Question

If the angular bisectors of the lines \[ 3x-4y-5=0 \] and \[ 8x-6y+1=0 \] are \[ x+y+c=0 \] and \[ x-y+k=0, \] then \[ 7(c+k)= \]

• A. \(2\)
• B. \(16\)
• C. \(-35\)
• D. \(34\) \bigskip

Hint:

For two lines \[ a_1x+b_1y+c_1=0 \quad \text{and} \quad a_2x+b_2y+c_2=0, \] their angle bisectors are obtained using \[ \frac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}} = \pm \frac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}}. \] Always simplify both resulting equations separately.

Answer 1

The Correct Option is D

Concept: The angle bisectors of the lines \[ L_1=0 \quad \text{and} \quad L_2=0 \] are obtained from \[ \frac{L_1}{\sqrt{a_1^2+b_1^2}} = \pm \frac{L_2}{\sqrt{a_2^2+b_2^2}}. \]

Step 1: Write the equation of the angle bisectors. Given \[ L_1=3x-4y-5, \] \[ L_2=8x-6y+1. \] Since \[ \sqrt{3^2+(-4)^2}=5, \] and \[ \sqrt{8^2+(-6)^2}=10, \] the angle bisectors are \[ \frac{3x-4y-5}{5} = \pm \frac{8x-6y+1}{10}. \] Multiplying by \(10\), \[ 2(3x-4y-5) = \pm(8x-6y+1). \]

Step 2: Take the positive sign. \[ 6x-8y-10 = 8x-6y+1. \] \[ -2x-2y-11=0. \] \[ x+y+\frac{11}{2}=0. \] Thus, \[ c=\frac{11}{2}. \]

Step 3: Take the negative sign. \[ 6x-8y-10 = -8x+6y-1. \] \[ 14x-14y-9=0. \] \[ x-y-\frac{9}{14}=0. \] Thus, \[ k=-\frac{9}{14}. \]

Step 4: Find \(7(c+k)\). \[ c+k = \frac{11}{2} - \frac{9}{14}. \] \[ = \frac{77-9}{14}. \] \[ = \frac{68}{14} = \frac{34}{7}. \] Therefore, \[ 7(c+k) = 7\cdot\frac{34}{7} = 34. \]

Step 5: Write the final answer. \[ \boxed{34} \]