$A$ 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

If the slope of a line passing through the point $A(3, 2)$ is $3/4$,then the points on the line which are $5$ units away from $A$ are:

$A$ line cuts the $X$ and $Y$ axes at the points $A$ and $B$ respectively. If the point $(5, 6)$ divides the line segment $AB$ internally in the ratio $3: 1$,then the equation of the line is:

If the line $6x - 7y + 8 + \lambda(3x - y + 5) = 0$ is parallel to the $y$-axis,then $\lambda$ is equal to

The line passing through the points $(1, 4)$ and $(-5, 1)$ intersects the line $4x + 3y - 5 = 0$ at the point:

$\beta$ 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 $a \tan \beta + p^2 =$