If the sum of the distances of a point from two perpendicular lines in a plane is $1$ unit, then its locus is

$A$ point moves in the $XY$-plane such that the sum of its distances from two mutually perpendicular lines is always equal to $3$. The area enclosed by the locus of that point is (in sq. units)

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

The set of all points that forms a triangle of area $15$ sq units with the points $(1, -2)$ and $(-5, 3)$ lies on

$A$ straight line through the point of intersection of the lines $x+2y=4$ and $2x+y=4$ meets the coordinate axes at $A$ and $B$. The locus of the mid-point of $AB$ is

Given $A(1, 1)$ and $AB$ is any line through it cutting the $x-$ axis in $B$. If $AC$ is perpendicular to $AB$ and meets the $y-$ axis in $C$,then the equation of the locus of the midpoint $P$ of $BC$ is