The equation of the locus of the point of intersection of the straight lines $$x\sin \theta + (1 - \cos \theta )y = a\sin \theta $$ and $$x\sin \theta - (1 + \cos \theta )y + a\sin \theta = 0$$ is

If $$(1,5)$$ be the midpoint of the segment of a line between the line $$5 x-y-4=0$$ and $$3 x+4 y-4=0$$, then the equation of the line will be

In $$\triangle \mathrm{ABC}$$, co-ordinates of $$\mathrm{A}$$ are $$(1,2)$$ and the equation of the medians through $$\mathrm{B}$$ and C are $$x+\mathrm{y}=5$$ and $$x=4$$ respectively. Then the midpoint of $$\mathrm{BC}$$ is

A, B are fixed points with coordinates (0, a) and (0, b) (a > 0, b > 0). P is variable point (x, 0) referred to rectangular axis. If the angle $$\angle$$APB is maximum, then