Find the equation of acute angled bisector of lines :
`3x-4y+7=0and12x-5y-8=0`

Answer

To find the equation of the acute angled bisector of the lines given by the equations \(3x - 4y + 7 = 0\) and \(12x - 5y - 8 = 0\), we can follow these steps:

### Step 1: Write the equations in standard form
The equations of the lines are already in standard form:
1. Line 1: \(3x - 4y + 7 = 0\)
2. Line 2: \(12x - 5y - 8 = 0\)

### Step 2: Identify coefficients
From the equations, we identify the coefficients:
- For Line 1: \(A_1 = 3\), \(B_1 = -4\), \(C_1 = 7\)
- For Line 2: \(A_2 = 12\), \(B_2 = -5\), \(C_2 = -8\)

### Step 3: Use the angle bisector formula
The equation of the angle bisector can be expressed as:
\[ \frac{A_1x + B_1y + C_1}{\sqrt{A_1^2 + B_1^2}} = \pm \frac{A_2x + B_2y + C_2}{\sqrt{A_2^2 + B_2^2}} \]

### Step 4: Calculate the magnitudes
Calculate the magnitudes of the coefficients:
- For Line 1:
\[ \sqrt{A_1^2 + B_1^2} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]
- For Line 2:
\[ \sqrt{A_2^2 + B_2^2} = \sqrt{12^2 + (-5)^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \]

### Step 5: Set up the equation for the acute angle bisector
Since we are looking for the acute angle bisector, we take the positive sign:
\[ \frac{3x - 4y + 7}{5} = \frac{12x - 5y - 8}{13} \]

### Step 6: Cross-multiply to eliminate the fractions
Cross-multiplying gives:
\[ 13(3x - 4y + 7) = 5(12x - 5y - 8) \]

### Step 7: Expand both sides
Expanding both sides:
\[ 39x - 52y + 91 = 60x - 25y - 40 \]

### Step 8: Rearrange the equation
Rearranging the terms to one side:
\[ 39x - 60x + 25y - 52y + 91 + 40 = 0 \]
This simplifies to:
\[ -21x - 27y + 131 = 0 \]
Multiplying through by -1 gives:
\[ 21x + 27y - 131 = 0 \]

### Final Equation
The equation of the acute angled bisector is:
\[ 21x + 27y - 131 = 0 \]