If the three points A(1, 6), B(3, $$-$$4) and C(x, y) are collinear then the equation satisfying by x and y is

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