What is the distance of the orthocenter from the vertex of a triangle with vertices $(2, 3), (4, 5),$ and $(-1, 10)$?

In $\Delta ABC$,the coordinates of $B$ are $(0, 0)$,$AB = 2$,$\angle ABC = \frac{\pi}{3}$,and the coordinates of the midpoint of $BC$ are $(2, 0)$. Find the centroid of the triangle.

Let $PQR$ be an acute-angled triangle in which $PQ < QR$. From the vertex $Q$,draw the altitude $QQ_1$,the angle bisector $QQ_2$,and the median $QQ_3$,with $Q_1, Q_2, Q_3$ lying on $PR$. Then,

Assertion: If $(0, 3), (1, 1)$ and $(-1, 2)$ are the midpoints of the sides of a triangle,then the centroid of the original triangle is $(0, 2)$.
Reason: The centroid of a triangle and the centroid of the triangle formed by joining the midpoints of the sides of the original triangle are the same.

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

Let $A(6,8)$,$B(10 \cos \alpha, -10 \sin \alpha)$ and $C(-10 \sin \alpha, 10 \cos \alpha)$ be the vertices of a triangle. If $L(a, 9)$ and $G(h, k)$ are its orthocenter and centroid respectively,then $(5a - 3h + 6k + 100 \sin 2\alpha)$ is equal to . . . . . .