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

The points $(1,3)$ and $(5,1)$ are opposite vertices of a diagonal of a rectangle. If the other two vertices lie on the line $y=2x+c$,then one of the vertices on the other diagonal is

If the orthocentre of the triangle formed by the lines $y = x + 1$,$y = 4x - 8$,and $y = mx + c$ is at $(3, -1)$,then $m - c$ is :

Let $A(1, 1)$,$B(-4, 3)$,and $C(-2, -5)$ be the vertices of a triangle $ABC$. Let $P$ be a point on the side $BC$,and let $\Delta_{1}$ and $\Delta_{2}$ be the areas of triangle $APB$ and triangle $ABC$,respectively. If $\Delta_{1} : \Delta_{2} = 4 : 7$,then find the area enclosed by the lines $AP$,$AC$,and the $x$-axis.

One side of a square is inclined at an acute angle $\alpha$ with the positive $x-$axis,and one of its extremities is at the origin. If the remaining three vertices of the square lie above the $x-$axis and the side of a square is $4$,then the equation of the diagonal of the square which is not passing through the origin is