The locus of a point such that the sum of its distances from two given perpendicular lines is equal to $2$ units in the first quadrant is

$A(a, 0)$ and $B(-a, 0)$ are two fixed points. If $\angle A - \angle B = \theta$,what is the locus of point $C$ of triangle $ABC$?

If a line $AB$ of length $r$ moves so that $A$ and $B$ always lie respectively on the $X$-axis and the line $y=6x$,then the locus of the mid-point of $AB$ is:

Let $A(5, -3)$,$B(3, -2)$,and $C(-1, 5)$ be three points. If $P$ is a point satisfying the condition $PA^2 + 2PB^2 = 3PC^2$,then a point that lies on the locus of $P$ is

The Cartesian form of the polar equation $\theta = \tan^{-1} 2$ is

If the algebraic sum of the perpendicular distances from the points $(2,0)$,$(0,2)$,and $(1,1)$ to a variable line is zero,then the variable line always passes through a fixed point. The coordinates of that point are