For $a > b > c > 0$,the distance between $(1,1)$ and the point of intersection of the lines $ax + by + c = 0$ and $bx + ay + c = 0$ is less than $2\sqrt{2}$. Then:

The base $BC$ of a triangle $ABC$ is bisected at the point $(p, q)$ and the equations to the sides $AB$ and $AC$ are respectively $px + qy = 1$ and $qx + py = 1$. Then the equation to the median through $A$ is

The area of the parallelogram formed by the lines $y = mx$,$y = mx + 1$,$y = nx$,and $y = nx + 1$ is equal to

$O(0,0), A(-3,-1)$ and $B(-1,-3)$ are the vertices of a $\triangle OAB$. $P$ is a point on the perpendicular $AD$ drawn from $A$ on $OB$ such that $\frac{AP}{PD}=\frac{3}{4}$. Then the equation of the line $L$ parallel to $OB$ and passing through $P$ is:

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:

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: