The given points are $$A\left( {h,\,0} \right),\,B\left( {a,\,b} \right),\,C\left( {0,\,k} \right),$$      they lie on the same plane.
$$\therefore $$  Slope of $$AB =$$  Slope of $$BC$$
$$\therefore $$  Slope of $$AB = \frac{{b - 0}}{{a - h}} = \frac{b}{{a - h}}\,;$$
Slope of $$BC = \frac{{k - b}}{{0 - a}} = \frac{{k - b}}{{ - a}}$$
$$\eqalign{ & \therefore \,\frac{b}{{a - h}} = \frac{{k - b}}{{ - a}}\,{\text{ by cross multiplication}} \cr & {\text{or }} - ab = \left( {a - h} \right)\left( {k - b} \right) \cr & {\text{or }} - ab = ak - ab - hk + hb \cr & {\text{or }}0 = ak - hk + hb \cr & {\text{or }}ak + hb = hk \cr & {\text{Dividing by }}hk \Rightarrow \frac{{ak}}{{hk}} + \frac{{hb}}{{hk}} = 1{\text{ or }}\frac{a}{h} + \frac{b}{k} = 1 \cr} $$

$$\eqalign{ & p = \sqrt {{{\left( {4 - 4} \right)}^2} + {{\left( {3 - 0} \right)}^2}} = 3 \cr & q = \sqrt {{{\left( {4 - 0} \right)}^2} + {{\left( {3 - 3} \right)}^2}} = 4 \cr & {\text{Now,}} \cr & 4p = 4 \times 3 = 12 \cr & 3q = 3 \times 4 = 12 \cr & \therefore \,4p = 3q \cr} $$

Let the equation of any line passing through $$A\left( {h,\,k} \right){\text{ be }}y - k = m\left( {x - h} \right).$$

Let this line cut the $$x$$-axis and $$y$$-axis at $$P$$ and $$Q.$$
Then $$P \equiv \left( {h - \frac{k}{m},\,0} \right){\text{ and }}Q \equiv \left( {0,\,k - mh} \right).$$
Let $$S$$ be the area of $$\Delta OPQ,$$   then
$$\eqalign{ & S = \frac{1}{2}OP \times OQ \cr & \Rightarrow S = \frac{1}{2}\left( {h - \frac{k}{m}} \right)\left( {k - mh} \right) \cr & \Rightarrow S = \frac{1}{2}\frac{{\left( {mh - k} \right)\left( {k - mh} \right)}}{m} \cr & \Rightarrow 2mS = hkm - {k^2} - {h^2}{m^2} + khm \cr & \Rightarrow {h^2}{m^2} - 2\left( {hk - S} \right)m + {k^2} = 0 \cr} $$
Since, $$m$$ is real,
$$\therefore $$  its discriminant $$D \geqslant 0$$
$$\eqalign{ & \therefore \,4{\left( {hk - S} \right)^2} - 4{h^2}{k^2} \geqslant 0 \cr & \Rightarrow S - 2hk \geqslant 0 \cr & \Rightarrow S \geqslant 2hk \cr} $$
Hence, minimum value of $$S$$ is $$2hk$$  square units.

Let the line be $$y=mx+c$$    or $$mx-y+c=0.$$
The algebraic sum of the distances $$ = \frac{{m - 1 + c}}{{\sqrt {1 + {m^2}} }} + \frac{{2m + c}}{{\sqrt {1 + {m^2}} }} + \frac{{ - 2 + c}}{{\sqrt {1 + {m^2}} }} = 0$$
$$ \Rightarrow 3m + 3c = 3\,\,{\text{or }}1 = m + c$$
So, $$\left( {1,\,1} \right)$$  satisfies $$y=mx+c$$    for all $$m,\,c.$$

Equation of $$AB$$  is $$x\,\cos \,\alpha + y\,\sin \,\alpha = p;$$
$$\eqalign{ & \Rightarrow \frac{{x\,\cos \,\alpha }}{p} + \frac{{y{\mkern 1mu} \sin \,\alpha }}{p} = 1; \cr & \Rightarrow \frac{x}{{\frac{p}{{\cos \,\alpha }}}} + \frac{y}{{\frac{p}{{\sin \,\alpha }}}} = 1 \cr} $$
So co-ordinates of $$A$$ and $$B$$ are $$\left( {\frac{p}{{\cos \,\alpha }},\,0} \right){\text{ and }}\left( {0,\,\frac{p}{{\sin \,\alpha }}} \right);$$
So coordinates of midpoint of $$AB$$  are
$$\eqalign{ & \left( {\frac{p}{{2\,\cos \,\alpha }},\,\,\frac{p}{{2\,\sin \,\alpha }}} \right) = \left( {{x_1},\,{y_1}} \right)\left( {{\text{let}}} \right); \cr & {x_1} = \frac{p}{{2\,\cos \,\alpha }}\,\,\& \,\,{y_1} = \frac{p}{{2\,\sin \,\alpha }}; \cr & \Rightarrow \cos \,\alpha = \frac{p}{{2{x_1}}}\,\,\,{\text{and}}\,\,{\text{sin}}\,\alpha = \frac{p}{{2{y_1}}};\, \cr & {\cos ^2}\alpha + {\sin ^2}\alpha = 1 \cr & \Rightarrow \frac{{{p^2}}}{4}\left( {\frac{1}{{{x_1}^2}} + \frac{1}{{{y_1}^2}}} \right) = 1 \cr & {\text{Locus of}}\left( {{x_1},\,{y_1}} \right)\,{\text{is }}\frac{1}{{{x^2}}} + \frac{1}{{{y^2}}} = \frac{4}{{{p^2}}}{\text{ }} \cr} $$

$$\eqalign{ & p_1^2 = 4{a^2}{\cos ^2}4\theta \cr & p_2^2 = \frac{{16{a^2}{{\cos }^2}2\theta }}{{{{\sec }^2}\theta + {\text{cose}}{{\text{c}}^2}\theta }} \cr & = 16{a^2}{\cos ^2}2\theta \,{\cos ^2}\theta \,{\sin ^2}\theta \cr & = {a^2}{\sin ^2}4\theta \cr & \therefore \,p_1^2 + 4p_2^2 = 4{a^2} \cr} $$

$$\eqalign{ & {\text{Given, points are }}A = \left( {1,\,0} \right){\text{ and }}B = \left( {2,\,1} \right) \cr & {\text{Slope of }}AB = \frac{{1 - 0}}{{2 - 1}} = 1 \cr & {\text{Then angle of }}AB{\text{ with }}x{\text{ - axis is}} \cr & \angle BAX = {45^ \circ } \cr & {\text{Hence, }}\angle B'AX = {45^ \circ } - {30^ \circ } = {15^ \circ } \cr & {\text{Therefore for }}B'\left( {h,\,k} \right) \cr & h = 1 + \sqrt 2 \,\cos \,{15^ \circ },\,k = \sqrt 2 \,\sin \,{15^ \circ } \cr & {\text{We have, }}\sin \left( {{{15}^ \circ }} \right) = \frac{{\sqrt 6 - \sqrt 2 }}{4} \cr & \Rightarrow \cos \left( {{{15}^ \circ }} \right) = \frac{{\sqrt 6 + \sqrt 2 }}{4} \cr & \Rightarrow h = \frac{{3 + \sqrt 3 }}{2},\,k = \frac{{\sqrt 3 - 1}}{2} \cr} $$

If none of the coordinates are irrational then
\[\Delta = \frac{1}{2}\left| \begin{array}{l} {x_1}\,\,\,\,\,{y_1}\,\,\,\,\,1\\ {x_2}\,\,\,\,\,{y_2}\,\,\,\,\,1\\ {x_3}\,\,\,\,\,{y_3}\,\,\,\,\,1 \end{array} \right| = \frac{1}{2} \times {\rm{rational}} = {\rm{rational}}\]
But the area of an equilateral triangle
$$ = \frac{{\sqrt 3 }}{4} \times {\left( {{\text{side}}} \right)^2} = \frac{{\sqrt 3 }}{4} \times {\text{rational}} = {\text{irrational}}$$
So, both the coordinates of the third vertex cannot be rational.

\[\begin{array}{l} \frac{1}{2}\left| \begin{array}{l} \,\, - 2\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\\ \,\, - 1\,\,\,\,\,\,\,\,\,\frac{1}{{\sqrt 3 }}\,\,\,\,\,\,\,\,\,\,\,1\\ \cos \,\theta \,\,\,\,\sin \,\theta \,\,\,\,\,\,\,1 \end{array} \right| = 0\\ \Rightarrow \sqrt 3 \sin \,\theta - \cos \,\theta = 2\\ \Rightarrow \sin \left( {\theta - \frac{\pi }{6}} \right) = 1\\ \Rightarrow \theta - \frac{\pi }{6} = \frac{\pi }{2} \end{array}\]

Suppose $$B$$ (0, 1) be any point on given line and co-ordinate of $$A$$ is $$\left( {\sqrt 3 ,\,0} \right).$$   So, equation of

Reflected Ray is $$\frac{{ - 1 - 0}}{{0 - \sqrt 3 }} = \frac{{y - 0}}{{x - \sqrt 3 }}$$
$$ \Rightarrow \sqrt 3 y = x - \sqrt 3 $$