Find the equation of the bisectors of the internal angles of the triangle whose sides are :
`3x+4y-6=0` , `12x-5y-3=0`, `4x-3y+12=0`

Answer

To find the equations of the internal angle bisectors of the triangle formed by the lines \(3x + 4y - 6 = 0\), \(12x - 5y - 3 = 0\), and \(4x - 3y + 12 = 0\), we will follow these steps:

### Step 1: Identify the equations of the sides of the triangle.
The equations of the sides of the triangle are:
1. \(L_1: 3x + 4y - 6 = 0\) (AB)
2. \(L_2: 12x - 5y - 3 = 0\) (BC)
3. \(L_3: 4x - 3y + 12 = 0\) (AC)

### Step 2: Find the internal angle bisector between lines \(L_1\) and \(L_2\).
For lines \(L_1\) and \(L_2\):
- \(a_1 = 3\), \(b_1 = 4\), \(c_1 = -6\)
- \(a_2 = 12\), \(b_2 = -5\), \(c_2 = -3\)

Calculate \(a_1 a_2 + b_1 b_2\):
\[ 3 \cdot 12 + 4 \cdot (-5) = 36 - 20 = 16 > 0 \]
Since this value is greater than 0, we use the formula for the internal angle bisector:
\[ \frac{a_1 x + b_1 y + c_1}{\sqrt{a_1^2 + b_1^2}} = -\frac{a_2 x + b_2 y + c_2}{\sqrt{a_2^2 + b_2^2}} \]

Substituting the values:
\[ \frac{3x + 4y - 6}{\sqrt{3^2 + 4^2}} = -\frac{12x - 5y - 3}{\sqrt{12^2 + (-5)^2}} \]
Calculating the denominators:
\[ \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5, \quad \sqrt{12^2 + (-5)^2} = \sqrt{144 + 25} = 13 \]
Thus, we have:
\[ \frac{3x + 4y - 6}{5} = -\frac{12x - 5y - 3}{13} \]
Cross-multiplying gives:
\[ 13(3x + 4y - 6) = -5(12x - 5y - 3) \]
Expanding both sides:
\[ 39x + 52y - 78 = -60x + 25y + 15 \]
Rearranging terms:
\[ 39x + 60x + 52y - 25y - 78 - 15 = 0 \]
Combining like terms:
\[ 99x + 27y - 93 = 0 \]
Dividing through by 3:
\[ 33x + 9y - 31 = 0 \]

### Step 3: Find the internal angle bisector between lines \(L_2\) and \(L_3\).
For lines \(L_2\) and \(L_3\):
- \(a_1 = 12\), \(b_1 = -5\), \(c_1 = -3\)
- \(a_2 = 4\), \(b_2 = -3\), \(c_2 = 12\)

Calculate \(a_1 a_2 + b_1 b_2\):
\[ 12 \cdot 4 + (-5) \cdot (-3) = 48 + 15 = 63 > 0 \]
Using the angle bisector formula:
\[ \frac{12x - 5y - 3}{\sqrt{12^2 + (-5)^2}} = -\frac{4x - 3y + 12}{\sqrt{4^2 + (-3)^2}} \]
Calculating the denominators:
\[ \sqrt{12^2 + (-5)^2} = 13, \quad \sqrt{4^2 + (-3)^2} = 5 \]
Thus, we have:
\[ \frac{12x - 5y - 3}{13} = -\frac{4x - 3y + 12}{5} \]
Cross-multiplying gives:
\[ 5(12x - 5y - 3) = -13(4x - 3y + 12) \]
Expanding both sides:
\[ 60x - 25y - 15 = -52x + 39y - 156 \]
Rearranging terms:
\[ 60x + 52x - 25y - 39y + 156 - 15 = 0 \]
Combining like terms:
\[ 112x - 64y + 141 = 0 \]

### Step 4: Find the internal angle bisector between lines \(L_3\) and \(L_1\).
For lines \(L_3\) and \(L_1\):
- \(a_1 = 4\), \(b_1 = -3\), \(c_1 = 12\)
- \(a_2 = 3\), \(b_2 = 4\), \(c_2 = -6\)

Calculate \(a_1 a_2 + b_1 b_2\):
\[ 4 \cdot 3 + (-3) \cdot 4 = 12 - 12 = 0 \]
Since this value is equal to 0, we use the second formula:
\[ \frac{4x - 3y + 12}{\sqrt{4^2 + (-3)^2}} = \frac{3x + 4y - 6}{\sqrt{3^2 + 4^2}} \]
Calculating the denominators:
\[ \sqrt{4^2 + (-3)^2} = 5, \quad \sqrt{3^2 + 4^2} = 5 \]
Thus, we have:
\[ \frac{4x - 3y + 12}{5} = \frac{3x + 4y - 6}{5} \]
Cross-multiplying gives:
\[ 4x - 3y + 12 = 3x + 4y - 6 \]
Rearranging terms:
\[ 4x - 3x - 3y - 4y + 12 + 6 = 0 \]
Combining like terms:
\[ x - 7y + 18 = 0 \]

### Final Result:
The equations of the internal angle bisectors of the triangle are:
1. \(33x + 9y - 31 = 0\)
2. \(112x - 64y + 141 = 0\)
3. \(x - 7y + 18 = 0\)