Consider the family of lines (x−y−6)+λ(2x+y+3)=0 and (x+2y−4)+μ(3x−2y−4)=0.If the lines of these two families are at right angle to each other, then the locus of their point of intersection, is

solving x−y−6=0 and 2x+y+3=0 we get (1,-5)  and  x+2y−4=0 , 3x−2y−4=0 we get (2,1) Given family of lines are concurrent at A(1, -5) and B(2, 1). Now point P(h, k) is such that AP ⊥ BP ⇒k+5h−1⋅k−1h−2=−1∴  Locus of (h,k) is x2+y2−3x+4y−3=0.