The coordinates of the vertices of a quadrilateral are $(2, -1)$,$(0, 2)$,$(2, 3)$,and $(4, 0)$. The angle between its diagonals is:

In a quadrilateral $ABCD$,which is not a trapezium,it is known that $\angle DAB = \angle ABC = 60^{\circ}$. Moreover,$\angle CAB = \angle CBD$. Then,

The points $A(2a, 4a)$,$B(2a, 6a)$,and $C(2a + \sqrt{3}a, 5a)$,where $a > 0$,are the vertices of:

$A$ 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.

The equation of a straight line,perpendicular to $3x - 4y = 6$ and forming a triangle of area $6 \text{ sq. units}$ with the coordinate axes,is

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