The equation of the line joining the centroid with the orthocentre of the triangle formed by the points $(-2, 3), (2, -1), (4, 0)$ is

If the equation of the median through vertex $A(3, k)$ of $\triangle ABC$ with vertices $B(2, 1)$ and $C(-4, 5)$ is $x + 4y = p$,then $k = ?$ where $p$ and $k$ are constants.

The circumcentre of the triangle with vertices $(-2, 3)$,$(2, -1)$,and $(4, 0)$ is:

If $P(6,1)$ is the orthocentre of the triangle whose vertices are $A(5,-2)$,$B(8,3)$,and $C(h, k)$,then the point $C$ lies on the circle:

Let the vertex of a triangle $ABC$ be $A(1, 2)$, and the mid-point of the side $AB$ be $M(5, -1)$. If the centroid of this triangle is $G(3, 4)$ and its circumcenter is $O(\alpha, \beta)$, then $21(\alpha + \beta)$ is equal to:

If $O, G, S$ are respectively the orthocentre,centroid and circumcentre of a triangle whose vertices are $A(2,3), B(2,4)$ and $C(4,3)$,then $AO^2 + 9BG^2 + 4CS^2 =$