If the vertices $P$ and $Q$ of a triangle $PQR$ are given by $(2, 5)$ and $(4, -11)$ respectively,and the point $R$ moves along the line $N: 9x + 7y + 4 = 0$,then the locus of the centroid of the triangle $PQR$ is a straight line parallel to

$A$ light ray emerging from the point source placed at $P(1, 3)$ is reflected at a point $Q$ on the $x$-axis. If the reflected ray passes through the point $R(6, 7)$,then the abscissa of $Q$ 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 $45^{\circ}$

$A$ variable line passing through a fixed point $(\alpha, \beta)$ intersects the coordinate axes at $A$ and $B$. If $O$ is the origin,then the locus of the centroid of the $\triangle OAB$ is

The coordinates of a point $P$ on the line $2x - y + 5 = 0$ such that $|PA - PB|$ is maximum,where $A$ is $(4, -2)$ and $B$ is $(2, -4)$,are:

If the distance of any point $P(x, y)$ from the points $A(a + b, a - b)$ and $B(a - b, a + b)$ are equal,then the locus of $P$ is: