The three points $(-2, 2)$,$(8, -2)$,and $(-4, -3)$ are the vertices of:

The lines $x+y+4=0$,$x-2y-4=0$,and $3x+4y-2=0$:

Two sides of a square with an area of $25$ square units are given by the equations $3x - 4y = 0$ and $4x + 3y = 0$. The equations of the other two sides are:

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

Suppose $ABCD$ $(AB \parallel CD)$ is a trapezium such that the diagonals $AC$ and $BD$ bisect the angles $\angle DAB$ and $\angle CBA$,respectively. Then

Let $A, B$ be points on the two half-lines $x - \sqrt{3}|y| = \alpha, \alpha > 0$ at a distance of $\alpha$ from their point of intersection $P$. The line segment $AB$ meets the angle bisector of the given half-lines at the point $Q$. If $PQ = \frac{9}{2}$ and $R$ is the radius of the circumcircle of $\triangle PAB$, then $\frac{\alpha^2}{R}$ is equal to . . . . . .