Orthocentre of triangle with vertices $$\left( {0,0} \right),\left( {3,4} \right)$$ and $$\left( {4,0} \right)$$ is

Let $$0 < \alpha < {\pi \over 2}$$ be fixed angle. If $$P = \left( {\cos \theta ,\,\sin \theta } \right)$$ and $$Q = \left( {\cos \left( {\alpha - \theta } \right),\,\sin \left( {\alpha - \theta } \right)} \right),$$ then $$Q$$ is obtained from $$P$$ by

Let $$P = \left( { - 1,\,0} \right),\,Q = \left( {0,\,0} \right)$$ and $$R = \left( {3,\,3\sqrt 3 } \right)$$ be three points.
Then the equation of the bisector of the angle $$PQR$$ is

A straight line through the origin $$O$$ meets the parallel lines $$4x+2y=9$$ and $$2x+y+6=0$$ at points $$P$$ and $$Q$$ respectively. Then the point $$O$$ divides the segemnt $$PQ$$ in the ratio