If $P$ is a point equidistant from all the vertices $A(-1, 3)$,$B(3, 5)$,and $C(5, 7)$ of a triangle $ABC$,then $PA=$

Two vertices of a triangle are $(5, -1)$ and $(-2, 3)$. If the orthocenter of the triangle is the origin $(0, 0)$,find the coordinates of the third vertex.

If a triangle $ABC$ has vertices $A(1, -6)$, $B(2, -3)$ and $C(3, -2)$, then the coordinates of its orthocenter are....

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:

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:

In a $\triangle ABC$,if the medians $AD$ and $BE$ are such that $AD=4$,$\angle DAB=\frac{\pi}{6}$ and $\angle ABE=\frac{\pi}{3}$,then the area of $\triangle ABC$ (in square units) is