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 

Two sides of a parallelogram are along the lines, $x + y = 3$ and $x -y + 3 = 0$. If its diagonals intersect at $(2, 4)$, then one of its vertex is

Let $A \equiv (3, 2)$ and $B \equiv (5, 1)$. $ABP$ is an equilateral triangle is constructed on the side of $AB$ remote from the origin then the orthocentre of triangle $ABP$ is

Two consecutive sides of a parallelogram are $4x + 5y = 0$ and $7x + 2y = 0.$ If the equation to one diagonal is $11x + 7y = 9,$ then the equation of the other diagonal 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 :

If in triangle $ABC$ ,$ A \equiv (1, 10) $, circumcentre $\equiv$ $\left( { - \,\,{\textstyle{1 \over 3}}\,\,,\,\,{\textstyle{2 \over 3}}} \right)$ and orthocentre $\equiv$ $\left( {{\textstyle{{11} \over 3}}\,\,,\,\,{\textstyle{4 \over 3}}} \right)$ then the co-ordinates of mid-point of side opposite to $A$ is :