The diagonals of a parallelogram $ABCD$ are along the lines $x+3y=4$ and $6x-2y=7$. Then $ABCD$ must be a

The area (in square units) of the triangle formed by the lines $x=0, y=0$ and $3x+4y=12$ is:

If $P(\sin \alpha, \cos \alpha)$ lies inside the triangle formed by the vertices $(0,0), \left(\frac{\sqrt{3}}{2}, 0\right)$ and $\left(0, \frac{\sqrt{3}}{2}\right)$,then $\alpha$ lies in the interval

If $ABCD$ is a quadrilateral,and the midpoints of consecutive sides $AB, BC, CD$,and $DA$ are joined by straight lines to form a quadrilateral $PQRS$,then $PQRS$ is always:

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

If the sum of the distances of a point from two perpendicular lines in a plane is $1$,then its locus is