Question

If line \(2y-3=0\) bisects angle between \(x+2y-k=0\) and \(x-2y+k=0\), then a point on the bisector of other angle is

• A. (k,k)
• B. (0,k)
• C. (k,0)
• D. (k,-k) \bigskip

Hint:

Pair of lines always gives two bisectors: internal and external.

Answer 1

The Correct Option is D

Concept: Angle bisectors: \[ \frac{L_1}{\sqrt{a_1^2+b_1^2}}=\pm \frac{L_2}{\sqrt{a_2^2+b_2^2}} \]

Step 1: Check bisector condition.
Given correct bisector implies consistency gives relation satisfied.

Step 2: Other bisector equation.
Other angle bisector passes through symmetric point: \[ (x,y)=(k,-k) \] \[ \boxed{(D)} \]