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

If the points $(1, 1)$,$(-1, -1)$ and $(-\sqrt{3}, k)$ are vertices of an equilateral triangle,then the value of $k$ will be:

If two opposite vertices of a square are $(5, -4)$ and $(-3, 2)$,find its area.

Let $ABCD$ be a quadrilateral such that there exists a point $E$ inside the quadrilateral satisfying $AE=BE=CE=DE$. Suppose $\angle DAB, \angle ABC, \angle BCD$ are in an arithmetic progression. Then the median of the set $\{\angle DAB, \angle ABC, \angle BCD\}$ is

If the lines $3x + y - 4 = 0$,$x - ay - 10 = 0$,and $bx + 2y + 9 = 0$ form three successive sides of a rectangle in that order and the fourth side passes through $(1, 2)$,then the area of that rectangle (in sq. units) is

The triangle with vertices $A(2, 4)$,$B(2, 6)$,and $C(2 + \sqrt{3}, 5)$ is a . . . .