The points $(0, 0)$,$(a, 0)$,and $\left( \frac{a}{2}, \frac{a\sqrt{3}}{2} \right)$ are vertices of:

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

If the points $(0, 0)$,$(2, 2\sqrt{3})$ and $(a, b)$ are the vertices of an equilateral triangle,then $(a, b) = $

Suppose $\triangle ABC$ is an isosceles triangle with $\angle C=90^{\circ}$,$A=(2,3)$ and $B=(4,5)$. Then the centroid of the triangle is

Two sides of a rhombus are along the lines $x-y+1=0$ and $7x-y-5=0$. If its diagonals intersect at $(-1,-2)$,then one of the vertices of this rhombus is

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