Consider the fourteen lines in the plane given by $y=x+r$ and $y=-x+r$,where $r \in \{0, 1, 2, 3, 4, 5, 6\}$. The number of squares formed by these lines,whose sides are of length $\sqrt{2}$,is:

Let $A(-1, 1)$,$B(3, 4)$,and $C(2, 0)$ be three given points. $A$ line $y = mx$,$m > 0$,intersects lines $AC$ and $BC$ at points $P$ and $Q$ respectively. Let $A_1$ and $A_2$ be the areas of $\Delta ABC$ and $\Delta PQC$ respectively,such that $A_1 = 3A_2$. Then the value of $m$ is equal to:

The line $x + y = p$ meets the $x$ and $y$ axes at $A$ and $B$ respectively. $A$ triangle $APQ$ is inscribed in the triangle $OAB$,where $O$ is the origin,with a right angle at $Q$. $P$ and $Q$ lie on $OB$ and $AB$ respectively. If the area of triangle $APQ$ is $3/8$ of the area of triangle $OAB$,then $\frac{AQ}{BQ}$ is equal to:

The lines $x + y = |a|$ and $ax - y = 1$ intersect each other in the first quadrant. Then,in which interval do all possible values of $a$ lie?

If $p_1$ and $p_2$ are the lengths of the perpendiculars from the origin to the lines $x \sec \alpha + y \csc \alpha = 2a$ and $x \cos \alpha - y \sin \alpha = a \cos 2\alpha$ respectively,then what is the value of $\left( \frac{p_1}{p_2} + \frac{p_2}{p_1} \right)^2$?

$A$ straight line $L$ is perpendicular to the line $5x - y = 1$ and the area of the triangle formed by the line $L$ and the coordinate axes is $5$ square units. The equation of the line $L$ can be