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 are given by \(\frac{L_1}{\sqrt{a_1^2+b_1^2}} = \pm \frac{L_2}{\sqrt{a_2^2+b_2^2}}\). Identify the two bisectors, extract the constant terms \(c\) and \(k\) as defined, and compute \(7(c+k)\).

Step 1: Write the bisector equations. \(\frac{3x-4y-5}{5} = \pm \frac{8x-6y+1}{10} \Rightarrow 2(3x-4y-5) = \pm(8x-6y+1)\).

Step 2: Positive sign: \(6x-8y-10 = 8x-6y+1 \Rightarrow -2x-2y-11=0 \Rightarrow x+y+\frac{11}{2}=0\). So \(c = \frac{11}{2}\).

Step 3: Negative sign: \(6x-8y-10 = -8x+6y-1 \Rightarrow 14x-14y-9=0 \Rightarrow x-y-\frac{9}{14}=0\). So \(k = -\frac{9}{14}\).

Step 4: Compute \(7(c+k)\). \(c+k = \frac{11}{2} - \frac{9}{14} = \frac{77-9}{14} = \frac{68}{14} = \frac{34}{7}\). \(7 \times \frac{34}{7} = 34\).

Step 5: Write the final answer. \(\boxed{34}\)