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(D) The pair of lines is given by $A x^2+2 H x y+B y^2=0$. Since the triangle formed by these lines and $a x+b y+c=0$ is equilateral,the angle between

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$4 H^2$

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(D) The pair of lines is given by $A x^2+2 H x y+B y^2=0$. Since the triangle formed by these lines and $a x+b y+c=0$ is equilateral,the angle between the lines of the pair must be $60^{\circ}$. The angle $	heta$ between the lines $A x^2+2 H x y+B y^2=0$ is given by $	an 	heta = \frac{2\sqrt{H^2-A B}}{|A+B|}$. For an equilateral triangle,$	heta = 60^{\circ}$,so $	an 60^{\circ} = \sqrt{3} = \frac{2\sqrt{H^2-A B}}{|A+B|}$. Squaring both sides,we get $3 = \frac{4(H^2-A B)}{(A+B)^2}$. $3(A+B)^2 = 4(H^2-A B)$. $3(A^2+B^2+2 A B) = 4 H^2-4 A B$. $3 A^2+3 B^2+6 A B = 4 H^2-4 A B$. $3 A^2+10 A B+3 B^2 = 4 H^2$. Factoring the left side,we get $(3 A+B)(A+3 B) = 4 H^2$.

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 the line $a x+b y+c=0$,then $(A+3 B)(3 A+B)$ is equal to :

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