If the axes are rotated through an angle $45^{\circ}$,then the coordinates of the point $(4 \sqrt{2}, -6 \sqrt{2})$ in the new system are . . . . . .

When the coordinate axes are rotated about the origin in the positive direction through an angle $\frac{\pi}{4}$,if the equation $25x^2+9y^2=225$ is transformed to $\alpha x^2+\beta xy+\gamma y^2=\delta$,then $(\alpha+\beta+\gamma-\sqrt{\delta})^2=$

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.

The point to which the origin should be shifted in order to eliminate the $x$ and $y$ terms from the equation $9x^2+4y^2+10x+12y+1=0$ is

When the coordinate axes are rotated through an angle $135^{\circ}$,the coordinates of a point $P$ in the new system are known to be $(4, -3)$. Find the coordinates of $P$ in the original system.

When the coordinate axes are rotated through an angle $\theta$ in anti-clockwise direction,if the transformed equation of $x^2+y^2+2xy+2x+6y+1=0$ is $(2+\sqrt{3})X^2+2XY+(2-\sqrt{3})Y^2+aX+bY+2=0$,then $3a-b=$