The correct option is A 6x−10y+4xy=0
Equation of line L is
(x+2y+1)+λ(2x+3y−1)=0⇒(1+2λ)x+(2+3λ)y+(1−λ)=0
The triangle formed is rightangle so the cirmcumcentre is the midpoint of hypotenuse
Let the midpoint be (h,k)
Now, the intercepts of the line is
x− intercept is λ−11+2λy− intercept is λ−12+3λ
Now, the midpoint is
(h,k)=(λ−12(1+2λ),λ−12(2+3λ))⇒2h=λ−11+2λ⇒λ=−2h+14h−1
Similarly,
λ=−4k+16k−1
Now, the locus is
2h+14h−1=4k+16k−1⇒12hk−2h+6k−1=16hk+4h−4k−1∴6x−10y+4xy=0