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:

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:

Two vertices of a triangle are $(4, -3)$ and $(-2, 5)$. If the orthocentre of the triangle is at $(1, 2)$,then the third vertex is

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

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 $(a, b)=$

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