$A$ straight line passes through a fixed point $(h, k)$. The locus of the foot of the perpendicular drawn from the origin to this line 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

$ABC$ is an isosceles triangle. If the coordinates of the base are $B(1, 3)$ and $C(-2, 7)$,the coordinates of vertex $A$ can be

If $P = (1, 1)$,$Q = (3, 2)$ and $R$ is a point on the $x$-axis,then the value of $PR + RQ$ will be minimum at

$O$ is the origin and $A$ is the point $(3,4)$. If a point $P(x,y)$ moves such that the line segment $OP$ is always parallel to the line segment $OA$,then the equation of the locus of $P$ is:

If $L : ax + by + c = 0$ is a variable straight line,where $a, b$ and $c$ are the second,fourth,and seventh terms of an $AP$ respectively,then $L$ passes through the fixed point: