$A$ line $AB$ passes through the origin $(0, 0)$ and is perpendicular to the line $CD$ given by $3x + 4y + 6 = 0$. The equation of line $AB$ is:

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

The equation of the line making an angle of $135^\circ$ with the positive $x$-axis and having a $y$-intercept equal to $-2$ is:

If $\alpha$ is the angle made by the perpendicular drawn from the origin to the line $12x - 5y + 13 = 0$ with the positive $X$-axis in the anti-clockwise direction,then $\alpha =$

The equations of the sides $AB$,$AC$,and $BC$ of a $\triangle ABC$ are respectively $x-3y=0$,$3x-y=0$,and $x+y+4=0$. If $P$ and $Q$ are points on the line $3x-y+k=0$ passing through $B$ such that $PB:BQ=1:1$,then $k=$

$A$ line passing through the point $A(2, 3)$ makes an angle of $45^{\circ}$ with the $x$-axis and intersects the line $x + y + 1 = 0$ at point $P$. Find the distance $AP$.