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

$A$ straight line makes equal negative intercepts on the coordinate axes. If the perpendicular distance from the origin to the line is $4 \text{ units}$, then the equation of the line is:

Let $A(a, b)$,$B(3, 4)$,and $C(-6, -8)$ respectively denote the centroid,circumcentre,and orthocentre of a triangle. Then,the distance of the point $P(2a+3, 7b+5)$ from the line $2x+3y-4=0$ measured parallel to the line $x-2y-1=0$ is

Let $R$ be the point $(3,7)$ and let $P$ and $Q$ be two points on the line $x+y=5$ such that $\triangle PQR$ is an equilateral triangle. Then the area of $\triangle PQR$ is

If the perpendicular distances from the points $(2, 3)$,$(4, a)$ and $(\alpha, \beta)$ to the line $3x + 4y - 3 = 0$ are equal and $4\alpha - 3\beta + 1 = 0$,then the sum of all possible values of $a$,$\alpha$,and $\beta$ is:

Find the perpendicular distance from the origin to the line joining the points $(\cos \theta, \sin \theta)$ and $(\cos \phi, \sin \phi)$.