Consider a square $ABCD$ of side $12$ and let $M, N$ be the midpoints of $AB, CD$ respectively. Take a point $P$ on $MN$ and let $AP=r, PC=s$. Then,the area of the triangle whose sides are $r, s, 12$ is

$A$ straight line passing through a fixed point $(3, 5)$ intersects the coordinate axes at two points $A$ and $B$. If the locus of $C(x, y)$,which forms a rectangle with the points $A, O$ (origin),and $B$,is $ax + 2hxy + by = 0$,then $a + b + h =$

$A$ line passing through the point $(1, 2)$ meets the axes at $P$ and $Q$ such that it forms a triangle $OPQ$,where $O$ is the origin. If the area of triangle $OPQ$ is minimum,then the slope of the line $PQ$ is:

$A \equiv (\cos \theta, \sin \theta)$ and $B \equiv (\sin \theta, -\cos \theta)$ are two points. The locus of the centroid of $\triangle OAB$,where $O$ is the origin,is

The equation of the locus of a point which is equidistant from the points $(2, 3)$ and $(4, 5)$ is

If $M$ is the foot of the perpendicular drawn from the origin $O$ to a variable line $L$ passing through a fixed point $Q(a, b)$,then the locus of the mid-point of $OM$ is