Reduce the following equation into slope-intercept form and find its slope and the $y$-intercept.
$x+7y=0$

If $x \cos \alpha + y \sin \alpha = p$ is the normal form of the equation of a straight line $x + \sqrt{3} y + 4 = 0$ and $a, b$ are respectively $X$ and $Y$ intercepts of this line,then $\sqrt{3} \pi b p - 3 a \alpha = $

If the length of the perpendicular to a line from the origin is $2 \sqrt{2}$ units,which makes an angle of $135^{\circ}$ with the $X$-axis,then the equation of the line is

$A$ ray of light coming from the point $(2, 2\sqrt{3})$ is incident at an angle $30^{\circ}$ with the normal to the line $x=1$ at the point $A$. The ray gets reflected on the line $x=1$ and meets the $x$-axis at the point $B$. Then,the line $AB$ passes through the point

The line passing through $(1, 0)$ and $(-2, \sqrt{3})$ makes an angle of ...... with the $x$-axis. (in $^o$)

If the line $\frac{x}{a} + \frac{y}{b} = 1$ passes through the points $(2, -3)$ and $(4, -5)$,then $(a, b) = $