If $Q$ and $R$ are the images of the point $P(2,3)$ with respect to the lines $x-y+2=0$ and $2x+y-2=0$ respectively,then $Q$ and $R$ lie on

If the line $2x - 3y + 5 = 0$ is the perpendicular bisector of the line segment joining $(1, -2)$ and $(\alpha, \beta)$,then $\alpha + \beta =$

If $(h, k)$ is the image of the point $(3, -4)$ with respect to the line $2x - 3y - 5 = 0$ and $(\ell, m)$ is the foot of the perpendicular from $(h, k)$ onto the line $3x + 2y + 12 = 0$,then $\ell h + mk + 1 =$ ?

The orthocentre and the centroid of $\triangle ABC$ are $(5,8)$ and $\left(3, \frac{14}{3}\right)$ respectively. The equation of the side $BC$ is $x-y=0$. Given that the image of the orthocentre of a triangle with respect to any side lies on the circumcircle of that triangle,then the diameter of the circumcircle of $\triangle ABC$ is

The image of the point $(4, -3)$ with respect to the line $y = x$ is

Let $0 < \alpha < \pi/2$ be a constant angle. If $P \equiv (\cos \theta, \sin \theta)$ and $Q \equiv (\cos(\alpha - \theta), \sin(\alpha - \theta))$,how is $Q$ obtained from $P$?