If the vertices of a triangle are $(0,0)$,$(6,0)$,and $(6,8)$,then its incentre will be:

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

If $A(x_1, y_1)$,$B(x_2, y_2)$,and $C(x_3, y_3)$ are the vertices of a triangle with side lengths $a, b, c$ opposite to vertices $A, B, C$ respectively,then the excentre with respect to vertex $B$ is:

The coordinates of the orthocentre of the triangle whose sides are given by the lines $x = 3$,$y = 4$,and $3x + 4y = 6$ are:

The circumcentre of a triangle lies at the origin and its centroid is the midpoint of the line segment joining the points $(a^2 + 1, a^2 + 1)$ and $(2a, -2a)$,where $a \ne 0$. Then for any $a$,the orthocentre of this triangle lies on the line:

$A(1, -2), B(-2, 3), C(-1, -3)$ are the vertices of a triangle $ABC$. $L_1$ is the perpendicular drawn from $A$ to $BC$ and $L_2$ is the perpendicular bisector of $AB$. If $(l, m)$ is the point of intersection of $L_1$ and $L_2$, then $26m - 3 =$ (in $l$)