The equation of the base of an equilateral triangle is $x + y = 2$ and the vertex is $(2, -1)$. The length of the side of the triangle is

The distance between the lines $3x + 4y = 9$ and $6x + 8y = 15$ is (in $\text{ units}$)

The points $(2,3)$ and $(-4,-4/3)$ lie on the opposite sides of the line $L \equiv 5x - 6y + k = 0$,where $k$ is an integer. If the points $(1,2)$ and $(4,5)$ lie on the same side of the line $L = 0$,then the perpendicular distance from the origin to the line $L = 0$ is:

The length of the perpendicular from the point $(b, a)$ to the line $\frac{x}{a} - \frac{y}{b} = 1$ is:

$L \equiv x \cos \alpha + y \sin \alpha - p = 0$ 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 distance of the point $(2, 3)$ from the line $2x - 3y + 28 = 0$,measured parallel to the line $\sqrt{3}x - y + 1 = 0$,is equal to