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

Let $$f\left( \theta \right) = \sin \theta \left( {\sin \theta + \sin 3\theta } \right)$$. Then $$f\left( \theta \right)$$ is

The triangle PQR is inscribed in the circle $${x^2}\, + \,\,{y^2} = \,25$$. If Q and R have co-ordinates (3, 4) and ( - 4, 3) respectively, then $$\angle \,Q\,P\,R$$ is equal to

If the circles $${x^2}\, + \,{y^2}\, + \,\,2x\, + \,2\,k\,y\,\, + \,6\,\, = \,\,0,\,\,{x^2}\, + \,\,{y^2}\, + \,2ky\, + \,k\, = \,0$$ intersect orthogonally, then k is