The area of the triangle bounded by the straight line $ax + by + c = 0$ $(a, b, c \neq 0)$ and the coordinate axes is

Let $O$ denote the origin. If $M(1, 2)$,$N(0, 1)$,and $A(x, y)$ are points such that $xy > 0$ and $x + y < 1$,then choose the correct statement.

If the vertices of a triangle $ABC$ are $A(1,7)$,$B(-5,-1)$,and $C(7,4)$,then the equation of the internal angle bisector of $\angle ABC$ is

$A$ line $L$ passes through the points $(1, 1)$ and $(2, 0)$ and another line $L'$ passes through $\left( \frac{1}{2}, 0 \right)$ and is perpendicular to $L$. Then the area of the triangle formed by the lines $L, L'$ and the $y$-axis is:

Three vertices of a triangle are $A(4, 3)$,$B(1, -1)$,and $C(7, k)$. The value$(s)$ of $k$ for which the centroid,orthocentre,incentre,and circumcentre of the $\Delta ABC$ lie on the same straight line is/are:

Let a triangle be bounded by the lines $L_{1}: 2x + 5y = 10$,$L_{2}: -4x + 3y = 12$,and the line $L_{3}$,which passes through the point $P(2, 3)$,intersects $L_{2}$ at $A$ and $L_{1}$ at $B$. If the point $P$ divides the line segment $AB$ internally in the ratio $1:3$,then the area of the triangle is equal to