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

The line passing through the points $(-2, 6)$ and $(4, 8)$ is perpendicular to the line passing through the points $(8, 12)$ and $(x, 24)$. Find the value of $x$.

$\beta$ 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 $x \cos \theta + y \sin \theta = p$ is the normal form and $y = mx + c$ is the slope-intercept form of the line $x + 2y + 1 = 0$,then $\tan^{-1}(\tan \theta + m + c) = $

$A$ line through $A(-5, -4)$ meets the lines $x + 3y + 2 = 0$,$2x + y + 4 = 0$,and $x - y - 5 = 0$ at $B$,$C$,and $D$ respectively. If $\left( \frac{15}{AB} \right)^2 + \left( \frac{10}{AC} \right)^2 = \left( \frac{6}{AD} \right)^2$,then the equation of the line is

$P(6,4)$ is a point on the line $x-y-2=0$. If $A(\alpha, \beta)$ and $B(\gamma, \delta)$ are two points on this line lying on either side of $P$ at a distance of $4$ units from $P$,then $\alpha^2+\beta^2+\gamma^2+\delta^2=$