If a straight line perpendicular to $2x - 3y + 7 = 0$ forms a triangle with the coordinate axes whose area is $3 \text{ sq. units}$,then the equation of the straight line is:

The triangle formed by the lines $x + y - 4 = 0,$ $3x + y = 4,$ and $x + 3y = 4$ is

Suppose the vertices of a triangle are given by $A(0,3)$,$B(-2,0)$,and $C(6,1)$. For the point $(\alpha, \alpha+1)$ to lie inside the triangle,$\alpha$ must lie in the interval:

$A$ square of side length $a$ has one vertex at the origin and one side along the $x$-axis. The side passing through the origin makes an angle $\alpha$ $(0 < \alpha < \pi/4)$ with the positive $x$-axis. Find the equation of the diagonal that does not pass through the origin.

One diagonal of a square is the $x$-axis. If one vertex of the square is $(1, 2)$,find the equations of the sides passing through this vertex.

If the vertices of a parallelogram are $(0, 0)$,$(1, 0)$,$(2, 2)$,and $(1, 2)$,then the angle between its diagonals is: