$A(a, 0)$ and $B(-a, 0)$ are two fixed points. If $\angle A - \angle B = \theta$,what is the locus of point $C$ of triangle $ABC$?

If the locus of a point which is equidistant from the coordinate axes forms a triangle with the line $y=3$,then the area of the triangle is

$A$ ray of light passing through the point $A(1, 2)$ is reflected at a point $B$ on the $x$-axis and then passes through $(5, 3)$. Then the equation of $AB$ is:

Suppose we have to cover the $XY$-plane with identical tiles such that no two tiles overlap and no gap is left between the tiles. Suppose that we can choose tiles of the following shapes: equilateral triangle,square,regular pentagon,regular hexagon. Then,the tiling can be done with tiles of

If the points $A(3, 4)$,$B(7, 7)$,and $C(a, b)$ are collinear and $AC = 10$,then $(a, b) =$

If $A(2, -3)$ and $B(-2, 1)$ are two vertices of a $\triangle ABC$ and if the centroid of $\triangle ABC$ lies on the line $2x + 3y = 1$,then the locus of vertex $C$ of $\triangle ABC$ is equal to