When the origin is shifted to the point $(2, b)$ by translation of axes,the coordinates of the point $(a, 4)$ change to $(6, 8)$. When the origin is shifted to $(a, b)$ by translation of axes,if the transformed equation of $x^2+4xy+y^2=0$ is $X^2+2HXY+Y^2+2GX+2FY+C=0$,then $2H(G+F)=$

If $2x^2+xy-6y^2+k=0$ is the transformed equation of $2x^2+xy-6y^2-13x+9y+15=0$ when the origin is shifted to the point $(a, b)$ by translation of axes,then $k=$

If the origin is shifted to the point $\left(\frac{3}{2},-2\right)$ by the translation of axes,then the transformed equation of $2x^2+4xy+y^2+2x-2y+1=0$ is

Statement $(A) :$ The area of the triangle formed by the points $A (20, 22), B (21, 24),$ and $C (22, 23)$ is equal to the area of the triangle formed by the points $P (0, 0), Q (1, 2),$ and $R (2, 1).$
Reason $(R) :$ The area of a triangle remains invariant under the translation of axes.

If $\theta_1, \theta_2, \theta_3$ are respectively the angles by which the coordinate axes are to be rotated to eliminate the $xy$ term from the following equations,then the descending order of these angles is:
$A_1 = 3x^2 + 5xy + 3y^2 + 2x + 3y + 4 = 0$
$A_2 = 5x^2 + 2\sqrt{3}xy + 3y^2 + 6 = 0$
$A_3 = 4x^2 + \sqrt{3}xy + 5y^2 - 4 = 0$

If $(h, k)$ is the new origin to be chosen to eliminate first degree terms from the equation $S \equiv 2x^2 - xy - y^2 - 3x + 3y = 0$ by translation and if $\theta$ is the angle with which the axes are to be rotated about the origin in anticlockwise direction to eliminate the $xy$-term from $S = 0$,then $\tan 2\theta =$