If the image of the point $(3, 8)$ in the line $x + 3y = 7$ is $(\alpha, \beta)$,then $\alpha + \beta =$

If the image of $\left(\frac{-7}{5}, \frac{-6}{5}\right)$ in a line is $(1, 2)$,then the equation of the line is

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

Two straight lines are drawn through the point $(0, 2)$ such that the lengths of the perpendiculars from the point $(4, 4)$ to these lines are each equal to $2$ units. The equation of the line joining the feet of these perpendiculars is

$A$ ray of light along $x + \sqrt{3}y = \sqrt{3}$ gets reflected upon reaching the $x$-axis. The equation of the reflected ray is:

The length of the perpendicular from the origin,on the normal to the curve $x^{2}+2xy-3y^{2}=0$ at the point $(2,2)$ is