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

If $A=(2,3)$ and $B=(-4,5)$ are two fixed points,then the locus of a point $P$ such that the area of $\triangle PAB$ is $12$ square units is

If $A=(5,3)$,$B=(3,-2)$ and a point $P$ is such that the area of the triangle $PAB$ is $9$,then the locus of $P$ represents

If for a variable line $\frac{x}{a} + \frac{y}{b} = 1$,the condition $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{c^2}$ ($c$ is a constant) is satisfied,then the locus of the foot of the perpendicular drawn from the origin to the line is:

Let $A(2,1)$ be a point and the equation of the straight line $L$ be $x-y=0$. Let $a$ and $b$ respectively represent the distances from a variable point $P(\alpha, \beta)$ to $A$ and to the line $L$. If $c$ is the distance of the point $A$ from the origin such that $a=bc$,then the locus of $P$ is

$A(1, 0)$,$B(0, 2)$,and $C(1, 2)$ are three points on the $XY$-plane. If a point $P(x, y)$ moves such that the area of $\triangle PAB$ is twice the area of $\triangle ABC$,then the locus of the point $P$ is: