Find the equations of the lines bisecting the angles between the following pairs of straight lines writing first the bisector of the angle in which the origin lies :
`12x-5y+3=0, 4x+3y-2=0`.

Answer

To find the equations of the angle bisectors for the lines given by the equations \(12x - 5y + 3 = 0\) and \(4x + 3y - 2 = 0\), we can follow these steps:

### Step 1: Identify the equations of the lines
Let:
- Line L: \(12x - 5y + 3 = 0\)
- Line M: \(4x + 3y - 2 = 0\)

### Step 2: Evaluate the lines at the origin (0, 0)
We need to evaluate both lines at the origin to determine which angle contains the origin.

For Line L:
\[ L(0, 0) = 12(0) - 5(0) + 3 = 3 \]

For Line M:
\[ M(0, 0) = 4(0) + 3(0) - 2 = -2 \]

### Step 3: Calculate the product of the evaluations
Now, we find the product of the evaluations:
\[ L(0, 0) \cdot M(0, 0) = 3 \cdot (-2) = -6 \]
Since the product is negative, the origin lies between the two lines.

### Step 4: Use the angle bisector formula
The formula for the angle bisectors of two lines given by \(a_1x + b_1y + c_1 = 0\) and \(a_2x + b_2y + c_2 = 0\) is:
\[ \frac{a_1x + b_1y + c_1}{\sqrt{a_1^2 + b_1^2}} = \pm \frac{a_2x + b_2y + c_2}{\sqrt{a_2^2 + b_2^2}} \]

### Step 5: Substitute values into the formula
For Line L:
- \(a_1 = 12\), \(b_1 = -5\), \(c_1 = 3\)

For Line M:
- \(a_2 = 4\), \(b_2 = 3\), \(c_2 = -2\)

Calculating the denominators:
\[ \sqrt{a_1^2 + b_1^2} = \sqrt{12^2 + (-5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \]
\[ \sqrt{a_2^2 + b_2^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]

### Step 6: Write the equations for the angle bisectors
Using the angle bisector formula, we first write the equation for the bisector in which the origin lies (using the negative sign):
\[ \frac{12x - 5y + 3}{13} = -\frac{4x + 3y - 2}{5} \]

### Step 7: Cross-multiply and simplify
Cross-multiplying gives:
\[ 5(12x - 5y + 3) = -13(4x + 3y - 2) \]
Expanding both sides:
\[ 60x - 25y + 15 = -52x - 39y + 26 \]

Bringing all terms to one side:
\[ 60x + 52x - 25y + 39y + 15 - 26 = 0 \]
\[ 112x + 14y - 11 = 0 \]

### Step 8: Write the second angle bisector
Now, for the other angle bisector (using the positive sign):
\[ \frac{12x - 5y + 3}{13} = \frac{4x + 3y - 2}{5} \]

Cross-multiplying gives:
\[ 5(12x - 5y + 3) = 13(4x + 3y - 2) \]
Expanding both sides:
\[ 60x - 25y + 15 = 52x + 39y - 26 \]

Bringing all terms to one side:
\[ 60x - 52x - 25y - 39y + 15 + 26 = 0 \]
\[ 8x - 64y + 41 = 0 \]

### Final Equations of the Angle Bisectors
1. The angle bisector in which the origin lies:
\[ 112x + 14y - 11 = 0 \]
2. The other angle bisector:
\[ 8x - 64y + 41 = 0 \]