If polygon has $$n$$ sides, then the number of diagonals
$$\eqalign{ & = \frac{{n\left( {n - 3} \right)}}{2} = 9 \cr & \therefore \,n = 6 \cr} $$

Now $$A$$ and $$B$$ can be adjacent vertices alternate vertices or opposite vertices.
If $$A$$ and $$B$$ are adjacent then side $$AB = 2,$$  then area $$ = 6 \times \Delta OAB$$
i.e area $$ = 6 \times \frac{{\sqrt 3 }}{4} \times {\left( 2 \right)^2} = 6\sqrt 3 $$

If $$A$$ and $$B$$ are alternate, then
$$\eqalign{ & 2\,\cos \,{30^ \circ } = a + a\,\cos \,{60^ \circ } \cr & \therefore \,{\text{side }}a = \frac{2}{{\sqrt 3 }} \cr & \therefore \,{\text{area}} = 6 \times \frac{{\sqrt 3 }}{4}{\left( {\frac{2}{{\sqrt 3 }}} \right)^2} = 2\sqrt 3 \cr} $$
Finally if $$A$$ and $$B$$ are opposite vertices then side $$a = \frac{1}{2}AB = 1$$
Then area $$ = 6 \times \frac{{\sqrt 3 }}{4}{\left( 1 \right)^2} = \frac{{3\sqrt 3 }}{2}$$

The diagram is self-explanatory.

From the diagram, $$PQRS$$   is a cyclic quadrilateral.

Let the required line be $$\frac{x}{a} + \frac{y}{b} = 1.....(1)$$
then $$a + b = - 1.....(2)$$
(1) passes through (4, 3), $$ \Rightarrow \frac{4}{a} + \frac{3}{b} = 1$$
$$ \Rightarrow 4b + 3a = ab.....(3)$$
Eliminating $$b$$ from (2) and (3), we get
$${a^2} - 4 = 0\,\, \Rightarrow a = \pm 2\,\, \Rightarrow b = - 3\,\,{\text{or}}\,1$$
$$\therefore $$ Equations of straight lines are
$$\frac{x}{2} + \frac{y}{{ - 3}} = 1\,\,{\text{and}}\,\,\frac{x}{{ - 2}} + \frac{y}{1} = 1$$

The given point lies on the lines $$x + y = 2,$$   if $$3{t^2} + 3t + 3 = 0.$$    Here discriminant $$9 - 12 < 0.$$
Therefore the value of $$t$$ is imaginary. Thus the given point cannot lie on the line.

Reflection about the line $$y =x,$$  changes the point (4, 1) to (1, 4).
On translation of (1, 4) through a distance of 2 units along $$+ve$$  direction of $$x$$-axis the point becomes (1 + 2, 4), i.e., (3, 4).

On rotation about origin through an angle $$\frac{\pi }{4}$$ the point $$P$$ takes the position $$P'$$ such that
$$OP= OP'$$
Also $$OP = 5 = OP'$$    and $$\cos \,\theta = \frac{3}{5},\,\,\,\sin \,\theta = \frac{4}{5}$$
$$\eqalign{ & {\text{Now, }}\,x = OP'\cos \left( {\frac{\pi }{4} + \theta } \right) \cr & = 5\left( {\cos \frac{\pi }{4}\cos \,\theta - \sin \frac{\pi }{4}\sin \,\theta } \right) \cr & = 5\left( {\frac{3}{{5\sqrt 2 }} - \frac{4}{{5\sqrt 2 }}} \right) \cr & = - \frac{1}{{\sqrt 2 }} \cr & y = OP'\sin \left( {\frac{\pi }{4} + \theta } \right) \cr & = 5\left( {\sin \frac{\pi }{4}\cos \,\theta + \cos \frac{\pi }{4}\sin \,\theta } \right) \cr & = 5\left( {\frac{3}{{5\sqrt 2 }} + \frac{4}{{5\sqrt 2 }}} \right) \cr & = \frac{7}{{\sqrt 2 }} \cr & \therefore P' = \left( { - \frac{1}{{\sqrt 2 }},\,\frac{7}{{\sqrt 2 }}} \right) \cr} $$

Let $$\frac{{AN}}{{BN}} = \lambda .$$
Then, the coordinates of $$N$$ are $$\left( {\frac{a}{{1 + \lambda }},\,\frac{{\lambda a}}{{1 + \lambda }}} \right).$$

Where $$\left( {a,\,0} \right)$$  and $$\left( {0,\,a} \right)$$  are the coordinates of $$A$$ and $$B$$ respectively. Now, equation of $$MN$$  perpendicular to $$AB$$  is
$$\eqalign{ & y - \frac{{\lambda a}}{{1 + \lambda }} = x - \frac{a}{{1 + \lambda }} \cr & \Rightarrow x - y = \frac{{1 - \lambda }}{{1 + \lambda }}a \cr} $$
So, the coordinates of $$M$$ are $$\left( {0,\,\frac{{1 - \lambda }}{{1 + \lambda }}a} \right)$$
Therefore, area of the $$\Delta AMN$$   is
$$\eqalign{ & = \frac{1}{2}\left| {\left[ {a\left( {\frac{{ - a}}{{\lambda + 1}}} \right) + \frac{{1 - \lambda }}{{{{\left( {1 + \lambda } \right)}^2}}}{a^2}} \right]} \right| \cr & = \frac{{\lambda {a^2}}}{{{{\left( {1 + \lambda } \right)}^2}}} \cr} $$
Also, area of $$\Delta OAB = \frac{{{a^2}}}{2}$$
So, that according to the given condition
$$\eqalign{ & \frac{{\lambda {a^2}}}{{{{\left( {1 + \lambda } \right)}^2}}} = \frac{3}{8}.\frac{1}{2}{a^2} \cr & \Rightarrow 3{\lambda ^2} - 10\lambda + 3 = 0 \cr & \Rightarrow \lambda = 3{\text{ or }}\lambda = \frac{1}{3} \cr} $$
For $$\lambda = \frac{1}{3},\,M$$   lies outside the segment $$OB$$  and hence the required value of $$\lambda $$ is $$3.$$

The sides of parallelogram are $$x=2, \,x=3, \,y=1, \,y=5.$$

$$\therefore $$ Diagonal $$AC$$  is $$\frac{{y - 1}}{{5 - 1}} = \frac{{x - 2}}{{3 - 2}}\,\,{\text{or}}\,\,y = 4x - 7$$
Equation diagonal $$BD$$  is $$\frac{{x - 2}}{{3 - 2}} = \frac{{y - 5}}{{1 - 5}}\,\,{\text{or}}\,\,4x + y = 13$$

$$\eqalign{ & \frac{{0 + x}}{2} = 1\,\,;\,\,\frac{{0 + y}}{2} = 2 \cr & x = 2\,\,;\,\,y = 4 \cr} $$
Equation of line passing through $$\left( {2,\,0} \right)$$  and $$\left( {0,\,4} \right)$$
$$\eqalign{ & y - 0 = \frac{{4 - 0}}{{0 - 2}}\left( {x - 2} \right) \cr & {\text{or }}y = - 2x + 4 \cr & {\text{or }}2x + y = 4 \cr} $$

Consider the first line
$$\eqalign{ & 2x - 3y = 1 \cr & \frac{x}{{\frac{1}{2}}} - \frac{y}{{\frac{1}{3}}} = 1 \cr} $$
Hence the intercepts are $$\frac{1}{2}$$ and $$\left( { - \frac{1}{3}} \right)......\left( {\text{i}} \right)$$
Similarly for the second line $$\frac{x}{{ - 3}} + \frac{y}{{\frac{{ - 3}}{2}}} = 1$$
Hence it has intercepts $$-3$$  and $$-1.5$$
Now both the lines intersect at $$\left( { - 1,\, - 1} \right)$$
This will be our origin.
Hence
Now consider the point $$\left( {2,\, - 1} \right)$$
Substituting the point in $$2x - 3y - 1 = 0$$
We get
$$8 > 0......\left( {\text{i}} \right)$$
Substituting it in $$x + 2y + 3 = 0$$
$$3 > 0......\left( {{\text{ii}}} \right)$$
Clearly from i and ii, it will lie in the new iv quadrant. Similarly applying the above steps for the point $$\left( {3,\,2} \right)$$  and $$\left( { - 1,\,2} \right)$$  we get that the point $$\left( {3,\,2} \right)$$  lies in the new first quadrant while the point $$\left( { - 1,\, - 2} \right)$$  lies in the new III^rd quadrant.

$${x^2} - {y^2} + 2y = 1{\text{ or }}x = \pm \left( {y - 1} \right)$$

The bisectors of the above lines are $$x = 0$$  and $$y = 1$$

So, the area between $$x = 0, y = 1$$   and $$x + y = 3$$   is the shaded region shown in the figure. The area is given by $$\left( {\frac{1}{2}} \right) \times 2 \times 2 = 2{\text{ sq}}{\text{. units}}.$$