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

If $A=(0,1), B=(1,2), C=(-2,1)$,then the equation of the locus of a point $P(x,y)$ such that the area of triangle $PAB$ equals the area of triangle $PAC$ is:

If the equation of the locus of a point equidistant from the points $(a_1, b_1)$ and $(a_2, b_2)$ is $(a_1 - a_2)x + (b_1 - b_2)y + c = 0$,then the value of $c$ is

Let $A$ be the point $(0,4)$ in the $xy$-plane and let $B$ be the point $(2t, 0)$. Let $L$ be the midpoint of $AB$ and let the perpendicular bisector of $AB$ meet the $y$-axis at $M$. Let $N$ be the midpoint of $LM$. Then the locus of $N$ is

$A$ quadrilateral $ABCD$ is divided by the diagonal $AC$ into two triangles of equal areas. If $A, B, C$ are respectively $(3, 4), (-3, 6), (-5, 1)$,then the locus of $D$ is

$A$ straight line passing through a fixed point $(2, 3)$ intersects the coordinate axes at distinct points $P$ and $Q$. If $O$ is the origin and the rectangle $OPRQ$ is completed,then the locus of $R$ is: