The medians $AD$ and $BE$ of a triangle with vertices $A(0, b)$,$B(0, 0)$,and $C(a, 0)$ are perpendicular to each other,if

Let $A(1,0), B(6,2)$ and $C(\frac{3}{2}, 6)$ be the vertices of a triangle $ABC$. If $P$ is a point inside the triangle $ABC$ such that the triangles $APC, APB$ and $BPC$ have equal areas,then the length of the line segment $PQ,$ where $Q$ is the point $(-\frac{7}{6}, -\frac{1}{3})$ is

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

Let $A \equiv (4,4), B \equiv (8,4), C \equiv (4,8)$. If $P, Q, R$ are the midpoints of sides $AB, BC, CA$ respectively and $(\alpha, \beta)$ are the coordinates of the orthocentre of $\Delta PQR$,then the value of $\alpha + \beta$ is

$(-2, -1)$ and $(2, 5)$ are two vertices of a triangle and $\left(2, \frac{5}{3}\right)$ is its orthocenter. If $(m, n)$ is the third vertex of that triangle,then $m+n=$

The circumcentre of the triangle formed by the points $A(1, \sqrt{3})$,$B(-1, -\sqrt{3})$,and $C(3, -\sqrt{3})$ is