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 $(-5, 1), (p, 5)$ and $(10, 7)$ are collinear,then the value of $p$ will be

The number of points,having both coordinates as integers,that lie in the interior of the triangle with vertices $(0,0)$,$(0,41)$,and $(41,0)$ is:

Let the area of the triangle with vertices $A(1, \alpha)$,$B(\alpha, 0)$,and $C(0, \alpha)$ be $4 \text{ sq. units}$. If the points $(\alpha, -\alpha)$,$(-\alpha, \alpha)$,and $(\alpha^2, \beta)$ are collinear,then $\beta$ is equal to:

If the mid-points of the sides $BC$,$CA$ and $AB$ of a triangle $ABC$ are respectively $(2,1)$,$(-1,-2)$ and $(3,3)$,then the equation of the side $BC$ is

Let the mirror image of point $A(\alpha, \beta)$ in the line mirror $x + 2y = 3$ be point $B$,and the image of $B$ in the line $3x - 2y = 5$ be $C$. If the origin is the orthocentre of triangle $ABC$ and $P(a, b)$ is a point inside the triangle such that triangles $PAB$,$PBC$,and $PCA$ have the same area,then $3(a + b)$ is: