Point $(-1, 2)$ is changed to $(a, b)$ when the origin is shifted to the point $(2, -1)$ by translation of axes. Point $(a, b)$ is changed to $(c, d)$ when the axes are rotated through an angle of $45^{\circ}$ about the new origin. Point $(c, d)$ is changed to $(e, f)$ when $(c, d)$ is reflected through the line $y = x$. Then $(e, f) =$

The origin is shifted to the point $(2,3)$ by translation of axes and then the coordinate axes are rotated about the origin through an angle $\theta$ in the counter-clockwise sense. Due to this,if the equation $3x^2+2xy+3y^2-18x-22y+50=0$ is transformed to $4x^2+2y^2-1=0$,then the angle $\theta=$

If the coordinate axes are rotated in the positive direction by $45^{\circ}$ without changing the origin,then the transformed equation of $3x^2 + 3y^2 + 2xy - 2 = 0$ 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.

When the origin is shifted to the point $(h, k)$ by translating the coordinate axes,the equation $S = 2x^2 - xy + y^2 + 2x + 3y + 1 = 0$ is changed to $S' = ax^2 + 2hxy + by^2 + C' = 0$. If the coordinate axes are then rotated about the new origin through an angle $\theta$ in the positive direction to eliminate the $xy$ term,the equation $S' = 0$ becomes $Ax^2 + By^2 + C = 0$. Find the value of $h + k + \tan 2\theta$.

If the transformed equation of the equation $2x^2+3xy-2y^2-17x+6y+8=0$ after translating the coordinate axes to a new origin $(\alpha, \beta)$ is $aX^2+2hXY+bY^2+c=0$,then $3\alpha+c=$