The ends of a rod of length $l$ move on two mutually perpendicular lines. The locus of the point on the rod which divides it in the ratio $1 : 2$ is

If the sum of the perpendicular distances of a variable point $P(x, y)$ from the lines $x + y - 5 = 0$ and $3x - 2y + 7 = 0$ is always $10$,show that $P$ must move on a line.

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

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 $45^{\circ}$

$P(x, y)$ moves such that the area of the triangle formed by $P, Q(a, 2a)$ and $R(-a, -2a)$ is equal to the area of the triangle formed by $P, S(a, 2a)$ and $T(2a, 3a)$. The locus of $P$ is a straight line given by:

If the sum of distances from a point $P$ to two mutually perpendicular straight lines is $1$ unit, then the locus of $P$ is