The equation of the locus of points which are equidistant from the points $(2,3)$ and $(4,5)$ is

$A$ point moves in the $XY$-plane such that the sum of its distances from two mutually perpendicular lines is always equal to $3$. The area enclosed by the locus of that point is (in sq. units)

The locus of the orthocentre of the triangle formed by the lines $(1+p) x-p y+p(1+p)=0$,$(1+q) x-q y+q(1+q)=0$,and $y=0$,where $p \neq q$,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

Let $a \neq 0, b \neq 0, c$ be three real numbers and $L(p, q) = \frac{ap + bq + c}{\sqrt{a^2 + b^2}}, \forall p, q \in \mathbb{R}$. If $L\left(\frac{2}{3}, \frac{1}{3}\right) + L\left(\frac{1}{3}, \frac{2}{3}\right) + L(2, 2) = 0$,then the line $ax + by + c = 0$ always passes through the fixed point:

$A$ point $P$ moves inside a square of area $4$ square units such that it is nearer to the point of intersection of its diagonals than any vertex. The area of the region traced by $P$ is