Consider the locus of the point $P(x, y)$ which is equidistant from $(3, 0)$ and $(0, 4)$. If $A$ and $B$ are two points on this locus that satisfy $4x = 3y$ and $x = y$ respectively,then the distance between $A$ and $B$ is

$A$ straight line $L$ with negative slope passes through the point $(1,1)$ and cuts the positive coordinate axes at the points $A$ and $B$. If $O$ is the origin,then the minimum value of $OA + OB$ as $L$ varies,is

Let $A(-1,1)$ and $B(2,3)$ be two points and $P(x,y)$ be a variable point above the line $AB$ such that the area of $\triangle PAB$ is $10$. If the locus of $P$ is $ax+by=15$,then $5a+2b$ is:

The Cartesian form of the polar equation $\theta = \tan^{-1} 2$ 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 distance from a variable point $P$ to the point $(4, 3)$ is equal to the perpendicular distance from $P$ to the line $x + 2y - 1 = 0$,then the equation of the locus of the point $P$ is