Let $A(0,0), B(3,0), C(0,-4)$ be the vertices of $\triangle ABC$. Then the coordinates of the incentre of $\triangle ABC$ are:

The vertices of a triangle are at $-\hat{i}+3 \hat{j}$ and $2 \hat{i}+5 \hat{j}$ and its orthocenter is at $\hat{i}+2 \hat{j}$. If the position vector of the third vertex is $a \hat{i}+b \hat{j}$,then $(a, b)=$

If the equation of the median through vertex $A(3, k)$ of $\triangle ABC$ with vertices $B(2, 1)$ and $C(-4, 5)$ is $x + 4y = p$,then $k = ?$ where $p$ and $k$ are constants.

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:

Two vertices of a triangle are $(5, 4)$ and $(-2, 4)$. If its centroid is $(5, 6)$,then the third vertex has the coordinates:

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