The centroid of the triangle $ABC$,where $A \equiv (2,3)$,$B \equiv (8,10)$,and $C \equiv (5,5)$ is

Let $PS$ be the median of the triangle with vertices $P(2,2)$,$Q(6,-1)$,and $R(7,3)$. The equation of the line passing through $(1,-1)$ and parallel to $PS$ is:

$A$ tower stands vertically inside an acute-angled triangular park $\Delta PQR$. If the angle of elevation of the top of the tower from each corner of the park is the same,then in $\Delta PQR$,the foot of the tower is at the

The orthocentre of a $\Delta ABC$ is $B$ and the circumcentre is $S(a, b)$. If $A$ is the origin,then the coordinates of $C$ are-

The points $(11,9), (2,1)$ and $(2,-1)$ are the mid-points of the sides of a triangle. Then,the centroid is

Let the area of a $\triangle PQR$ with vertices $P(5, 4)$,$Q(-2, 4)$,and $R(a, b)$ be $35$ square units. If its orthocenter and centroid are $O\left(2, \frac{14}{5}\right)$ and $C(c, d)$ respectively,then $c+2d$ is equal to: