The point $P(2, 1)$ is translated to a point $Q$ parallel to the line $L \equiv x-y-4=0$ by $2 \sqrt{3}$ units. If the point $Q$ lies in the third quadrant,then the equation of the line passing through $Q$ and perpendicular to $L$ is

Let $A$ be the point $(0,4)$ and $B$ be a moving point on the $x$-axis. Let $M$ be the midpoint of $AB$ and let the perpendicular bisector of $AB$ meet the $y$-axis at $R$. The locus of the midpoint $P$ of $MR$ 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 the equation of the locus of a point equidistant from the points $(a_1, b_1)$ and $(a_2, b_2)$ is $(a_1 - a_2)x + (b_1 - b_2)y + c = 0$,then the value of $c$ is

$ABC$ is an isosceles triangle. If the coordinates of the base are $B(1, 3)$ and $C(-2, 7)$,the coordinates of vertex $A$ can be

$A$ rod of length $2l$ slides with its ends on two perpendicular lines. The locus of its mid-point is