The pair of lines $x^2 - 8x + 12 = 0$ and $y^2 - 14y + 45 = 0$ form a square. What is the center of the circle inscribed in the square?

The triangle with vertices at $(-2, 2)$,$(2, -2)$,and $(1, 1)$ is:

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 . . . . . .

The points $(0, -1), (2, 1), (0, 3),$ and $(-2, 1)$ are the vertices of which figure?

If $A(6, 3)$,$B(-3, 5)$,$C(4, -2)$,and $D(x, 3x)$ are four points. If the ratio of the area of $\Delta DBC$ and $\Delta ABC$ is $1 : 2$,then the value of $x$ is:

The area (in sq units) of the quadrilateral formed by the lines $2x + 3y + 6 = 0$,$2x - 3y + 6 = 0$,$2x + 3y - 6 = 0$,and $2x - 3y - 6 = 0$ is