Find the equation of thebisector of the angle between the lines `2x-3y - 5 = 0 and 6x - 4y + 7 = 0 ` which is the supplement of the angle containing the point `(2,-1)`

Answer

To find the equation of the bisector of the angle between the lines \(2x - 3y - 5 = 0\) and \(6x - 4y + 7 = 0\) that is the supplement of the angle containing the point \((2, -1)\), we can follow these steps:

### Step 1: Identify the equations of the lines
We have two lines given by:
1. \(L_1: 2x - 3y - 5 = 0\)
2. \(L_2: 6x - 4y + 7 = 0\)

### Step 2: Write the angle bisector formula
The angle bisector of two lines can be found using the formula:
\[ \frac{A_1x + B_1y + C_1}{\sqrt{A_1^2 + B_1^2}} = \pm \frac{A_2x + B_2y + C_2}{\sqrt{A_2^2 + B_2^2}} \]
where \(A_1, B_1, C_1\) are the coefficients from the first line, and \(A_2, B_2, C_2\) are the coefficients from the second line.

For our lines:
- From \(L_1\): \(A_1 = 2\), \(B_1 = -3\), \(C_1 = -5\)
- From \(L_2\): \(A_2 = 6\), \(B_2 = -4\), \(C_2 = 7\)

### Step 3: Substitute into the angle bisector formula
Substituting the values into the formula gives:
\[ \frac{2x - 3y - 5}{\sqrt{2^2 + (-3)^2}} = \pm \frac{6x - 4y + 7}{\sqrt{6^2 + (-4)^2}} \]
Calculating the denominators:
\[ \sqrt{2^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13} \]
\[ \sqrt{6^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13} \]
Thus, the equation becomes:
\[ \frac{2x - 3y - 5}{\sqrt{13}} = \pm \frac{6x - 4y + 7}{2\sqrt{13}} \]

### Step 4: Clear the denominators
Multiplying through by \(2\sqrt{13}\) gives:
\[ 2(2x - 3y - 5) = \pm (6x - 4y + 7) \]
This simplifies to:
\[ 4x - 6y - 10 = \pm (6x - 4y + 7) \]

### Step 5: Solve for both cases (positive and negative)
1. **Positive case**:
\[ 4x - 6y - 10 = 6x - 4y + 7 \]
Rearranging gives:
\[ -2x - 2y - 17 = 0 \implies 2x + 2y + 17 = 0 \implies x + y + \frac{17}{2} = 0 \]

2. **Negative case**:
\[ 4x - 6y - 10 = - (6x - 4y + 7) \]
Rearranging gives:
\[ 4x - 6y - 10 = -6x + 4y - 7 \]
Simplifying leads to:
\[ 10x - 10y - 3 = 0 \implies 10x - 10y = 3 \implies x - y = \frac{3}{10} \]

### Step 6: Determine which bisector contains the point (2, -1)
We check which equation contains the point \((2, -1)\):
1. For \(x + y + \frac{17}{2} = 0\):
\[ 2 - 1 + \frac{17}{2} = 1 + \frac{17}{2} = \frac{19}{2} \quad (\text{not } 0) \]
2. For \(10x - 10y - 3 = 0\):
\[ 10(2) - 10(-1) - 3 = 20 + 10 - 3 = 27 \quad (\text{not } 0) \]

Since neither equation contains the point, we need to check the sign of the left-hand side for both equations. The correct bisector will be the one whose left-hand side is negative when substituting the point \((2, -1)\).

### Final Answer
The equation of the angle bisector that is the supplement of the angle containing the point \((2, -1)\) is:
\[ 10x - 10y - 3 = 0 \quad \text{or} \quad x - y = \frac{3}{10} \]