If a vertex of an equilateral triangle is at the origin and the second vertex is $(4, 0)$,then its third vertex is

If a straight line $L$ is perpendicular to the line $4x - 2y = 1$ and forms a triangle of area $4 \text{ sq unit}$ with the coordinate axes,then the equation of the line $L$ is

Let a point $A$ lie between the parallel lines $L_1$ and $L_2$ such that its distances from $L_1$ and $L_2$ are $6$ and $3$ units, respectively. Then the area (in sq. units) of the equilateral triangle $ABC$, where the points $B$ and $C$ lie on the lines $L_1$ and $L_2$ respectively, is:

Let $A(1, 1)$,$B(-4, 3)$,and $C(-2, -5)$ be the vertices of a triangle $ABC$. Let $P$ be a point on the side $BC$,and let $\Delta_{1}$ and $\Delta_{2}$ be the areas of triangle $APB$ and triangle $ABC$,respectively. If $\Delta_{1} : \Delta_{2} = 4 : 7$,then find the area enclosed by the lines $AP$,$AC$,and the $x$-axis.

If $A(6, 3)$,$B(-3, 5)$,$C(4, -2)$,and $D(x, 3x)$ are four points. If the ratio of the area of $\Delta DBC$ and $\Delta ABC$ is $1 : 2$,then the value of $x$ is:

The diagonals $AC$ and $BD$ of a rhombus $ABCD$ intersect at the point $(3,4)$. If $BD=2 \sqrt{2}$,$A=(1,2)$,$B=(\alpha, \beta)$,$D=(\gamma, \delta)$ and $\alpha < \delta < \gamma < \beta$,then $\beta+\gamma-\delta=$