In a $\triangle ABC$,$2x+3y+1=0$ and $x+2y-2=0$ are the perpendicular bisectors of its sides $AB$ and $AC$ respectively. If $A=(3,2)$,then the equation of the side $BC$ is

The area of the parallelogram formed by the lines $x \cos \alpha + y \sin \alpha = p$,$x \cos \alpha + y \sin \alpha = q$,$x \cos \beta + y \sin \beta = r$,and $x \cos \beta + y \sin \beta = s$ is:

If the equation of the base of an equilateral triangle is $x+y=6$ and the opposite vertex is the point $(-1, -1)$,then the area of the triangle is equal to $k$ sq. units. Find the value of $k$.

The area of the parallelogram formed by the lines $y = mx, y = mx + 1, y = nx, y = nx + 1$ is:

Let $A, B$ be points on the two half-lines $x - \sqrt{3}|y| = \alpha, \alpha > 0$ at a distance of $\alpha$ from their point of intersection $P$. The line segment $AB$ meets the angle bisector of the given half-lines at the point $Q$. If $PQ = \frac{9}{2}$ and $R$ is the radius of the circumcircle of $\triangle PAB$, then $\frac{\alpha^2}{R}$ is equal to . . . . . .

The set of values that $\beta$ can assume so that the point $(0, \beta)$ lies on or inside the triangle formed by the lines $3x+y+2=0$,$2x-3y+5=0$,and $x+4y-14=0$ is