If $(2, 3)$,$(4, 5)$,and $(-2, 11)$ are the vertices of a triangle,what is the distance between the vertex $(4, 5)$ and the circumcenter?

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 the vertices of a triangle are $(a, 1), (b, 3),$ and $(4, c),$ then the centroid of the triangle will lie on the $x$-axis if

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 . . . . . .

The vertices of $\Delta PQR$ are $P(2, 1)$,$Q(-2, 3)$,and $R(4, 5)$. Find the equation of the median through the vertex $R$.

Show that the area of the triangle formed by the lines $y=m_{1}x+c_{1}$,$y=m_{2}x+c_{2}$ and $x=0$ is $\frac{(c_{1}-c_{2})^{2}}{2|m_{1}-m_{2}|}$.