We have, x−3y+3=0 kx+y+k=0 2x+y−8=0 On solving Eqs. (i) and (iii), we get x=3,y=2 From Eq. (ii), we get 3k+2+k=0⇒K=−‌
1
2
( ∵ Eqs. (i), (ii) and (iii) are concurrent) and a=3,b=2 [point of concurrency is ( 3,2 )] So, L≡3x−2y−1=0 given, distance from origin to L is P P=‌
|0−0−1|
√32+(−2)2
=‌
1
√13
Hence, perpendicular distance from the point (2,3) to L is =‌
