$A$ triangle $ABC$ lying in the first quadrant has two vertices as $A(1, 2)$ and $B(3, 1)$. If $\angle BAC = 90^{\circ}$ and $\text{ar}(\Delta ABC) = 5\sqrt{5}$ sq. units,then the abscissa of the vertex $C$ is

Three points are $A(6, 3)$,$B(-3, 5)$,and $C(4, -2)$. If $P(x, y)$ is a point,then the ratio of the area of $\Delta PBC$ to the area of $\Delta ABC$ is:

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

If the points $(k, 3), (2, k),$ and $(-k, 3)$ are collinear,then the values of $k$ 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