The locus of the mid-points of the perpendiculars drawn from points on the line, $\mathrm{x}=2 \mathrm{y}$ to the line $\mathrm{x}=\mathrm{y}$ is 

Let $b, d>0$. The locus of all points $P(r, \theta)$ for which the line $P$ (where, $O$ is the origin) cuts the line $r \sin \theta=b$ in $Q$ such that $P Q=d$ is

Let $PS$ be the median of the triangle with vertices $P(2,2) , Q(6,-1) $ and $R(7,3) $. The equation of the line passing through $(1,-1) $ and parallel to $PS $ is :

The locus of a point $P$ which divides the line joining $(1, 0)$ and $(2\cos \theta ,2\sin \theta )$ internally in the ratio $2 : 3$ for all $\theta $, is a

Let $PQR$ be a right angled isosceles triangle, right angled at $P\, (2, 1)$. If the equation of the line $QR$ is $2x + y = 3$, then the equation representing the pair of lines $PQ$ and $PR$ is

The line $\frac{x}{a} + \frac{y}{b} = 1$ moves in such a way that $\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{2{c^2}}}$, where $a, b, c \in R_0$ and $c$ is constant, then locus of the foot of the perpendicular from the origin on the given line is -