.

### Final Result
The equations of the angle bisectors are:
1. \( (1 - mm')y - (1 - m^2)x + (b(1 - m^2) - 2ma) + (b(1 - m'^2) - 2m'a) = 0 \)
2. \( (1 - mm')y - (1 - m^2)x + (b(1 - m^2) - 2ma) - (b(1 - m'^2) - 2m'a) = 0 \)