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:

The equations of two sides $AB$ and $AC$ of a triangle $ABC$ are $4x + y = 14$ and $3x - 2y = 5$,respectively. The point $\left(2, -\frac{4}{3}\right)$ divides the third side $BC$ internally in the ratio $2:1$. The equation of the side $BC$ is:

If the straight lines $2x + 3y - 1 = 0$,$x + 2y - 1 = 0$,and $ax + by - 1 = 0$ form a triangle with the origin as the orthocentre,then $(a, b)$ is equal to

If the equations of three sides of a triangle are $x = 2$,$y + 1 = 0$,and $x + 2y = 4$,then the coordinates of the circumcentre of this triangle are:

Suppose a triangle is formed by $x+y=10$ and the coordinate axes. Then the number of points $(x, y)$ where $x$ and $y$ are natural numbers,lying inside the triangle is

$L \equiv 7x - y + 8 = 0$ is one of the diagonals of a square for which $(-4, 5)$ and $(3, 4)$ are two vertices. Find the coordinates of the two vertices lying on the diagonal $L = 0$.