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

Let $A=(2,3)$ and $B=(3,-5)$ be two vertices of $\triangle ABC$ such that $C$ is a point on the line $L \equiv 3x + 4y - 5 = 0$. Then the locus of the centroid of $\triangle ABC$ is a line parallel to

$A (a, 0)$ and $B (-a, 0)$ are two fixed points of $\Delta ABC$. If the vertex $C$ moves such that $\cot A + \cot B = \lambda$,where $\lambda$ is a constant,then what is the locus of point $C$?

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 $45^{\circ}$

The locus of a point which is at a distance of $2$ units from the line $2x - 3y + 4 = 0$ and at a distance of $\sqrt{13}$ units from the point $(5, 0)$ 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: