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 $PA=$

$(-2, -1)$ 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 $m+n=$

Consider a $\triangle PQR$ in which the relation $QR^2 + PR^2 = 5PQ^2$ holds. Let $G$ be the point of intersection of medians $PM$ and $QN$. Then,$\angle QGM$ is always

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

The equation of the line joining the centroid with the orthocentre of the triangle formed by the points $(-2, 3), (2, -1), (4, 0)$ is

If the midpoints of the sides of a triangle are $(0, 1)$,$(1, 1)$,and $(1, 0)$,find the $x$-coordinate of its incentre.