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(D) The given pair of straight lines is $(ax+by)^2 - 3(bx-ay)^2 = 0$. This can be written as $(ax+by)^2 = 3(bx-ay)^2$,which implies $ax+by = \pm \sqrt

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$\frac{c^2}{\sqrt{3}(a^2+b^2)}$

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(D) The given pair of straight lines is $(ax+by)^2 - 3(bx-ay)^2 = 0$. This can be written as $(ax+by)^2 = 3(bx-ay)^2$,which implies $ax+by = \pm \sqrt{3}(bx-ay)$. This gives two lines: $L_1: (a - \sqrt{3}b)x + (b + \sqrt{3}a)y = 0$ $L_2: (a + \sqrt{3}b)x + (b - \sqrt{3}a)y = 0$ Let the third line be $L_3: ax+by+c = 0$. The area of the triangle formed by $Ax^2 + 2Hxy + By^2 = 0$ and $lx+my+n = 0$ is given by $\frac{n^2 \sqrt{H^2-AB}}{|Al^2 - 2Hlm + Bm^2|}$. Here,the pair of lines is $(ax+by)^2 - 3(bx-ay)^2 = 0$,which expands to $(a^2-3b^2)x^2 + 8abxy - (b^2-3a^2)y^2 = 0$. So,$A = a^2-3b^2$,$H = 4ab$,$B = -(b^2-3a^2) = 3a^2-b^2$. Also,$l = a$,$m = b$,$n = c$. $H^2 - AB = (4ab)^2 - (a^2-3b^2)(3a^2-b^2) = 16a^2b^2 - (3a^4 - a^2b^2 - 9a^2b^2 + 3b^4) = 16a^2b^2 - 3a^4 + 10a^2b^2 - 3b^4 = -3a^4 + 26a^2b^2 - 3b^4$. Wait,using the standard formula for area $\Delta = \frac{c^2 \sqrt{h^2-ab}}{a b^2 - 2h ab + b a^2}$ is complex. Alternatively,the lines are $ax+by = \pm \sqrt{3}(bx-ay)$. The angle between the lines is $	an 	heta = \frac{2\sqrt{h^2-ab}}{a+b} = \frac{2\sqrt{3(a^2+b^2)}}{a^2+b^2} = \frac{2\sqrt{3}}{\sqrt{a^2+b^2}}$. The area is $\frac{c^2 	an 	heta}{4(a^2+b^2)} 	imes \dots$ Correct formula for area is $\frac{c^2 \sqrt{h^2-ab}}{|a b^2 - 2h ab + b a^2|}$. For the given equation,the area simplifies to $\frac{c^2}{\sqrt{3}(a^2+b^2)}$.

The area of the triangle formed by the pair of straight lines $(ax+by)^2 - 3(bx-ay)^2 = 0$ and the line $ax+by+c = 0$ is

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