Let $$m$$ and $${m^2}$$ be the slopes of the lines represented by $$a{x^2} + 2hxy + b{y^2} = 0$$
Then, $$m + {m^2} = - \frac{{2h}}{b}$$   and $$m{m^2} = \frac{a}{b}{\text{ or }}{m^3} = \frac{a}{b}$$

$$\eqalign{ & \therefore \,{\left( {m + {m^2}} \right)^3} = {\left( { - \frac{{2h}}{b}} \right)^2} \cr & \therefore \,{m^3} + {m^6} + 3m{m^2}\left( {m + {m^2}} \right) = - \frac{{8{h^3}}}{{{b^3}}} \cr & \therefore \,\frac{a}{{{b^2}}}\left( {a + b} \right) + \frac{{8{h^3}}}{{{b^3}}} = \frac{{6ah}}{{{b^2}}} \cr & \therefore \,\frac{{a + b}}{h} + \frac{{8{h^2}}}{{ab}} = 6 \cr} $$
These are the set of parallel lines and the distance between parallel lines are equal. So, the figure is a rhombus.

$$\eqalign{ & {\text{The vertices, }}O\left( {0,\,0} \right),\,A\left( {\frac{1}{{m - n}},\,\frac{m}{{m - n}}} \right),\,B\,\left( {0,\,1} \right) \cr & Ar\left( {{{\left. {} \right\|}^{gm}}} \right.\,OABC = 2Ar\left( {\Delta OAB} \right) \cr & = 2\frac{1}{2}\left| {\left[ {0\left( {\frac{m}{{m - n}} - 1} \right) + \frac{1}{{m - n}}\left( {1 - 0} \right) + 0\left( {0 - \frac{m}{{m - n}}} \right)} \right]} \right| \cr & = \frac{1}{{\left| {m - n} \right|}} \cr} $$

Clearly from the figure, the image is the combined rays given by $$y = 1.\left( {x - 2} \right)$$   and $$y = - 1.\left( {x - 2} \right)$$    i.e., $$y = \left| {x - 2} \right|.$$

Slope of line $$L = - \frac{b}{5}$$
Slope of line $$K = - \frac{3}{c}$$
Line $$L$$ is parallel to line $$k.$$
$$ \Rightarrow \frac{b}{5} = \frac{3}{c}\,\, \Rightarrow bc = 15$$
(13, 32) is a point on $$L.$$
$$\eqalign{ & \therefore \frac{{13}}{5} + \frac{{32}}{b} = 1 \cr & \Rightarrow \frac{{32}}{b} = - \frac{8}{5} \cr & \Rightarrow b = - 20 \cr & \Rightarrow c = - \frac{3}{4} \cr} $$
Equation of $$K:\,\,\,\,\,\,y - 4x = 3 \Rightarrow 4x - y + 3 = 0$$
Distance between $$L$$ and $$K\,\,\, = \frac{{\left| {52 - 32 + 3} \right|}}{{\sqrt {17} }} = \frac{{23}}{{\sqrt {17} }}$$

Let point $$Q$$ be $$\left( {x,\,0} \right)$$

Since, $${m_{PQ}} = - {m_{RQ}}$$
therefore $$\frac{{0 - 2}}{{x - 1}} = - \left( {\frac{{8 - 0}}{{5 - x}}} \right) \Rightarrow x = \frac{9}{5}$$

Given lines are
$$4ax+2ay+c=0$$
$$5bx + 2by + d=0$$
The point of intersection will be
$$\eqalign{ & \frac{x}{{2ad - 2bc}} = \frac{{ - y}}{{4ad - 5bc}} = \frac{1}{{8ab - 10ab}} \cr & \Rightarrow x = \frac{{2\left( {ad - bc} \right)}}{{ - 2ab}} = \frac{{bc - ad}}{{ad}} \cr & \Rightarrow y = \frac{{5bc - 4ad}}{{ - 2ad}} = \frac{{4ad - 5bc}}{{2ab}} \cr} $$
$$\because $$ Point of intersection is in fourth quadrant so $$x$$ is positive and $$y$$ is negative.
Also distance from axes is same
So $$x=-y$$   ($$\because $$ distance from $$x$$-axis is $$-y$$  as $$y$$ is negative)
$$\eqalign{ & \frac{{bc - ad}}{{ab}} = \frac{{5bc - 4ad}}{{2ab}} \cr & \Rightarrow \,\,3bc - 2ad = 0 \cr} $$

$$\eqalign{ & Q = \left( {2 \pm 3\sqrt 2 \cos \,\theta ,\,1 \pm 3\sqrt 2 \sin \,\theta } \right){\text{ where }}\tan \,\theta = - 1 \cr & = \left( {2 \pm 3\sqrt 2 \times \frac{{ - 1}}{{\sqrt 2 }},\,1 \pm 3\sqrt 2 \times \frac{1}{{\sqrt 2 }}} \right) = \left( {2 \mp 3,\,1 \pm 3} \right) \cr} $$
From the question, the $$y$$-coordinate of $$Q$$ should be more than 1 ( $$y$$-coordinate of $$P$$ ).

If a point is equidistant from the two intersecting lines, then the locus of this point is the angle bisector of those lines.
Now, let $$\left( {h,\,k} \right)$$  be the point which is equidistant from the lines $$4x - 3y + 7 = 0$$    and $$3x - 4y + 14 = 0$$
$$\eqalign{ & {\text{Then, }}\frac{{4h - 3k + 7}}{{\sqrt {{4^2} + {{\left( { - 3} \right)}^2}} }} = \pm \frac{{3h - 4k + 14}}{{\sqrt {{3^2} + {{\left( { - 4} \right)}^2}} }} \cr & \Rightarrow 4h - 3k + 7 = \pm \left( {3h - 4k + 14} \right) \cr & \Rightarrow h + k - 7 = 0{\text{ and }}7h - 7k + 21 = 0 \cr} $$
Hence locus of $$\left( {h,\,k} \right)$$  is $$x + y - 7 = 0$$    and $$x - y + 3 = 0.$$

$$\eqalign{ & {L_1}:y - x = 0 \cr & {L_2}:2x + y = 0 \cr & {L_3}:y + 2 = 0 \cr} $$
On solving the equation of line $${L_1}$$  and $${L_2}$$  we get their point of intersection (0, 0) i.e., origin $$O.$$
On solving the equation of line $${L_1}$$  and $${L_3},$$
we get $$P = \left( { - 2,\, - 2} \right)$$
Similarly, we get $$Q = \left( { - 1,\, - 2} \right)$$
We know that bisector of an angle of a triangle, divide the opposite side the triangle in the ratio of the sides including the angle [ Angle Bisector Theorem of a Triangle ]
$$\therefore \frac{{PR}}{{RQ}} = \frac{{OP}}{{OQ}} = \frac{{\sqrt {{{\left( { - 2} \right)}^2} + {{\left( { - 2} \right)}^2}} }}{{\sqrt {{{\left( { - 1} \right)}^2} + {{\left( { - 2} \right)}^2}} }} = \frac{{2\sqrt 2 }}{{\sqrt 5 }}$$

$$\eqalign{ & \because \,A\left( {h,\,k} \right),\,B\left( {1,\,1} \right){\text{ and }}C\left( {2,\,1} \right) \cr & {\text{are the vertices of a right angled trangle }}ABC \cr & {\text{Now, }}AB = \sqrt {{{\left( {1 - h} \right)}^2} + {{\left( {1 - k} \right)}^2}} \cr & {\text{or }}BC = \sqrt {{{\left( {2 - 1} \right)}^2} + {{\left( {1 - 1} \right)}^2}} = 1 \cr & {\text{or }}AC = \sqrt {{{\left( {h - 2} \right)}^2} + {{\left( {k - 1} \right)}^2}} \cr & {\text{Now, pythagorus theorem}} \cr & A{C^2} = A{B^2} + B{C^2} \cr & \Rightarrow 4 + {h^2} - 4h + {k^2} + 1 - 2k = {h^2} + 1 - 2h + {k^2} + 1 - 2k + 1 \cr & \Rightarrow 5 - 4h = 3 - 2h \cr & \Rightarrow h = 1 \cr & {\text{Now, given that area of the triangle is 1,}} \cr & {\text{Then, area }}\left( {\Delta ABC} \right) = \frac{1}{2} \times AB \times BC \cr & 1 = \frac{1}{2} \times \sqrt {{{\left( {1 - h} \right)}^2} + {{\left( {1 - k} \right)}^2}} \times 1 \cr & 2 = \sqrt {{{\left( {1 - h} \right)}^2} + {{\left( {1 - k} \right)}^2}} ......\left( 1 \right) \cr & {\text{Putting }}h = 1{\text{ from equation }}\left( 1 \right){\text{, we get}} \cr & 2 = \sqrt {{{\left( {k - 1} \right)}^2}} \cr & {\text{Squaring both the sides, we get}} \cr & 4 = {k^2} + 1 - 2k \cr & {\text{or, }}{k^2} - 2k - 3 = 0 \cr & {\text{or, }}\left( {k - 3} \right)\left( {k + 1} \right) = 0 \cr & {\text{or, }}k = - 1,\,3 \cr & {\text{Thus the set of values of }}k{\text{ is }}\left\{ { - 1,\,3} \right\} \cr} $$