If the equation of the pair of straight lines passing through the point (1,1) and perpendicular to the pair of lines 3 x ^2+11 x y-4 y^2=0 is a ^2+2 ~h x y+ b y^2+2 ~g x+2 f y+12=0 , then 2(a-h+b-g+f-12)=

Step-by-step solution

3 x^2+11 x y-4 y^2=0 (3 x-y)(x+4 y)=0 Equation of lines are 3 x-y=0 (i); x+4 y=0 (ii) m₁=3, m₂= -1 4 So lines perpendicular to (i) and (ii) will have slope -1 2 , 4 Also the lines passes through (1,1) y-1=- 1 3 (x-1) x+3 y=4 ....(iii) and y-1=4(x-1) 4 x-y=3 ....(iv) Combined equation of (iii) and (iv) is (x+3 y-4)(4 x-y-3)=0 4 x^2-3 y^2+11 x y-19 x-5 y+12=0 a=4, b=-3, h= 11 2 , g= -19 2 , f= -5 2 2(a-h+b-g+f-12)=-19