If the pair of straight lines given by A x^2+2 H x y+B y^2=0 (H^2>A B ) forms an equilateral triangle with line a x+b y+c=0 , then (A+3 B)(3 A+B) is equal to :

Step-by-step solution

Given that, A x^2+2 H x y+B y^2=0 ...(i) and a x+b y+c=0 ...(ii) Since, triangle is equilateral, then angle between the two lines is 60^ . Angle between pair of lines is given by 60^ = A+B (A-B)^2+4 H^2 A+B (A-B)^2+4 H^2 = 1 2 (A-B)^2+4 H^2=4(A+B)^2 4 (A^2+B^2+2 A B )- (A^2+B^2-2 A B )=4 H^2 3 (A^2+B^2 )+10 A B=4 H^2 3 A^2+10 A B+3 B^2=4 H^2 (3 A+B)(A+3 B)=4 H^2