The equation of the line bisecting perpendicularly the segment joining the points $(-4, 6)$ and $(8, 8)$ is

The coordinates of the foot of the perpendicular drawn from $(0,0)$ to the line joining $(a \cos \alpha, a \sin \alpha)$ and $(a \cos \beta, a \sin \beta)$ are

The image of the point $(3,8)$ in the line $x+3y=7$ is

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

Find the coordinates of the foot of the perpendicular from the point $(-1, 3)$ to the line $3x - 4y - 16 = 0$.