Given points $A(6,0)$,$B(0,4)$,and $O$ as the origin,find the locus of a point $P(x, y)$ such that the area of $\triangle POB$ is $2$ times the area of $\triangle POA$.

Consider the locus of the point $P(x, y)$ which is equidistant from $(3, 0)$ and $(0, 4)$. If $A$ and $B$ are two points on this locus that satisfy $4x = 3y$ and $x = y$ respectively,then the distance between $A$ and $B$ is

$A$ line is drawn through the point $(1, 2)$ to meet the coordinate axes at $P$ and $Q$ such that it forms a $\triangle OPQ$,where $O$ is the origin. If the area of $\triangle OPQ$ is least,then the slope of the line $PQ$ is

If the equation of a curve remains unchanged by replacing $x$ with $y$ and $y$ with $x$,then the curve is

If the algebraic sum of the distances from the points $(2,0)$, $(0,2)$, and $(1,1)$ to a variable straight line is zero, then the line passes through the fixed point:

$A$ light ray emerging from the point source placed at $P(1, 3)$ is reflected at a point $Q$ on the $x$-axis. If the reflected ray passes through the point $R(6, 7)$,then the abscissa of $Q$ is