locus of the point (2,3) in the line (2x-3y+4) +k(x-2y+3)=0 is a

Any fixed point doesn`t form a locus... SO plzzzz recheck your question and ask it again.... Hope U like it

2x − 3y + 4 + k(x − 2y + 3) = 0 is family of lines passing through (1, 2) Now look at the given figure:Taking triangle ABM and AB`M,AM=AM (common)∠AMB=∠AMB`=90°BM=MB` (Given AM is perpendicular bisector of BB`)∆ABM≅∆AMB` (SAS)hence,AB=AB` so,Let M is mid point of BB` and AM is ⊥ bisector of BB` (where A is the point of intersection given lines) (x – 2)(x – 1) + (y – 2)(y – 3) = 0M will lie on the above line:so,(h+22 – 2)(h+22 – 1) + (k+32 – 2)(k+32– 3) = 0⇒(h−2)(h)+(k−1)(k−3)=0⇒h^2−2h+k^2−4k+3=0⇒(h−1)2+(k−2)2=2replacing h by x and k by y.(x−1)^2+(y−2)^2=

Approved