$(a, b)$ is the point to which the origin has to be shifted by translation of axes so as to remove the first-degree terms from the equation $2 x^{2}-3 x y+4 y^{2}+5 y-6=0$. If the angle by which the axes are to be rotated in positive direction about the origin to remove the $x y$-term from the equation $a x^{2}+23 a b x y+b y^{2}=0$ is $\theta$, then $\tan 2 \theta=$

$A(1,-2), B(-2,3), C(-1,-3)$ are the vertices of a $\triangle A B C . L_{1}$ is the perpendicular drawn from $A$ to $B C$ and $L_{2}$ is the perpendicular bisector of $A B$. If $(l, m)$ is the point of intersection of $L_{1}$ and $L_{2}$, then $26 m-3=$

The area of the parallelogram formed by the lines $L_{1} \equiv \lambda x+4 y+2=0, L_{2} \equiv 3 x+4 y-3=0$, $L_{3} \equiv 2 x+\mu y+6=0, L_{4} \equiv 2 x+y+3=0$, where $L_{1}$ is parallel to $L_{2}$ and $L_{3}$ is parallel to $L_{4}$ is

If the angle between the pair of lines given by the equation $a x^{2}+4 x y+2 y^{2}=0$ is $45^{\circ}$, then the possible values of $a$