L1:3x−4y+12=0 L2:8x+6y+11=0 Equation of angle bisector of L1 and L2 of angle containing origin 2(3x−4y+12)=8x+6y+11 2x+14y−13=0.......‌ (i) ‌ ‌
3α−4β+12
5
=1 ⇒3α−4β+7=0‌..... (ii) ‌ Solution of 2x+14y−13=0 and 3x−4y+7=0 gives the required point P(α,β),α=‌