If the midpoints of the sides $BC, CA$ and $AB$ of the triangle $ABC$ are $(1, 3), (5, 7)$ and $(-5, 7)$ respectively,then the equation of the side $AB$ is

$A$ straight line passing through the point $(2, 2)$ intersects the lines $\sqrt{3}x + y = 0$ and $\sqrt{3}x - y = 0$ at points $A$ and $B$ respectively. Find the equation of the line $AB$ such that the triangle $OAB$ is an equilateral triangle,where $O$ is the origin.

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

$A$ line $L$ passes through the point $P(1, 2)$ and makes an angle of $60^{\circ}$ with the positive $X$-axis. $A$ and $B$ are two points lying on $L$ at a distance of $4$ units from $P$. If $O$ is the origin,then the area of $\triangle OAB$ is

$\beta$ is the angle made by the perpendicular drawn from the origin to the line $L \equiv x+y-2=0$ with the positive $X$-axis in the anticlockwise direction. If '$a$' is the $X$-intercept of the line $L=0$ and $p$ is the perpendicular distance from the origin to the line $L=0$,then $a \tan \beta + p^2 =$

If the line passing through $(4, 3)$ and $(2, k)$ is perpendicular to the line $y = 2x + 3$,then the value of $k$ is: