Choose the correct statement which describes the position of the point $(-6, 2)$ relative to the straight lines $2x + 3y - 4 = 0$ and $6x + 9y + 8 = 0$.

$L \equiv x \cos \alpha + y \sin \alpha - p = 0$ represents a line perpendicular to the line $x + y + 1 = 0$. If $p$ is positive,$\alpha$ lies in the fourth quadrant,and the perpendicular distance from $(\sqrt{2}, \sqrt{2})$ to the line $L = 0$ is $5$ units,then $p =$

The base of an equilateral triangle is along the line given by $3x + 4y = 9$. If a vertex of the triangle is $(1, 2)$,then the length of a side of the triangle is

If $p$ is the length of the perpendicular from the origin to the line whose intercepts on the axes are $a$ and $b$,then show that $\frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}}$.

If ${p_1}, {p_2}$ and ${p_3}$ are the perpendicular distances from the points $({m^2}, 2m)$,$(mm', m + m')$ and $(m'^2, 2m')$ respectively to the line $x \cos \alpha + y \sin \alpha + \frac{\sin^2 \alpha}{\cos \alpha} = 0$,then ${p_1}, {p_2}$ and ${p_3}$ are in:

The equation of the line passing through $(1, 2)$ and having a distance equal to $7$ units from the point $(8, 9)$ is: