If the pair of straight lines given by $Ax^{2} + 2Hxy + By^{2} = 0(H^{2} AB)$ forms an equilateral triangle with line $ax + by + c = 0,$ then $(A + 3B)(3A + B)$ is equal to

Question

If the pair of straight lines given by $Ax^{2} + 2Hxy + By^{2} = 0(H^{2} > AB)$ forms an equilateral triangle with line $ax + by + c = 0,$ then $(A + 3B)(3A + B)$ is equal to

Dr. Abubacker · Accepted Answer

(d)
Given, $Ax^{2} + 2Hxy + {By}^{2} = 0\$ …(i)
and $ax + by + c = 0$ …(ii)
Since, triangle is equilateral, then angle between the two lines is $60{^\circ}$
Angle between pair of lines is given by$\$
$\cos{60{^\circ} = \frac{A + B}{\sqrt{(A - B)^{2} + {4H}^{2}}}}$
$\Rightarrow \ \ \frac{A + B}{\sqrt{(A - B)^{2} + {4H}^{2}}} = \frac{1}{2}$
$\Rightarrow \ (A - B)^{2} + {4H}^{2} = 4(A + B)^{2}$
$\Rightarrow \ 4\left( A^{2} + B^{2} + 2AB \right) - \left( A^{2} + B^{2} - 2AB \right) = {4H}^{2}$
$\Rightarrow \ 3A^{2} + 10AB + 3B^{2} = 4H^{2}$
$\Rightarrow \ (3A + B)(A + 3B) = 4H^{2}$

Q. If the pair of straight lines given by $Ax^{2} + 2Hxy + By^{2} = 0(H^{2} > AB)$ forms an equilateral triangle with line $ax + by + c = 0,$ then $(A + 3B)(3A + B)$ is equal to

Explanation

Reviewed by Dr. Abubacker

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