The equation of the lines on which the perpendiculars from the origin make ${30^o}$ angle with $x$-axis and which form a triangle of area $\frac{{50}}{{\sqrt 3 }}$ with axes, are

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 diagonals of the parallelogram whose sides are $lx + my + n = 0,$ $lx + my + n' = 0$,$mx + ly + n = 0$, $mx + ly + n' = 0$ include an angle

The medians $AD$ and $BE$ of a triangle with vertices $A\;(0,\;b),\;B\;(0,\;0)$ and $C\;(a,\;0)$ are perpendicular to each other, if

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 -

The area of the parallelogram formed by the lines $y = mx,\,y = mx + 1,\,y = nx$ and $y = nx + 1$ equals