The required line is the perpendicular bisector of the line segment joining the given points.

Equation of bisectors of second pair of straight lines is,
$$q{x^2} + 2xy - q{y^2} = 0.....(1)$$
It must be identical to the first pair
$${x^2} - 2pxy - {y^2} = 0.....(2)$$
From equation (1) and (2), we get
$$\eqalign{ & \frac{q}{1} = \frac{2}{{ - 2p}} = \frac{{ - q}}{{ - 1}} \cr & \Rightarrow pq = - 1 \cr} $$

Here $$\left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right) = 0.$$
Clearly, $$x+y=0$$   is a real straight line. But $${x^2} - xy + {y^2} = 0$$    represents two imaginary straight lines because $$D = {\left( { - 1} \right)^2} - 4.1.1 < 0.$$

Solving the given equations of lines pairwise, we get the vertices of $$\Delta $$ as
$$\eqalign{ & A\left( { - 2,\,2} \right)\,,B\left( {2,\, - 2} \right),\,C\left( {1,\,1} \right) \cr & {\text{Then}}\,AB = \sqrt {16 + 16} = 4\sqrt 2 \cr & BC = \sqrt {1 + 9} = \sqrt {10} \cr & CA = \sqrt {9 + 1} = \sqrt {10} \cr & \therefore \Delta {\text{ is isosceles}}{\text{.}} \cr} $$

Let the parametric equation of drawn line is $$\frac{x}{{\cos \,\theta }} = \frac{y}{{\sin \,\theta }} = r\, \Rightarrow x = r\,\cos \,\theta ,\,\,y = r\,\sin \,\theta $$
Putting it in $$'{L_1}',$$  we get
$$\eqalign{ & r\,\sin \,\theta = r\,\cos \,\theta + 10 \cr & \Rightarrow \frac{1}{{OA}} = \frac{{\sin \,\theta - \cos \,\theta }}{{10}} \cr} $$

Similarly, putting the general point of drawn line is the equation of $${L_2},$$ we get
$$\eqalign{ & \frac{1}{{OB}} = \frac{{\sin \,\theta - \cos \,\theta }}{{20}} \cr & {\text{Let }}P = \left( {h,\,k} \right){\text{ and }}OP = r \cr & \Rightarrow r\,\cos \,\theta = h,\,\,r\,\sin \,\theta = k,{\text{ we have}} \cr & \frac{2}{r} = \frac{{\sin \,\theta - \cos \,\theta }}{{10}} + \frac{{\sin \,\theta - \cos \,\theta }}{{20}} \cr & \Rightarrow 40 = 3r\,\sin \,\theta - 3r\,\cos \,\theta \Rightarrow 3y - 3x = 40 \cr} $$

Since the point $$A\left( {2,\,1} \right)$$   is translated parallel to $$x - y = 3,\,AA'$$    has the same slope as that of $$x - y = 3.$$   Therefore, $$AA'$$  passes through $$\left( {2,\,1} \right)$$  and has slope $$1$$. Here, $$\tan \,\theta = 1{\text{ or }}\cos \,\theta = \frac{1}{{\sqrt 2 }},\,\sin \,\theta = \frac{1}{{\sqrt 2 }}.$$

Thus, the equation of $$AA'$$  is $$\frac{{x - 2}}{{\cos \left( {\frac{\pi }{4}} \right)}} = \frac{{y - 1}}{{\sin \left( {\frac{\pi }{4}} \right)}}$$
Since $$AA' = 4,$$   the coordinates of $$A'$$ are given by
$$\eqalign{ & \frac{{x - 2}}{{\cos \left( {\frac{\pi }{4}} \right)}} = \frac{{y - 1}}{{\sin \left( {\frac{\pi }{4}} \right)}} = - 4 \cr & {\text{or }}x = 2 - 4\,\cos \frac{\pi }{4},\,\,y = 1 - 4\,\sin \frac{\pi }{4} \cr & {\text{or }}x = 2 - 2\sqrt 2 ,\,\,y = 1 - 2\sqrt 2 \cr} $$
Hence, the coordinates of $$A'$$ are $$\left( {2 - 2\sqrt 2 ,\,1 - 2\sqrt 2 } \right)$$

Let other two sides of rhombus are
$$x - y + \lambda = 0$$
and $$7x - y + \mu = 0$$
then $$O$$ is equidistant from $$AB$$  and $$DC$$  and from $$AD$$  and $$BC$$
$$\eqalign{ & \therefore \left| { - 1 + 2 + 1} \right| = \left| { - 1 + 2 + \lambda } \right| \Rightarrow \lambda = - 3 \cr & {\text{and }}\left| { - 7 + 2 - 5} \right| = \left| { - 7 + 2 + \mu } \right| \Rightarrow \mu = 15 \cr} $$
$$\therefore $$ Other two sides are $$x-y-3 =0$$   and $$7x-y+15=0$$
On solving the equations of sides pairwise, we get the vertices as $$\left( {\frac{1}{3},\frac{{ - 8}}{3}} \right),\,\left( {1,\,2} \right),\,\left( {\frac{{ - 7}}{3},\frac{{ - 4}}{3}} \right),\,\left( { - 3,\, - 6} \right)$$

$$\because Ar\left( {\Delta OPR} \right) = Ar\left( {\Delta PQR} \right) = Ar\left( {\Delta OQR} \right)$$

$$\therefore $$ By simply geometry $$R$$ should be the centroid of $$\Delta PQO$$
$$ \Rightarrow R\left( {\frac{{3 + 6 + 0}}{3},\,\frac{{4 + 0 + 0}}{3}} \right) = \left( {3,\,\frac{4}{3}} \right)$$

$$\eqalign{ & {\text{Let the origin be }}O. \cr & {\text{So, }}O = \left( {0,\,0} \right) \cr & {\text{Now }}A{B^2} = {\left( {a - c} \right)^2} + {\left( {b - d} \right)^2} \cr & O{A^2} = {\left( {a - 0} \right)^2} + {\left( {b - 0} \right)^2} = {a^2} + {b^2} \cr & {\text{and }}O{B^2} = {\left( {c - 0} \right)^2} + {\left( {d - 0} \right)^2} = {c^2} + {d^2} \cr & {\text{Now from the }}\Delta AOB: \cr & \cos \,\theta = \frac{{O{A^2} + O{B^2} - A{B^2}}}{{2\,OA.\,OB}} \cr & = \frac{{{a^2} + {b^2} + {c^2} + {d^2} - \left\{ {{{\left( {a - c} \right)}^2} + {{\left( {b - d} \right)}^2}} \right\}}}{{2\sqrt {{a^2} + {b^2}} \sqrt {{c^2} + {d^2}} }} \cr & = \frac{{2\left( {ac + bd} \right)}}{{2\sqrt {\left( {{a^2} + {b^2}} \right)\left( {{c^2} + {d^2}} \right)} }} \cr & = \frac{{ac + bd}}{{\sqrt {\left( {{a^2} + {b^2}} \right)\left( {{c^2} + {d^2}} \right)} }} \cr} $$

The given lines are
$$\eqalign{ & \,2x + y = \frac{9}{2}.....(1) \cr & {\text{and }}2x + y = - 6.....(2) \cr} $$
Signs of constants on R.H.S. show that two lines lie on opposite sides of origin. Let any line through origin meets these lines in $$P$$ and $$Q$$ respectively then required ratio is $$OP: OQ$$

$$\eqalign{ & {\text{Now in }}\Delta OPA\,\,{\text{and }}\Delta OQC, \cr & \angle POA = \angle QOC\,\,\left( {{\text{ver}}{\text{.}}\,\,{\text{opp}}{\text{.}}\,\,\angle 's} \right) \cr & \angle PAO = \angle OCQ\,\,\left( {{\text{alt}}{\text{. int}}{\text{.}}\,\,\angle 's} \right) \cr & \therefore \Delta OPA \sim \Delta OQC\,\,\left( {{\text{by }}AA{\text{ similarly}}} \right) \cr & \therefore \frac{{OP}}{{OQ}} = \frac{{OA}}{{OC}} = \frac{{\frac{9}{4}}}{3} = \frac{3}{4} \cr & \therefore \,{\text{Required ratio is 3 : 4}} \cr} $$