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

Starting at time $t=0$ from the origin with speed $1 \text{ m/s}$,a particle follows a two-dimensional trajectory in the $x-y$ plane so that its coordinates are related by the equation $y=\frac{x^2}{2}$. The $x$ and $y$ components of its acceleration are denoted by $a_x$ and $a_y$,respectively. Then:
$(A)$ $a_x=1 \text{ m/s}^2$ implies that when the particle is at the origin,$a_y=1 \text{ m/s}^2$
$(B)$ $a_x=0$ implies $a_y=1 \text{ m/s}^2$ at all times
$(C)$ at $t=0$,the particle's velocity points in the $x$-direction
$(D)$ $a_x=0$ implies that at $t=1 \text{ s}$,the angle between the particle's velocity and the $x$-axis is $45^{\circ}$

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 $P_1, P_2, P_3, \ldots, P_n$ are $n$ points on the line $y=x$ all lying in the first quadrant,such that $OP_n = n(OP_{n-1})$ ($O$ is origin),$OP_1 = 1$ and $P_n = (2520 \sqrt{2}, 2520 \sqrt{2})$,then $n=$

The locus of the incentre of the triangle formed by the lines $xy-4x-4y+16=0$ and $x+y=5$ is

Let $A(2, -3)$ and $B(-2, 1)$ be two vertices of $\Delta ABC$. If the centroid of the triangle moves on the line $2x + 3y = 1$,then the locus of the vertex $C$ is given by