The diagonals of the parallelogram whose sides are $lx + my + n = 0$,$lx + my + n' = 0$,$mx + ly + n = 0$,and $mx + ly + n' = 0$ include an angle of:

Let one of the sides of a triangle be $17 \text{ cm}$ and the sum of all the sides of the triangle be $40 \text{ cm}$. If the sum of two adjacent sides is $35 \text{ cm}$, then the area (in $\text{cm}^2$) of the triangle is (in $\sqrt{2}$)

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 . . . . . .

Two sides of a parallelogram are along the lines $x + y = 3$ and $x - y + 3 = 0$. If its diagonals intersect at $(2, 4)$,then one of its vertices is

Straight lines $x-y=7$ and $x+4y=2$ intersect at $B$. Points $A$ and $C$ are so chosen on these two lines such that $AB=AC$. The equation of line $AC$ passing through $(2,-7)$ is

If the points $(1, 1)$,$(-1, -1)$,and $(-\sqrt{3}, \sqrt{3})$ are the vertices of a triangle,then this triangle is: