The area of the triangle formed by the pair of straight lines (a x+b y)^2-3(b x-a y)^2=0 and a x+b y+c=0 is

Step-by-step solution

Given that, equation of pair of straight lines is aligned &(a x+b y)^2-3(b x-a y)^2=0 & a^2 x^2+b^2 y^2+2 a b x y &-3 (b^2 x^2+a^2 y^2-2 a b x y )=0 & (a^2-3 b^2 ) x^2- (b^2-3 a^2 ) y^2+8 a b x y=0 aligned Let the equation of line be y=m₁ x, y=m₂ x . m₁+m₂= -4 a b b^2-3 a^2 [ array c m₁+m₂=- 2 h b m₁ m₂= a b array ] and m₁ m₂= a^2-3 b^2 b^2-3 a^2 Now, (m₁-m₂ )^2= (m₁+m₂ )^2-4 m₁ m₂ aligned & = 16 a^2 b^2 (b^2-3 a^2 )^2 - 4 (a^2-3 b^2 ) (b^2-3 a^2 ) & = 4 (b^2-3 a^2 )^2 [4 a^2 b^2- (a^2-3 b^2 ) (b^2-3 a^2 ) ] & = 4