Find the area of the quadrilateral formed by the lines $4y - 3x = 1, 4y - 3x - 3 = 0, 3y - 4x + 1 = 0,$ and $3y - 4x + 2 = 0$.

If vertices of a quadrilateral are $A(0,0), B(3,4), C(7,7)$ and $D(4,3)$,then the quadrilateral $ABCD$ is

Let $ABC$ be a triangle with $A(-3, 1)$ and $\angle ACB = \theta$,where $0 < \theta < \frac{\pi}{2}$. If the equation of the median through $B$ is $2x + y - 3 = 0$ and the equation of the angle bisector of $C$ is $7x - 4y - 1 = 0$,then $\tan \theta$ is equal to:

Let $ABC$ be an isosceles triangle in which $A$ is at $(-1, 0)$,$\angle A = \frac{2\pi}{3}$,$AB = AC$ and $B$ is on the positive $x$-axis. If $BC = 4\sqrt{3}$ and the line $BC$ intersects the line $y = x + 3$ at $(\alpha, \beta)$,then $\frac{\beta^4}{\alpha^2}$ is :

If the mid-points of the sides $BC$,$CA$ and $AB$ of a triangle $ABC$ are respectively $(2,1)$,$(-1,-2)$ and $(3,3)$,then the equation of the side $BC$ is

If the vertices of a triangle are $(0, 2)$,$(1, 0)$,and $(3, 1)$,then the triangle is