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

The ends $A$, $B$ of a straight line segment of constant length $c$ slide upon the fixed rectangular axes $O X, O Y$ respectively. If the rectangle $O A P B$ completed, then the locus of the foot of perpendicular drawn from $P$ to $A B$ is

Consider three points $P(\cos \alpha, \sin \beta), Q(\sin \alpha, \cos \beta)$ and $R(0,0)$, where $0<\alpha, \beta<\frac{\pi}{4}$. Then