$$\eqalign{ & {\text{Let }}A = \left( {a,\,0} \right),\,B = \left( {0,\,b} \right) \cr & {\text{Then }}\sqrt {{a^2} + {b^2}} = AB = k\,\left( {{\text{constant}}} \right) \cr & {\text{If the middle point of }}AB{\text{ be }}P = \left( {{x_1},\,{y_1}} \right){\text{ then }}{x_1} = \frac{a}{2},\,{y_1} = \frac{b}{2} \cr & \therefore {\left( {2{x_1}} \right)^2} + {\left( {2{y_1}} \right)^2} = {k^2}, \cr & {\text{i}}{\text{.e}}{\text{., the equation of the locus of }}P{\text{ is }}{x^2} + {y^2} = {\left( {\frac{k}{2}} \right)^2} \cr} $$

Here the lines are coincident. So, $$5{\left( {\frac{y}{x}} \right)^2} + 4\left( {\frac{y}{x}} \right) + k = 0$$      will have equal roots.
$$\therefore \,\,{4^2} - 4.5.k = 0\,\,\,\, \Rightarrow k = \frac{4}{5}$$

The equation of $$AB$$  is
$$\eqalign{ & y - 1 = \frac{{0 - 1}}{{2 - 1}}x \cr & {\text{or }}x + 2y - 2 = 0 \cr} $$

$$\left| {PA - PB} \right| \leqslant AB$$
Thus, $$\left| {PA - PB} \right|$$   is maximum if the points $$A,\,B,$$  and $$P$$ are collinear.
Hence, solving $$x + 2y - 2 = 0$$    and $$4x + 3y + 9 = 0,$$
we get point $$P\,\left( { - \frac{{84}}{5},\,\frac{{13}}{5}} \right)$$

It is clear from the figure $$m < 5$$  and $$n < 4.$$  so, $$1 \leqslant m < 5$$   and $$2 \leqslant n < 4.$$   Also $$2m + n - 2 > 0$$    and $$4m + 5n - 20 < 0.$$    Under these conditions possible coordinates are $$\left( {1,\,1} \right),\,\left( {1,\,2} \right),\,\left( {1,\,3} \right),\,\left( {2,\,1} \right),\,\left( {2,\,2} \right),\,\left( {3,\,2} \right)$$

The point of coincidence on the $$y$$-axis is $$Q\left( {0,\,\lambda } \right)$$

The image of $$P\left( {2,\,3} \right)$$   on the $$y$$-axis is $${P_1}\left( { - 2,\,3} \right)$$
$${P_1},\,Q$$  and $$R$$ are collinear.
Therefore, Slope of $${P_1}Q = $$   Slope of $${P_1}R$$
$$\eqalign{ & {\text{or }}\frac{{\lambda - 3}}{{0 - \left( { - 2} \right)}} = \frac{{10 - 3}}{{5 - \left( { - 2} \right)}} \cr & {\text{or }}\lambda - 3 = 2 \cr & {\text{or }}\lambda = 5 \cr} $$
Therefore, the point $$Q$$ is $$\left( {0,\,5} \right).$$

The relations between polar coordinates and cartesian coordinates are $$x = r\cos \,\theta ,\,y = r\sin \,\theta .$$      So, the cartesian coordinates of the points are $$\left( {0,\,0} \right),\,\left( {0,\,3} \right)$$   and $$\left( {\frac{{3\sqrt 3 }}{2},\,\frac{3}{2}} \right)$$   respectively. Hence the sides are equal.

Let $$P\left( {x,\,y} \right)$$   be a point and $$A = \left( {a,\,0} \right),\,B = \left( { - a,\,0} \right)$$
Now, $$P{A^2} = {\left( {x - a} \right)^2} + {y^2}$$
$$P{B^2} = {\left( {x + a} \right)^2} + {y^2}$$
Since the sum of the distances of the point $$P\left( {x,\,y} \right)$$   from the points $$A = \left( {a,\,0} \right)$$   and $$B = \left( { - a,\,0} \right)$$   is $$2{b^2}$$
$$\eqalign{ & \therefore \,P{A^2} + P{B^2} = 2{b^2} \cr & {\left( {x - a} \right)^2} + {\left( {y - 0} \right)^2} + {\left( {x + a} \right)^2} + {\left( {y - 0} \right)^2} = 2{b^2} \cr & \Rightarrow {x^2} + {a^2} - 2ax + {y^2} + {x^2} + {a^2} + 2ax + {y^2} = 2{b^2} \cr & \Rightarrow {x^2} + {a^2} + {y^2} = {b^2} \cr & \Rightarrow {x^2} + {a^2} = {b^2} - {y^2} \cr} $$

Given, $$x + y = 3$$
∴ Required area
$$\eqalign{ & = 4 \times \frac{1}{2} \times 3 \times 3 \cr & = 18{\text{ sq}}{\text{. units}} \cr} $$

Intersection of $$3x + 4y=9$$    and $$y=mx+1.$$
For $$x$$ co-ordinate
$$\eqalign{ & 3x + 4\left( {mx + 1} \right) = 9 \cr & \Rightarrow \left( {3 + 4m} \right)x = 5 \cr & \Rightarrow x = \frac{5}{{3 + 4m}} \cr} $$
For $$x$$ to be an integer $$3 + 4m$$   should be a divisor of 5 i.e., $$1,\, - 1,\,5\,\,{\text{or }}\, - 5$$
$$\eqalign{ & 3 + 4m = 1\,\,\, \Rightarrow m = - \frac{1}{2}\,\,\left( {{\text{not an integer}}} \right) \cr & 3 + 4m = - 1\,\,\, \Rightarrow m = - 1\,\,\,\left( {{\text{integer}}} \right) \cr & 3 + 4m = 5\,\,\, \Rightarrow m = \frac{1}{2}\,\,\left( {{\text{not an integer}}} \right) \cr & 3 + 4m = - 5\,\,\, \Rightarrow m = - 2\,\,\left( {{\text{integer}}} \right) \cr} $$
$$\therefore $$ There are 2 integral values of $$m.$$
$$\therefore $$ (A) is the correct alternative.

$$\eqalign{ & {\text{From the choice of the axes,}} \cr & A = \left( {0,\,0} \right),\,B = \left( {2\cot \,\theta ,\,2} \right),\,C = \left( {4\cot \,\overline {{{60}^ \circ } - \theta } ,\, - 4} \right) \cr & \therefore {\left( {{\text{side of the equilateral triangle}}} \right)^2} \cr & \,\,\,\,\,\,\,\,\,\,\, = 4{\cot ^2}\theta + 4 \cr & \,\,\,\,\,\,\,\,\,\,\, = 16{\cot ^2}\left( {{{60}^ \circ } - \theta } \right) + 16 \cr & \Rightarrow 4{\text{cose}}{{\text{c}}^2}\,\theta = 16{\text{cose}}{{\text{c}}^2}\left( {{{60}^ \circ } - \theta } \right) \cr & \therefore {\text{cosec}}\,\theta = 2{\text{cosec}}\left( {{{60}^ \circ } - \theta } \right) \cr & {\text{or}}\,\,2\sin \,\theta = \sin \left( {{{60}^ \circ } - \theta } \right) \cr & \therefore 2\sin \,\theta = \frac{{\sqrt 3 }}{2}\cos \,\theta - \frac{1}{2}\sin \,\theta \cr & {\text{or}}\,\,5\sin \,\theta = \sqrt 3 \cos \,\theta \,\,\,\therefore \tan \,\theta = \frac{{\sqrt 3 }}{5} \cr & \therefore {\text{the required length}} = 2{\text{cosec}}\,\theta = 2 \times \frac{{\sqrt {25 + 3} }}{{\sqrt 3 }} = \frac{{2\sqrt {28} }}{{\sqrt 3 }} \cr} $$