The coordinates of three points $A(-4, 0)$,$B(2, 1)$,and $C(3, 1)$ determine the vertices of an isosceles trapezium $ABCD$. The coordinates of the vertex $D$ are:

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

The area of the parallelogram formed by the lines $x \cos \alpha + y \sin \alpha = p$,$x \cos \alpha + y \sin \alpha = q$,$x \cos \beta + y \sin \beta = r$,and $x \cos \beta + y \sin \beta = s$ 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 . . . . . .

If the orthocenter of the triangle formed by the lines $2x + 3y - 1 = 0$,$x + 2y + 1 = 0$,and $ax + by - 1 = 0$ lies at the origin,then $\frac{1}{a} + \frac{1}{b} =$

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?