Find the slope of the line making an inclination of $60^{\circ}$ with the positive direction of the $x$-axis.

Let $O=(0,0)$. Let $A$ and $B$ be points on the $X$-axis and $Y$-axis respectively such that $\angle OBA = 60^{\circ}$. Let $D$ be a point in the first quadrant such that $\triangle OAD$ is an equilateral triangle. Then,the slope of $DB$ is

$A$ line $L$ is perpendicular to the line $5x - y = 1$ and the area of the triangle formed by the line $L$ and the coordinate axes is $5$. The equation of the line $L$ is

The lines $2x + 3y = 6$ and $2x + 3y = 8$ cut the $X$-axis at $A$ and $B$,respectively. $A$ line $L$ drawn through the point $(2, 2)$ meets the $X$-axis at $C$ in such a way that the abscissae of $A, B,$ and $C$ are in arithmetic progression. Then,the equation of the line $L$ is

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

The straight line passing through $(-3, 6)$ and the midpoint of the line segment joining the points $(4, -5)$ and $(-2, 9)$ has an inclination of: