Consider a $\triangle ABC$ in the $XY$-plane with vertices $A=(0,0)$,$B=(1,1)$,and $C=(9,1)$. If the line $x=a$ divides the triangle into two parts of equal area,then $a$ equals

The points $(0, 8/3), (1, 3)$ and $(82, 30)$ are:

In a triangle $ABC$,$AD$ and $BE$ are medians. If $AD=4$,$\angle DAB = \frac{\pi}{6}$ and $\angle ABE = \frac{\pi}{3}$,then the area of $\triangle ABC$ is

$A$ square of side length $a$ has one vertex at the origin and one side along the $x$-axis. The side passing through the origin makes an angle $\alpha$ $(0 < \alpha < \pi/4)$ with the positive $x$-axis. Find the equation of the diagonal that does not pass through the origin.

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 $\left(5, \frac{a}{4}\right)$ be the circumcenter of a triangle with vertices $A(a, -2)$,$B(a, 6)$,and $C\left(\frac{a}{4}, -2\right)$. Let $\alpha$ denote the circumradius,$\beta$ denote the area,and $\gamma$ denote the perimeter of the triangle. Then $\alpha + \beta + \gamma$ is