Maths M.C.Q (1 Marks)

==> $3x - 4y + 3 = 0 \Rightarrow \frac{x}{{ - 1}} + \frac{y}{{3/4}} = 1$

Hence required length is $\sqrt {{{( - 1)}^2} + {{\left( {\frac{3}{4}} \right)}^2}} = \frac{5}{4}$.

Thus the points of intersection of the three lines on the transversal are $\left( {\frac{1}{{1 + {m_1}}},\frac{{{m_1}}}{{1 + {m_1}}}} \right),$ $\left( {\frac{1}{{1 + {m_2}}},\frac{{{m_2}}}{{1 + {m_2}}}} \right)$ and$\left( {\frac{1}{{1 + {m_3}}},\frac{{{m_3}}}{{1 + {m_3}}}} \right)$

By hypothesis, ${\left( {\frac{1}{{1 + {m_1}}} - \frac{1}{{1 + {m_2}}}} \right)^2} + {\left( {\frac{{{m_1}}}{{1 + {m_1}}} - \frac{{{m_2}}}{{1 + {m_2}}}} \right)^2}$

= ${\left( {\frac{1}{{1 + {m_2}}} - \frac{1}{{1 + {m_3}}}} \right)^2} + {\left( {\frac{{{m_2}}}{{1 + {m_2}}} - \frac{{{m_3}}}{{1 + {m_2}}}} \right)^2}$

==> $\frac{{{m_2} - {m_1}}}{{1 + {m_1}}} = \frac{{{m_3} - {m_2}}}{{1 + {m_3}}}$or $\frac{{1 + {m_2}}}{{1 + {m_1}}} - 1 = 1 - \frac{{1 + {m_2}}}{{1 + {m_3}}}$

==> $\frac{{1 + {m_2}}}{{1 + {m_1}}} + \frac{{1 + {m_2}}}{{1 + {m_3}}} = 2$==> $1 + {m_2} = \frac{{2(1 + {m_1})(1 + {m_3})}}{{(1 + {m_1}) + (1 + {m_3})}}$

==> $1 + {m_1},1 + {m_2},1 + {m_3}$ are in $H.P.$

Hence the equation of straight line is $\frac{x}{{14 - b}} + \frac{y}{b} = 1$.

Also, it passes through $(3,4)$

$\therefore $ $\frac{3}{{14 - b}} + \frac{4}{b} = 1 \Rightarrow b = 8$ or $7$

Therefore equations are $4x + 3y = 24$ and $x + y = 7$.

Trick : This question can be checked with the options as the line $4x + 3y = 24$ passes through $(3, 4)$ and also cuts the intercepts from the axes whose sum is $14$.

.${\tan ^{ - 1}}m \pm {\tan ^{ - 1}}m = {\tan ^{ - 1}}\frac{{2m}}{{1 - {m^2}}}$ or ${\tan ^{ - 1}}0$

Therefore equation of lines are $y = 0$, $y = \frac{{2mx}}{{1 - {m^2}}}$.

Therefore the equation of line $AB$ is $\frac{x}{a} + \frac{y}{b} = 1$

==> $\frac{x}{{2{x_1}}} + \frac{y}{{2{y_1}}} = 1 $

$\Rightarrow x{y_1} + y{x_1} = 2{x_1}{y_1}$.

$⇒$$\left| {\frac{{ - m}}{{\sqrt {1 + {m^2}} }}} \right| = \frac{{\sqrt 3 }}{2}$

$⇒$ $m = \pm \sqrt 3 $.

Hence the lines are $\sqrt 3 x + y - \sqrt 3 = 0$ and $\sqrt 3 x - y - \sqrt 3 = 0$.

Hence the required line is $3x - 2y = 30$.

$\therefore $ Line is $y = - \sqrt 3 x + c$

==> $\sqrt 3 x + y = c$

Now $\frac{c}{2} = |4|$

$\Rightarrow c = \pm {\rm{ }}8 $

$\Rightarrow x\sqrt 3 + y = \pm \,8$.

Thus $\tan {45^o} = \pm \frac{{m - \frac{1}{2}}}{{1 + m.\frac{1}{2}}}$ ==> $m = \pm 3$.

Hence option $(b)$ is correct.

$y - a = m{\rm{ }}(x - 0)$ or $mx - y + a = 0$ …..$(i)$

If the length of perpendicular from $(2a, 2a)$ to the line $(i)$ is ‘a’, then $a = \pm\frac{{m(2a) - 2a + a}}{{\sqrt {{m^2} + 1} }} \Rightarrow m = 0,\frac{4}{3}$.

Hence the required equations of lines are $y - a = 0$, $4x - 3y + 3a = 0$.

$\frac{x}{{ - (32/3)}} + \frac{y}{{(24/5)}} = 1$

$ \Rightarrow 9x - 20y + 96 = 0$.

Therefore, the required lines will make angles of ${90^o}$and ${210^o}$ i.e., ${30^o}$with the positive direction of $x$-axis.

Hence the lines are $x = 0$ and $y = \frac{1}{{\sqrt 3 }}x$.

So, $3a + 2b = 13$.....$(i)$

Point $Q(b,a)$is on $4x - y = 5$

So, $4b - a = 5$.....$(ii)$

By solving $(i)$ and $(ii)$, $a = 3,b = 2$

$P(a,b) \to (3,\,2)$and $Q(b,a) \to (2,\,3)$

Now, equation of $PQ$ $y - {y_1} = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}(x - {x_1}) $

$\Rightarrow y - 2 = \frac{{3 - 2}}{{2 - 3}}(x - 3)$

==> $y - 2 = - (x - 3)$

$\Rightarrow x + y = 5$.

$\frac{x}{{ - 3}} + \frac{y}{2} = 1$....$.(ii)$

According to the condition, $ - \frac{8}{a} = - ( - 3)$

$ \Rightarrow a = - \frac{8}{3}$ and $ - \frac{8}{b} = - (2) \Rightarrow b = 4$.

${b^3} - {c^3} = k(b - c),$ ${c^3} - {a^3} = k(c - a)$,${a^3} - {b^3} = k(a - b)$

==> $b - c = 0$or ${b^2} + {c^2} + bc = k$

==> $c - a = 0$or ${c^2} + {a^2} + ac = k$

==> $a - b = 0$or ${a^2} + {b^2} + ab = k$

==> $b = c,c = a,a = b$ or ${b^2} + {c^2} + bc = {c^2} + {a^2} + ca$

==> ${b^2} - {a^2} = c(a - b)$

$\Rightarrow b = a$ or $a + b + c = 0$.

$y = \lambda $ .....$(i)$

Solving $(i)$ with the cuve $y = \sqrt x $ .....$(ii)$

We get $P({\lambda ^2},\,\,\lambda )$ the point of intersection at $P$

$\therefore $ Slope of $(ii)$ is, m= ${\left( {\frac{{dy}}{{dx}}} \right)_{{\rm{at }}P}} = \frac{1}{{2\lambda }}$

$\therefore $ $(i)$ and $(ii)$ intersect at $P$, at $45^\circ $

$\therefore $ ${\tan ^{ - 1}}\,\left( {\frac{{m - 0}}{{1 + m.0}}} \right) = \pm 45^\circ $.

==> $m = \left( {\frac{1}{{2\lambda }}} \right) = \pm \,1$ ==> $\lambda = \pm \,\frac{1}{2}$

$\therefore $ The equation of line is $y = \frac{1}{2}$ or $y = \frac{{ - 1}}{2}$ but $y = \frac{{ - 1}}{2}$ is not given, hence the required line is $y = \frac{1}{2}$.

then intersecting point $\left( {\frac{{\lambda {x_2} + {x_1}}}{{\lambda + 1}},\,\frac{{\lambda {y_2} + {y_1}}}{{\lambda + 1}}} \right)$ lies on $ax + by + c = 0$.

Thus any point on the line joining $(3,\, - 1)$ and $(8,\,9)$ dividing it in the ratio $\lambda :1$ is $\left( {\frac{{8\lambda + 3}}{{\lambda + 1}},\,\frac{{9\lambda - 1}}{{\lambda + 1}}} \right)$ and if it lies on $y - x + 2 = 0,$ then $\frac{{9\lambda - 1}}{{\lambda + 1}} - \frac{{8\lambda + 3}}{{\lambda + 1}} + 2 = 0$

or $9\lambda - 1 - (8\lambda + 3) + 2(\lambda + 1) = 0$

or $3\lambda - 2 = 0,\,\lambda = \frac{2}{3}$ i.e. ratio is $2:3$.

and equation of $BC$ is $2x + y - 22 = 0$.....$(ii)$

Thus angle between $(i)$ and $(ii)$ is given by

${\tan ^{ - 1}}\frac{{ - \frac{1}{4} + 2}}{{1 + \left( { - \frac{1}{4}} \right){\rm{ }}( - 2)}} = {\tan ^{ - 1}}\frac{7}{6}$.

Let the angle between first and third line is ${\theta _1}$and between second and third is ${\theta _2}$, then

$\tan {\theta _1} = \frac{{3 - m}}{{1 + 3m}}$ and $\tan {\theta _2} = \frac{{m - \frac{1}{2}}}{{1 + \frac{m}{2}}}$

But ${\theta _1} = {\theta _2} \Rightarrow \frac{{3 - m}}{{1 + 3m}} = \frac{{m - \frac{1}{2}}}{{1 + \frac{m}{2}}}$

==> $7{m^2} - 2m - 7 = 0$

==> $m = \frac{{1 \pm 5\sqrt 2 }}{7}$.

 $\sin \theta + \cos \theta - 1 > $as $3 + 2 - 1 > 0$

==> $\sqrt 2 \sin \left( {\theta + \frac{\pi }{4}} \right) > 1$

==> $\sin \left( {\theta + \frac{\pi }{4}} \right) > \frac{1}{{\sqrt 2 }} $

$\Rightarrow 0 < \theta < \frac{\pi }{4}$.

${L_{(3,7)}} = 3 \times 3 - 8 \times 7 - 7 < 0$

Hence $(-1, -1)$ and $(3, 7)$ lie on the same side of line.

Distance between lines $4x - 3y = 5$ and $6y - 8x = 1$ is,

$\beta = \frac{1}{{10}} + \frac{5}{5} = \frac{{11}}{{10}}$

Therefore $\frac{\alpha }{\beta } = \frac{{2\sqrt 2 }}{{11/10}} \Rightarrow 20\sqrt 2 \beta = 11\alpha $.

or $a{\rm{ }}(x + y + 3) + b{\rm{ }}( - 2x + 3y + 4) = 0$, which represents a family of  straight lines through point of intersection of $x + y + 3 = 0$ and $ - 2x + 3y + 4 = 0 $i.e $, (-1, -2).$

Trick :Point $(-1, -2)$ satisfies the given equation of straight line.

==> $ - a + 2b - c = 0 $.

$\Rightarrow 2b = a + c$

==> $a,b,c$ are in $A. P.$

$2x + y = 18$.....$(ii)$

and $4x - 3y = 26$....$(iii)$

are equations of three lines respectively

Solving of equation $(i)$ and $(ii)$, we get $x = 8$ and $y = 2$

Put the values of $x$ and $y$ in equation $(iii)$,

L.H.S $ = 4x - 2y$ $ = 4 \times 8 - 3 \times 2$ $ = 32 - 6$= 26 = R.H.S.

$\therefore $ Point $(8,2)$ lies on line $4x - 3y = 26$, so these three lines are concurrent. Hence, these three equations have one and only one solution.

$\left( {\frac{{{1^2} \times 2 - 1 \times 1 \times 4 + 1}}{{{1^2} + {1^2}}},\frac{{{1^2} \times 4 - 1 \times 1 \times 2 + 1}}{{{1^2} + {1^2}}}} \right) = \,\left( { - \frac{1}{2},\frac{3}{2}} \right)$.

Trick : Here, in options only point $\left( { - \frac{1}{2},\frac{3}{2}} \right)$ is satisfying the given equation of line.

Equation of perpendicular is $x - y + 1 = 0$. Now the coordinates of the foot of the perpendicular are the intersection point of the lines, hence point is $(5, 6)$.

Aliter : Apply the formula given in the theory part of this book, we get required foot as
$\left( {\frac{{{1^2} \times 2 - 1 \times 1 \times 3 - 1 \times ( - 11)}}{{{1^2} + {1^2}}},\frac{{{1^2} \times 3 - 1 \times 1 \times 2 - 1( - 11)}}{{{1^2} + {1^2}}}} \right)$
$ = \left( {\frac{{2 - 3 + 11}}{2},\frac{{3 - 2 + 11}}{2}} \right) = (5,\,6)$.

Now let the image of $A(3,8)$ be $A'({x_1},{y_1}),$then point $(1, 2)$ will be the mid point of $AA'$.

==> $\frac{{{x_1} + 3}}{2} = 1$

$\Rightarrow {x_1} = - 1$ and $\frac{{{y_1} + 8}}{2} = 2$

==> ${y_1} = - 4$.

Hence the image is $(-1, -4)$.

Then the mid-point $R{\rm{ }}\left( {\frac{{a + 4}}{2},\frac{{b - 13}}{2}} \right)$ lies on $5x + y + 6 = 0$.

 $5\left( {\frac{{a + 4}}{2}} \right) + \frac{{b - 13}}{2} + 6 = 0$

$\Rightarrow 5a + b + 19 = 0$ ......$(i)$

Also $PQ$is perpendicular to $5x + y + 6 = 0$.

Therefore $\frac{{b + 13}}{{a - 4}} \times \left( { - \frac{5}{1}} \right) = - 1$

$\Rightarrow a - 5b - 69 = 0$ .....$(ii)$

Solving $(i)$ and $(ii)$, we get $a = - 1,\,\,b = - 14$.

$PQ = \frac{{2 - 6}}{{4 + 2}} = \frac{{ - 2}}{3}$

 The slope of $L = \frac{3}{2}$

Hence the equation of line which passes through point $(1,\,4)$ is $y - 4 = \frac{3}{2}(x - 1)$

==> $3x - 2y + 5 = 0$ .

${x^2} + {y^2} - hx - ky = 0$, which is the required locus of the point.

Let m be the slope of a line inclined at an angle of ${45^o}$ to $AC$, then $\tan {45^o} = \pm \frac{{m - \frac{5}{2}}}{{1 + m.\frac{5}{2}}} \Rightarrow m = - \frac{7}{3},\frac{3}{7}$.

Thus, let the slope of $AB$ or $DC$ be $3/7$ and that of $AD$ or $BC$ be $ - \frac{7}{3}$ . Then equation of $AB$ is $3x - 7y + 19 = 0$.

Also the equation of $BC$ is $7x + 3y - 4 = 0$.

On solving these equations, we get, $B\,\,\,\left( { - \frac{1}{2},\frac{5}{2}} \right)$.

Now let the coordinates of the vertex $D$ be $(h, k)$. Since the middle points of $AC$ and $BD$ are same,

therefore $\frac{1}{2}\left( {h - \frac{1}{2}} \right)\, = \frac{1}{2}(3 + 1) \Rightarrow h = \frac{9}{2}$, $\frac{1}{2}\left( {k + \frac{5}{2}} \right) = \frac{1}{2}(4 - 1)$

==> $k = \frac{1}{2}$. Hence, $D = \left( {\frac{9}{2},\,\frac{1}{2}} \right)$.

Area of $\Delta OAB = \frac{1}{2}\left( {\frac{{2p}}{{\sqrt 3 }}} \right){\rm{ }}2p = \frac{{2{p^2}}}{{\sqrt 3 }}$

By hypothesis $\frac{{2{p^2}}}{{\sqrt 3 }} = \frac{{50}}{{\sqrt 3 }} \Rightarrow p = \pm 5$.
Hence the lines are $\sqrt 3 x + y \pm 10 = 0$.

==> $\lambda = - \frac{{{p^2} + {q^2} - 1}}{{2pq - 1}}$ and the equation of median through $A$ is $(px + qy - 1) - \frac{{{p^2} + {q^2} - 1}}{{2pq - 1}}(qx + py - 1) = 0$

==> $(2pq - 1)(px + qy - 1) = ({p^2} + {q^2} - 1)(qx + py - 1)$.

Therefore the area is $\frac{1}{2}|a \times b| = 6 \Rightarrow |ab| = 12$ …..$(i)$

and hypotenuse is $5$, therefore ${a^2} + {b^2} = 25$ …..$(ii)$

On solving $(i)$ and $(ii)$, we get

$a = \pm 4$or $ \pm 3$and $b = \pm 3$or $ \pm 4$

Hence equation of line is $ \pm \frac{x}{4} \pm \frac{y}{3} = 1$or $ \pm \frac{x}{3} \pm \frac{y}{4} = 1$.

Trick: Check with options. Obviously, the line $\frac{x}{4} + \frac{y}{3} = \pm 1$ satisfies both the conditions.

(Here m is the gradient of line $AC$)

 Equation of line AC is,$y - 1 = \frac{1}{2}(x - 2)$ ==> $y = \frac{x}{2}$.

$\tan 60^\circ = \frac{{AD}}{{BD}}$

==> $\sqrt 3 = \frac{{1\,/\sqrt 5 }}{{BD}}$

==> $BD = \frac{1}{{\sqrt {15} }}$

$BC = 2BD = 2\,/\sqrt {15} $.

==> $y - 4 = \frac{{1 \pm \tan 45^\circ }}{{1 \mp \tan 45^\circ }}(x - {x_1})$

==> $y - 4 = \frac{{1 \pm 1}}{{1 \mp 1}}(x - 3)$ ==> $y = 4$ or $x = 3$

Hence, the lines which make the triangle are $x - y = 2,$ $x = 3$ and $y = 4$. The intersection points of these lines are $(6,\,4),\,(3,\,1)$ and $(3,\,4)$

 $\Delta = \frac{1}{2}[6( - 3) + 3(0) + 3(3)]$ $ = \frac{9}{2}$.

Replace $(h,k)$ by $(x,\,y)$, we get

$13{x^2} + 13{y^2} - 83x + 64y + 182 = 0$, which is the required equation of the locus of the point.

${y_1} = \frac{{5 - 11 + y}}{3} \Rightarrow y = 3{y_1} + 6$

$\therefore $ $9(3{x_1} - 6) + 7(3{y_1} + 6) + 4 = 0$

Hence locus is $27x + 21y - 8 = 0$, which is parallel to $9x+7y+4 = 0.$

 

$\left( {\frac{{b/2}}{{a/2}}} \right){\rm{ }}\left( {\frac{b}{{ - a/2}}} \right) = - 1 $

$\Rightarrow {a^2} = 2{b^2}$

$\Rightarrow a = \pm \sqrt 2 b$.

==> $\frac{1}{2}.a.OB = T \Rightarrow OB = \frac{{2T}}{a}$

Hence the equation of line is $\frac{x}{{ - a}} + \frac{y}{{2T/a}} = 1$

$ \Rightarrow 2Tx - {a^2}y + 2aT = 0$.

$\tan {60^o}$ $ = \frac{{AD}}{{BD}} \Rightarrow \sqrt 3 = \frac{{\sqrt 5 }}{{BD}}$==> $BD = \sqrt {\frac{5}{3}} $

$\therefore $ $BC = 2BD = 2\sqrt {\frac{5}{3}} = \sqrt {\frac{{20}}{3}} $.

Hence the angle between diagonals $AC$ and $BD$ is $90^\circ $.

Equation of $y$-axis is $x = 0$

Hence the vertices of the triangle are $A(0,\,2),B\left( {0, - \frac{1}{2}} \right)$ and $C\left( {\frac{5}{4},\frac{3}{4}} \right)$. Therefore, the area of the triangle is $\frac{1}{2}\left| {\,\begin{array}{*{20}{c}}0&2&1\\0&{ - \frac{1}{2}}&1\\{\frac{5}{4}}&{\frac{3}{4}}&1\end{array}\,} \right| = \frac{{25}}{{16}}$.

$3 x-2 y=-1$ and $x+y=3$

solving above two equations we get $y=2$ and $x=1$

$B=(1,2)$

To find $C$

$3 x-2 y=-1$ and $x-3 y=-9$

solving above two equations we get $y=27 / 6$ and $x=15 / 7$

$C=\left(\frac{15}{7}, \frac{27}{6}\right)$

$s=\frac{\text {perimeteroftriangle}}{2}$

$a=A B=\sqrt{(1-0)^{2}+(2-3)^{2}}$

$=\sqrt{1+1}$

$=\sqrt{2}$

$b=B C=\sqrt{\left(\frac{15}{7}-1\right)^{2}+\left(\frac{26}{7}-2\right)^{2}}$

$=\sqrt{\left(\frac{8}{7}\right)^{2}+\left(\frac{12}{7}\right)^{2}}$

$\frac{4 \sqrt{13}}{7}$

$c=C A=\sqrt{\left(\frac{15}{7}-0\right)^{2}+\left(\frac{26}{7}-3\right)^{2}}$

$=\sqrt{\left(\frac{15}{7}\right)^{2}+\left(\frac{5}{7}\right)^{2}}$

$=\frac{5 \sqrt{10}}{7}$

$s=\frac{a+b+c}{2}$

$s=\frac{\sqrt{2}+\frac{4 \sqrt{13}}{7}+\frac{5 \sqrt{10}}{7}}{2}$

$s=\frac{14 \sqrt{2}+4 \sqrt{13}+5 \sqrt{10}}{14}$

$s-a=\frac{-7 \sqrt{2}+4 \sqrt{13}+5 \sqrt{10}}{14}$

$s-b=\frac{14 \sqrt{2}-4 \sqrt{13}+5 \sqrt{10}}{14}$

$s-c=\frac{14 \sqrt{2}+4 \sqrt{13}-5 \sqrt{10}}{14}$

$A=\sqrt{s(s-a)(s-b)(s-c)}$

$A=\sqrt{\left(\frac{-7 \sqrt{2}+4 \sqrt{13}+5 \sqrt{10}}{14}\right)\left(\frac{14 \sqrt{2}-4 \sqrt{13}+5 \sqrt{10}}{14}\right)\left(\frac{14 \sqrt{2}+4 \sqrt{13}-5 \sqrt{10}}{14}\right)}$

Multiplying and solving

$A=\frac{16}{7}$ sq unit

Thus, the equation of $A B$ is $(y-1)=-3(x-3)$ $\Rightarrow A=\left(\frac{10}{3}, 0\right), B=(0,10)$

$\Rightarrow \Delta O A B=\frac{1}{2}(O A)(O B)=\frac{1}{2} \times \frac{10}{3} \times 10=\frac{100}{6}=\frac{50}{3}$

 $\frac{x}{{2a}} + \frac{y}{a} = 1 \Rightarrow x + 2y = 2a$.....$(i)$

According to question,

Line $(i)$ also passes through midpoint of $(3, - 4)$ and $(5,2)$ i.e., $(4, - 1)$.

 $4 + 2( - 1) = 2a \Rightarrow a = 1$

Hence the equation of required line is, $x + 2y = 2$.

$\frac{{3x + 4y - 7}}{{\sqrt {{3^2} + {4^2}} }} = \pm \frac{{12x + 5y + 17}}{{\sqrt {{{(12)}^2} + {{(5)}^2}} }}$

$\frac{{3x + 4y - 7}}{5} = \pm \frac{{12x + 5y + 17}}{{13}}$.

==>$7x - 7y + 21 = 0$

==> $x - y + 3 = 0$.

Slope of perpendicular = $\frac{4}{3}$

==> $x = \pm 1.\cos \theta = \pm \frac{3}{5}$and $y = \pm 1.\sin \theta = \pm \frac{4}{5}$

Hence $\left( {\frac{3}{5},\frac{4}{5}} \right)$ lies on straight line.

and $\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}} = c$ (Let)

==> ${a_2} = \frac{{{a_1}}}{c},{b_2} = \frac{{{b_1}}}{c},{c_2} = \frac{{{c_1}}}{c}$
Given that $u + kv = 0$

==> ${a_1}x + {b_1}y + {c_1} + k({a_2}x + {b_2}y + {c_2}) = 0$

==> ${a_1}x + {b_1}y + {c_1} + k\frac{{{a_1}}}{c}x + k\frac{{{b_1}}}{c}y + k\frac{{{c_1}}}{c} = 0$

==> ${a_1}x\left( {1 + \frac{k}{c}} \right) + {b_1}y\left( {1 + \frac{k}{c}} \right) + {c_1}\left( {1 + \frac{k}{c}} \right) = 0$

==> ${a_1}x + {b_1}y + {c_1} = 0 = u$.

All these lines are parallel. Hence they do not intersect in finite plane.

Now point of intersection of $x - 2y = 0$ and $2x + y + 6 = 0$is $\left( {\frac{{ - 12}}{5},\frac{{ - 6}}{5}} \right)$ and point of intersection of $x - 2y = 0$ and $4x + 2y - 9 = 0$ is $\left( {\frac{9}{5},\frac{9}{{10}}} \right)$.

Now say origin divide the line $x - 2y = 0$ in the ratio $\lambda :1$

 $x = \frac{{\frac{9}{5}\lambda - \frac{{12}}{5}}}{{\lambda + 1}} = 0 \Rightarrow \frac{9}{5}\lambda = \frac{{12}}{5}$, $\therefore \lambda = \frac{4}{3}$

Thus origin divides the line $x = 2y$, in the ratio $4 : 3$.
 

==> $\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} = \frac{2}{{8{p^2}}} \Rightarrow \frac{1}{{{a^2}}},\frac{1}{{8{p^2}}},\frac{1}{{{b^2}}}$are in A. P.

==> ${a^2},8{p^2},{p^2}$are in H.P .

Where $‘r$’ represents the distance of any point $Q$ on this line from the given point $P \,(3, 4)$.

The coordinates $(x, y)$ of any point $Q$ on line $(i)$ are $(3 + r\cos {30^o},4 + r\sin {30^o})\,\,{\rm{ }}i.e.,\,{\rm{ }}\left( {3 + \frac{{r\sqrt 3 }}{2},4 + \frac{r}{2}} \right)$

If the point lies on the line $12x + 5y + 10 = 0$, then

$12\left( {3 + \frac{{r\sqrt 3 }}{2}} \right) + 5\left( {4 + \frac{r}{2}} \right) + 10 = 0$ 

$\Rightarrow r = \frac{{132}}{{12\sqrt 3 + 5}}$.

$y = mx +3$,$m \in[0, \sqrt{3}]$

$y = nx +3$,$n \in[-\sqrt{3}, 0]$

min area $=$ are $(\Delta PAB )=\frac{1}{2} \times 2 \sqrt{3} \times 3=3 \sqrt{3}$

Given, $P Q R$ is an acute angle triangle.

$\angle Q R P < \angle Q P R$

$P Q_3 =\frac{1}{2} P R$

$P Q_2: Q_2 R=r: p$

$P Q_2=\left(\frac{r}{r+p}\right) P R$

$r < p$

$P Q_2 < \frac{1}{2} P R$

Comparison between altitude and angle bisector

$\angle Q P Q_2+\angle P Q \cdot Q+\angle P Q Q_2=\angle R Q Q_2 +\angle Q Q_2 R+\angle Q R Q_2$

$\therefore \angle P Q Q_2=\angle R Q Q_2$

[ since, $Q Q_2$ is angle bisector of $\angle Q$ ]

$\angle Q P Q_2+\angle P Q_2 Q=\angle Q Q_2 R+\angle Q R Q_2$

$\therefore P Q$

Hence, $\angle Q Q_2 P<\angle Q Q_2 R$

But $\angle Q Q_2 P+\angle Q Q_2 R=180^{\circ}$

Hence, $\angle Q Q_2 P<90^{\circ}$ and $\angle Q Q_2 R>90^{\circ}$

$\because$ Foot from $Q$ to side $P R$ lie inside $\triangle P Q Q_2$

$P Q_1 < P Q_2 < P Q_3$

$\angle O B A=60^{\circ}$

$\angle O A B=30^{\circ}$

$O A D$ is an equilateral triangle.

$\therefore \quad \angle A O D=\angle O D A=\angle O A D=60^{\circ}$

$O A=A D=O D$

Let $B(0, e )$

$\therefore \quad A(\sqrt{3} a, 0)$

D $\left(\frac{\sqrt{3}}{2} a, \frac{3}{2} a\right)$

Slope of $B D=\frac{\frac{3}{2}-a}{\frac{\sqrt{3}}{2} a}=\frac{1}{\sqrt{3}}$

We have, length and size of two candles are same. Let $L$ be the length of candles.

Given, first candle burns in $5 h$ and second candle burns in $3 h$.

In one hours length of candles are $\frac{L}{5}$ and $\frac{L}{3}$, respectively.

Let after time $t h$ the length of candles are $L_1$ and $L_2$

$\therefore \quad L_1=L-\frac{L}{5} t$ and $L_2=L-\frac{L}{3} t$

According to the problem,

$L_1=3 L_2$

$\therefore \quad L-\frac{L}{5} t=3\left(L-\frac{L}{3} t\right)$

$\Rightarrow 1-\frac{1}{5} t=3-t \Rightarrow t\left(1-\frac{1}{5}\right)=3-1$

$\Rightarrow \quad \frac{4 t}{5}=2 \Rightarrow t=\frac{5}{2} h$

$\Rightarrow \quad t=\frac{5}{2} \times 60=150\,min$

$A B C$ is a triangle points $X$ and $Y$ on $A B$ and $A C$ respectively.

$X Y$ is parallel to $B C$.

$I$. Area of $B C X:$ Area of $B C Y$

It is true because same base between same parallels.

$II$. Area of $\triangle A C X=\frac{1}{2}(A X)(A C) \sin A$

Area of $\triangle A B Y=\frac{1}{2}(A Y)(A B) \sin A$

$\therefore$ (Area of $\triangle A C X)$ (Area of $\triangle A B Y)$

$=\frac{1}{2}(A X)(A C) \sin A \times \frac{1}{2}(A Y)(A B) \sin A$

$=\frac{1}{2}(A X)(A Y) \sin A \times \frac{1}{2}(A B)(A C) \sin A$

$=(\text { Area of } \triangle A X Y) \text { (Area of } \triangle A B C)$

Hence,$I$ and $II$ both are true.

$A B C D$ is square

$A B=B C=C D=A D=1$

$P Q$ is perpendicular to $R S$

$\because$ Slope of $P Q \times$ Slope of $R S=-1$

$\therefore \quad q-p \times 1-0=-1$

$ \Rightarrow  q-p =r-s $$ \Rightarrow (P Q)^2 =(1-0)^2+(q-p)^2 $

$\Rightarrow \quad(P Q)^2=(1-0)^2+(q-p)^2$

$\Rightarrow \quad\left(\frac{3 \sqrt{3}}{4}\right)^2=1+(q-p)^2$

$\Rightarrow \quad(q-p)^2=\frac{27}{16}-1=\frac{11}{16}$

$\Rightarrow \quad(r-s)^2=\frac{11}{16} \quad[\because q-p=r-s]$

$\Rightarrow \quad R S=\sqrt{(1-0)^2+(r-s)^2}$

$\Rightarrow \quad R S=\sqrt{1+\frac{11}{16}}$

$\therefore \quad R S=\sqrt{\frac{27}{16}}=\frac{3 \sqrt{3}}{4}$

We have, $P \cdot(r, \theta)=(x, y)$

Equation of line $O P$ is

$y=x \tan \theta$

$O P$ cut the line $r \sin \theta=b$

i.e. $y=b$

Given $P Q=d$

$\therefore$ point $P$ is $y=b \pm d \sin \theta$

$\Rightarrow r \sin \theta=b \pm d \sin \theta \Rightarrow(r \pm d) \sin \theta=b$

Hence, locus of $P(r, \theta)$ is $(r \pm d) \sin \theta=b$

Given,

$A B C D$ is a rectangle $A(1,2)$ and $B(3,6)$ equation of one of the diameter of circle circumscribing the rectangle is

$2 x-y+4=0 \text {. }$

Slope of diameter $=2$

and slope of line $A B=\frac{6-2}{3-1}=\frac{4}{2}=2$

$\therefore$ Side $A B$ is parallel to diameter of circle equation of line $A B$

$y-2=2(x-1) \Rightarrow 2 x-y=0$

Distance between diameter of circle and line $A B$

$d=\left|\frac{4}{\sqrt{2^2+1}}\right|=\frac{4}{\sqrt{5}}$

$\therefore \quad B C=2 d=\frac{8}{\sqrt{5}}$

Now, $\quad A B=\sqrt{(3-1)^2+(6-2)^2}$

$=\sqrt{4+16}=2 \sqrt{\overline{5}}$

Area of rectangle $=A B \times B C$

$=2 \sqrt{5} \times \frac{8}{\sqrt{5}}=16$

The line $2 x+3 y=18$

Cut the $Y$-axis at $B$

$\therefore \quad B=(0,6), C=(a, b)$

$\because P D$ is perpendicular bisector of line

segment $B C$

Equation of line $P D$ is perpendicular to $B C$.

$\therefore \quad 3 x-2 y=\lambda$ Since, $P D$ is passing through $P(10,10)$.

Since, $P D$ is passing through $F$ $\therefore \quad 30-20=\lambda \Rightarrow \lambda=10$

Equation of line $P D$ is $3 x-2 y=10$

Solving the equation $2 x+3 y=18$ and

$3 x-2 y=10$

we get $D\left(\frac{66}{13}, \frac{34}{13}\right)$

$D$ is mid-point of $B C$

$\therefore \frac{a+0}{2} =\frac{66}{13}, \frac{b+6}{2}=\frac{34}{13}$

$\Rightarrow \quad =\frac{132}{13}, b=\frac{-10}{13}$

$\therefore \quad 3 a+2 b =\frac{8 \times 132}{13}-\frac{20}{13}$

$= \frac{1056-20}{13}=\frac{1036}{13}$

Co-ordinates of $\mathrm{B}=\left(\frac{10}{3}, \frac{4}{3}\right)$

Slope of $O A=m_1=\frac{8}{5}$

Slope of $O B=m_2=\frac{2}{5}$

$\tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right| $

$ \tan \theta=\frac{\frac{6}{5}}{1+\frac{16}{25}}=\frac{30}{41} $
$\tan \theta=\frac{30}{41}$

$ \cos \theta, 3+\mathrm{r} \sin \theta \text { as } \tan \theta=\sqrt{3} $

$ \mathrm{P}\left(2+\frac{\mathrm{r}}{2}, 3+\frac{\sqrt{3} \mathrm{r}}{2}\right)$

$P$ must satisfy $2 \mathrm{x}-3 \mathrm{y}+28=0$

So, $2\left(2+\frac{\mathrm{r}}{2}\right)-3\left(3+\frac{\sqrt{3} \mathrm{r}}{2}\right)+28=0$

We find $\mathrm{r}=4+6 \sqrt{3}$

Finding Eqn. of $A B: 4 x-3 y+4=0$

Calculate distance $PM$

$\Rightarrow\left|\frac{4(5)-3(-2)+4}{5}\right|=6$

$\frac{\mathrm{x}-\mathrm{x}_1}{\mathrm{a}}=\frac{\mathrm{y}-\mathrm{y}_1}{\mathrm{~b}}=-2\left(\frac{\mathrm{ax_{1 } + \mathrm { by } _ { 1 } + \mathrm { c }}}{\mathrm{a}^2+\mathrm{b}^2}\right)$

Line : $x+2 y-2=0$

$ \frac{x+4}{1}=\frac{y-5}{2}=-2\left(\frac{-4+10-2}{1^2+2^2}\right) $

$ =\frac{-8}{5} $

$ \mathrm{x}=-4-\frac{8}{5}=-\frac{28}{5} $

$ y=-\frac{16}{5}+5=\frac{9}{5} $

Point lies on circle $(x+4)^2+(y-3)^2=r^2$

$ \frac{64}{25}+\left(\frac{9}{5}-3\right)^2=r^2 $

$ \frac{100}{25}=r^2, r=2$

$ \frac{4-\alpha}{6-\alpha}=\frac{10}{8} $

$ \alpha=\beta $

$ \alpha=14 \text { and } \beta=14$

$ \frac{3}{a}+\frac{5}{b}=1 \Rightarrow b=\frac{5 a}{a-3}, a>3$

$Image$

$\mathrm{A}=\frac{1}{2} \mathrm{ab}=\frac{1}{2} \mathrm{a} \frac{5 \mathrm{a}}{(\mathrm{a}-3)}=\frac{5}{2} \cdot \frac{\mathrm{a}^2}{\mathrm{a}-3}$

$ =\frac{5}{2}\left(\frac{a^2-9+9}{a-3}\right) $

$ =\frac{5}{2}\left(a+3+\frac{9}{a-3}\right) $

$ =\frac{5}{2}\left(a-3+\frac{9}{a-3}+6\right) \geq 30$

$ y+9=m(x-4) $

$ \therefore \quad A=\left(\frac{9+4 m}{m}, 0\right) $

$ \quad B=(0,-9-4 m) $

$ \therefore \quad O A+O B=\frac{9+4 m}{m}+9+4 m$

$ \because \mathrm{m}>0 $

$ =13+\frac{9}{\mathrm{~m}}+4 \mathrm{~m} $

$ \because \frac{4 \mathrm{~m}+\frac{9}{\mathrm{~m}}}{2} \geq \sqrt{36} \Rightarrow 4 \mathrm{~m}+\frac{9}{\mathrm{~m}} \geq 12 $

$ \therefore \mathrm{OA}+\mathrm{OB} \geq 25$

$ \mathrm{m}_{\mathrm{OA}}=\frac{2}{5} $

$ \mathrm{~m}_{\mathrm{OB}}=\frac{\mathrm{a}}{2}$

$\tan \frac{\pi}{4}=\left|\frac{2}{5}-\frac{a}{2}\right|$

$1=\left|\frac{4-5 a}{10+2 a}\right|$

$4-5 \mathrm{a}= \pm(10+2 \mathrm{a}) $

$4-5 \mathrm{a}=10+2 \mathrm{a} $    $ 4-5 \mathrm{a}=-10-2 \mathrm{a} $

$\Rightarrow 7 \mathrm{a}+6=0 $    $ 3 \mathrm{a}=14 $

$\Rightarrow \mathrm{a}=-\frac{6}{7} $    $ \mathrm{a}=+\frac{14}{3}$

$-\frac{6}{7} \times \frac{14}{3}=-4$

Gives

$\mathrm{X}+2 \mathrm{y}+7=0 \& 2 \mathrm{x}-\mathrm{y}+8=0$ are equal is

$\frac{x+2 y+7}{\sqrt{5}}= \pm \frac{2 x-y+8}{\sqrt{5}}$

$(x+2 y+7)^2-(2 x-y+8)^2=0$

Combined equation of lines

$(x-3 y+1)(3 x+y+15)=0$

$3 x^2-3 y^2-8 x y+18 x-44 y+15=0$

$x^2-y^2-\frac{8}{3}xy+6x-\frac{44}{3} y+5=0$

$x^2-y^2+2 h x y+2 g x 2+2 f y+c=0$

$h=\frac{4}{3}, g=3, f=-\frac{22}{3}, c=5$

$g+c+h-f=3+5-\frac{4}{3}+\frac{22}{3}=8+6=14$

$\begin{array}{ll}\frac{\alpha+\gamma}{2}=\frac{1+1}{2} & \& \frac{\beta+\delta}{2}=\frac{2+0}{2} \\ \alpha+\gamma=2 & \beta+\delta=2 \\ 2(\alpha+\beta+\gamma+\delta)=2(2+2)=8\end{array}$

$\Rightarrow \mathrm{a}=0, \mathrm{~b}=0 \quad \Rightarrow \mathrm{P}(3,5)$

Distance from $P$ measured along $x-2 y-1=0$

$\Rightarrow x=3+r \cos \theta, \quad y=5+r \sin \theta$

 Where $ \tan \theta=\frac{1}{2} $

$ r(2 \cos \theta+3 \sin \theta)=-17 $

$ \Rightarrow r=\left|\frac{-17 \sqrt{5}}{7}\right|=\frac{17 \sqrt{5}}{7}$

$ \frac{\alpha-2}{2}=\frac{\gamma+1}{2} \text { and } \frac{\beta-1}{2}=\frac{\delta}{2} $

$\Rightarrow \alpha-\gamma=3 \ldots . .(1), \beta-\delta=1 \ldots ....(2)$

Also, $(\gamma, \delta)$ lies on $3 x-2 y=6$

$3 \gamma-2 \delta=6$

and $(\alpha, \beta)$ lies on $2 x-y=5$

$\Rightarrow 2 \alpha-\beta=5 \text {......(4) }$

Solving $(1), (2), (3), (4)$

$\alpha=-3, \beta=-11, \gamma=-6, \delta=-12$

$|\alpha+\beta+\gamma+\delta|=32$

$2 c=8 \Rightarrow c=4$

$ \mathrm{AB}=|(\mathrm{b}+1)|=4 $

$ \mathrm{~b}=3, \mathrm{~m}_{\mathrm{AB}}=0 $

$ \mathrm{~m}_{\mathrm{BC}}=\frac{-1}{\sqrt{3}} $

$ B C:-y=\frac{-1}{\sqrt{3}}(x-3) $

$ \sqrt{3} \mathrm{y}+\mathrm{x}=3 $

$ \text { Point of intersection : } y=x+3, \sqrt{3} y+x=3 $

$ (\sqrt{3}+1) \mathrm{y}=6 $

$ \mathrm{y}=\frac{6}{\sqrt{3}+1} $

$ x=\frac{6}{\sqrt{3}+1}-3 $

$ =\frac{6-3 \sqrt{3}-3}{\sqrt{3}+1} $

$ =3 \frac{(1-\sqrt{3})}{(1+\sqrt{3})}=\frac{-6}{(1+\sqrt{3})^2} $

$ \frac{\beta^4}{\alpha^2}=36 $

equation of $\mathrm{AC} \rightarrow \mathrm{x}+\mathrm{y}=2$

equation of line parallel to $\mathrm{AC} \mathrm{x}+\mathrm{y}=\mathrm{d}$

$ \left|\frac{\mathrm{d}-2}{\sqrt{2}}\right|=1 $

$ \mathrm{~d}=2-\sqrt{2}$

$\mathrm{eq}^{\mathrm{n}}$ of new required line

$x+y=2-\sqrt{2}$

$ \mathrm{y}-2=\frac{1 \pm \tan \frac{\pi}{6}}{1 \mp \tan \frac{\pi}{6}}(\mathrm{x}-1) $

$ \mathrm{y}-2=\frac{\sqrt{3} \pm 1}{\sqrt{3} \mp 1}(\mathrm{x}-1)$

$\oplus$                                                           

$y-2=(2+\sqrt{3})(x-1) $

solve with $ y=x+4 $

$x+2=(2+\sqrt{3}) x-2-\sqrt{3} $

$x=\frac{4+\sqrt{3}}{1+\sqrt{3}}$

$\Theta$

$y-2=(2-\sqrt{3})(x-1)$

solve with $\mathrm{y}=\mathrm{x}+4$

$x+2=(2-\sqrt{3}) x-2+\sqrt{3} $

$x=\frac{4-\sqrt{3}}{1-\sqrt{3}}$

$ \alpha^2+\gamma^2=\left(\frac{4+\sqrt{3}}{1+\sqrt{3}}\right)^2+\left(\frac{4-\sqrt{3}}{1-\sqrt{3}}\right)^2 $

$ \alpha^2+\gamma^2=14$

$ \frac{(x-2)^2+(y-1)^2}{(x-1)^2+(y-3)^2}=\frac{25}{16} $

$ 9 x^2+9 y^2+14 x-118 y+170=0 $

$ a^2+2 b+3 c+4 d+e $

$ =81+18+0+56-118 $

$ =155-118 $

$ =37$

$ 2 x_2+x_1=6,3 x_2-4 x_1=-13 $

$ x_2=1, x_1=4$

So, $\mathrm{C}(1,-1), \mathrm{B}(4,-2)$

$\mathrm{m}=\frac{-1}{3}$

Equation of $B C: y+1=\frac{-1}{3}(x-1)$

$ 3 y+3=-x+1$

$ x+3 y+2=0$

$ \text { Area }=(4 \cos \theta+2 \sin \theta)(2 \cos \theta+4 \sin \theta) $

$ =8 \cos ^2 \theta+16 \sin \theta \cos \theta+4 \sin \theta \cos \theta+8 \sin ^2 \theta $

$ =8+20 \sin \theta \cos \theta $

$ =8+10 \sin 2 \theta $

$ \text { Max Area }=8+10=18(\sin 2 \theta=1) \theta=45^{\circ} $

$ (a+b)^2=(4 \cos \theta+2 \sin \theta+2 \cos \theta+4 \sin \theta)^2 $

$ =(6 \cos \theta+6 \sin \theta)^2 $

$ =36(\sin \theta+\cos \theta)^2 $

$ =36(\sqrt{2})^2 $

$ =72$

$ 3(\mathrm{r} \cos \theta)+4(\mathrm{r} \sin \theta)=12 $

$ \mathrm{r}(3 \cos \theta+4 \sin \theta)=12 \ldots(1) $

$ 3(-\mathrm{r} \sin \theta)+4(\mathrm{r} \cos \theta)=12 $

$ \mathrm{r}(-3 \sin \theta+4 \cos \theta)=12 \ldots(2) $

$ \left(\frac{12}{\mathrm{r}}\right)^2+\left(\frac{12}{\mathrm{r}}\right)^2=(3 \cos \theta+4 \sin \theta)^2+(-3 \sin \theta+4 \cos \theta)^2 $

$ 2\left(\frac{12}{\mathrm{r}}\right)^2=9+16 $

$ \frac{2 \times 144}{\mathrm{r}^2}=25 \Rightarrow 288=25 \mathrm{r}^2 $

$ \Rightarrow \frac{288}{25}=\mathrm{r}^2 $

$ \Rightarrow \sqrt{2}\left(\frac{12}{5}\right)=\mathrm{r} $

$ \ell=\mathrm{OP}^2+\mathrm{PQ}^2+\mathrm{QO} \mathrm{O}^2 $

$ \ell=\mathrm{r}^2+\mathrm{r}^2 \quad+\mathrm{r}^2(\cos \theta+\sin \theta)^2+\mathrm{r}^2(\sin \theta+\cos \theta)^2 $

$ =2 \mathrm{r}^2+\mathrm{r}^2(1+\sin 2 \theta+1-2 \sin 2 \theta) $

$ =2 \mathrm{r}^2+2 \mathrm{r}^2 $

$ =4 \mathrm{r}^2 $

$ =4\left(\frac{288}{25}\right)=\frac{1152}{25}=46.08 $

$ {[\ell]=46}$

$Image$

$ \therefore A=r \sqrt{2}=2 \sqrt{6} $

Area $=m=A^2=24 $

Perimeter $=n=4 A=8 \sqrt{6} $

$ \therefore m+n^2=24+384 $

$ =408$

$ \frac{x-7}{2}=\frac{y-2}{1}=-4 $

$ x=-1 \quad y=-2 \quad B^{\prime}(-1,-2)$

incident ray $\mathrm{AB}^{\prime}$

$ \mathrm{M}_{\mathrm{AB}^{\prime}}=3 $

$ \mathrm{y}+2=3(\mathrm{x}+1) $

$ 3 \mathrm{x}-\mathrm{y}+1=0 $

$ \mathrm{a}=3 \mathrm{~b}=-1 $

$ \mathrm{a}^2+\mathrm{b}^2+3 \mathrm{ab}=9+1-9=1$

$\mathrm{y}-3=\frac{5}{3}(\mathrm{x}-4)$

Above line will meet $\mathrm{x}$-axis at $\mathrm{Q}$ where

$ \mathrm{y}=0 \Rightarrow \mathrm{x}=\frac{11}{5} $

$ \therefore \mathrm{Q}\left(\frac{11}{5}, 0\right)$

$\because \mathrm{PQRS}$ is parallelogram so their diagonals will bisects each other

$ \Rightarrow \frac{4+1}{2}=\frac{\frac{11}{5}+h}{2} \& \frac{2+3}{2}=\frac{\mathrm{k}+0}{2} $

$ \Rightarrow \mathrm{h}=\frac{14}{5} \& \mathrm{k}=5 $

$ \therefore \mathrm{hk}^2=\frac{14}{5} \times 5^2=70$

Equation of $PC$ is $y-1=\frac{5}{4}(x-1)$...............($1$)

$\mathrm{M}_{\mathrm{AP}}=\frac{2}{-2}=-1 \Rightarrow \mathrm{M}_{\mathrm{BC}}=+1$

Equation of $\mathrm{BC}$ is $\mathrm{y}-3=(\mathrm{x}+2)$............($2$)

On solving ($1$) and ($2$)

$ x+4=\frac{5}{4}(x-1) \Rightarrow 4 x+16=5 x-5 \Rightarrow \alpha=21 $

$ \Rightarrow \beta=y=x+5=26 $

$ \alpha+\beta=47$

Equation of $\perp$ bisector of $AP$

$y-0=(x-2) $................($3$)

Equation of $\perp$ bisector of $A B$

$\mathrm{y}-1=\frac{5}{4}\left(\mathrm{x}-\frac{1}{2}\right)$.................($4$)

On solving ($3$) & ($4$)

$ (x-3) 4=5 x-\frac{5}{2} $

$ x=\frac{-19}{2}=h $

$ y=\frac{-23}{2}=k $

$ \Rightarrow 2(h+k)=-42$

$m _{ L _2}=\frac{6.3}{5}=\frac{18}{5}$

$y =\left( m _1+ m _2\right) x$

$y =\frac{9}{2} x$

Point of intersection with $L$ is $\left(\frac{10}{7}, \frac{45}{7}\right)$

Line $y=\frac{x}{\sqrt{3}}$ intersect $y+x=1$ at $\left(\frac{\sqrt{3}}{\sqrt{3}+1}, \frac{1}{\sqrt{3}+1}\right)$

Equation of reflected ray is

$y-\frac{1}{\sqrt{3}+1}=\sqrt{3}\left(x-\frac{\sqrt{3}}{\sqrt{3}+1}\right)$

Put $y=0 \Rightarrow x=\frac{2}{3+\sqrt{3}}$

$(1,1)(1,2)-(1,14) \quad \Rightarrow 14$ pts.

If $x =2, y =\frac{27}{2}=13.5$

$(2,2)(2,4) \ldots(2,12) \quad \Rightarrow 6$ pts.

If $x=3, y=\frac{51}{4}=12.75$

$(3,3)(3,6)-(3,12) \quad \Rightarrow 4$ pts.

If $x=4, y=12$

$(4,4)(4,8) \quad \Rightarrow 2$ pts.

If $x=5 . y=\frac{45}{4}=11.25$

$(5,5),(5,10) \quad \Rightarrow 2$ pts.

If $x=6, y=\frac{21}{2}=10.5$

$(6,6) \quad \Rightarrow 1 pt$.

If $x=7, y=\frac{39}{4}=9.75$

$(7,7) \quad \Rightarrow 1 pt$.

If $x=8, y=9$

$(8,8) \quad \Rightarrow 1 pt$.

If $x =9 y =\frac{33}{4}=8.25 \Rightarrow$ no pt.

Total $=31$ pts.

$Y-2 x=2$

$(-2,-2)$

Height from line $x + y =0$

$h=\frac{4}{\sqrt{2}}$

Area of $\Delta=\frac{\sqrt{3}}{4} \frac{ h ^2}{\sin ^2 60}=\frac{8}{\sqrt{3}}$

Or $x \cos \frac{\pi}{3}+y \sin \frac{\pi}{3}=p$

$\frac{x}{2}+\frac{y \sqrt{3}}{2}=p$

$\frac{x}{3 p}+\frac{y}{2 p}=1$

Comparing both $: a =2 p , b =\frac{2 p }{\sqrt{3}}$

Now area of $\triangle OAB =\frac{1}{2} \cdot ab =\frac{98}{3} \cdot \sqrt{3}$

$\frac{1}{2} \cdot 2 p \cdot \frac{2 p }{\sqrt{3}}=\frac{98}{3} \cdot \sqrt{3}$

$p^2=49$

$a^2-b^2=4 p^2-\frac{4 p^2}{3}=\frac{2}{3} 4 p^2$

$=\frac{8}{3} \cdot 49=\frac{392}{3}$

$y-\frac{7}{2}=\frac{\frac{7}{2}}{-\frac{1}{2}} \quad\left(x+\frac{1}{2}\right)$

$7 x+y=0$

Distance of $A$ from diagonal $BD$

$= d =\frac{23}{\sqrt{50}}$

$50 d ^2=(23)^2$

$50 d ^2=529$

$x -\text { intercept }=\frac{7}{\cos \theta}$

$y -\text { intercept }=\frac{7}{\sin \theta}$

$A:\left(\frac{7}{\cos \theta}, 0\right) B :\left(0, \frac{7}{\sin \theta}\right)$

Locus of mid pt $M :( h , k )$

$h =\frac{7}{2 \cos \theta}, k =\frac{7}{2 \sin \theta}$

$\frac{7}{2 \sin \theta}=\frac{7 \sqrt{3}}{3} \Rightarrow \sin \theta=\frac{\sqrt{3}}{2} \Rightarrow \theta=\frac{\pi}{3}$

$\alpha=\frac{7}{2 \cos \theta}=7$

$\ell_2: x - y +1=0 \text { is } P \equiv(0,1)$

Which lies on $\ell_3: \alpha x +\beta y +17=0$,

$\Rightarrow \beta=-17$

Consider a random point $Q \equiv(-1,0)$ on $\ell_2: x - y +1=0$, image of $Q$ about

$\ell_2: x-y+1=0 \quad$ is $\quad Q^{\prime} \equiv\left(\frac{-17}{13}, \frac{6}{13}\right)$

which is calculated by formulae

$\frac{x-(-1)}{2}=\frac{y-0}{-3}=-2\left(\frac{-2+3}{13}\right)$

Now, $Q^{\prime}$ lies on $\ell_3: \alpha x +\beta y +17=0$

$\Rightarrow \alpha=7$

Now, $\alpha^2+\beta^2-\alpha-\beta=348$

$B _1(0,8) \equiv \text { image of } B \text { w.r.t. } y=4$

for $AC + BC + AB$ to be minimum.

$m _{ AB ^{\prime}}=\frac{-7}{2}$

$\text { line } AB _1 \equiv 7 x +2 y =16$

$C\left(\frac{8}{7}, 4\right)$

$\beta=\frac{8}{7}$

For max. perimeter

$AB =\sqrt{5}: BC =4 \sqrt{2}, AC =\sqrt{13}$

$6 \alpha+21 \beta=24+24=48$

Points $( p , p +1)$ lies on $y = x +1$

So point of intersection between

$y = x +1  y =3- x \text { is } x =1, y =2$

and point of intersection between

$x+1=\sqrt{9-x^2}$ is $x=\frac{-1+\sqrt{17}}{2}$

Hence $p \in\left(1, \frac{-1+\sqrt{17}}{2}\right)$

Hence $b ^2+ b - a ^2=3$

$\tan \pi / 4=\left|\frac{3 / 2-2 / x}{1+6 / 2 x}\right|=1$

$\Rightarrow x_{1}=10, \quad x_{2}=-2 / 5$

$\Rightarrow AA ^{1}=52 / 5$

$\sqrt{\left(x_{1}-4\right)^{2}+\left(x_{1}-2-3\right)^{2}}=\frac{\sqrt{29}}{3}$

Squaring on both side

$18 x_{1}^{2}-162 x_{1}+340=0$

$x_{1}=\frac{51}{9}  \text { or }  x_{1}=\frac{10}{3}$ $y_{1}=\frac{33}{9}  \text { or }  y_{1}=\frac{4}{3}$

Option $(C)$ will satisfy $\left(\frac{10}{3}, \frac{4}{3}\right)$

$A \left(\alpha, 4+\frac{4}{3} \alpha\right)$

Points $B$ lies on $L _{1}$

$B \left(\beta, 2-\frac{2}{5} \beta\right)$

Points $P$ divides $AB$ internally in the ratio $1: 3$

$\Rightarrow P(2,3)=P\left(\frac{3 \alpha+\beta}{4}, \frac{3\left(4+\frac{4}{3} \alpha\right)+1\left(2-\frac{2}{5} \beta\right)}{4}\right)$

$\Rightarrow \alpha=\frac{3}{13}, \beta=\frac{95}{13}$

Point $A \left(\frac{3}{13}, \frac{56}{13}\right), B \left(\frac{95}{13},-\frac{12}{13}\right)$

Vertex $C$ of triangle is the point of intersection of $L _{1} and L _{2}$

$\Rightarrow C \left(-\frac{15}{13}, \frac{32}{13}\right)$

area $\triangle ABC =\frac{1}{2}\left\|\begin{array}{ccc}\frac{3}{13} & \frac{56}{13} & 1 \\ \frac{95}{13} & -\frac{12}{13} & 1 \\ -\frac{15}{13} & \frac{32}{13} & 1\end{array}\right\|$

$\frac{1}{2 \times 13^{3}}\left\|\begin{array}{ccc} 3 & 56 & 13 \\ 95 & -12 & 13 \\ -15 & 32 & 13 \end{array}\right\|$

area $\triangle ABC =\frac{132}{13}$ sq. units.

$m _{ PD }=0$

$D \left(\frac{ a + a }{2}, \frac{ b +3}{2}\right)$

$D \left( a , \frac{ b +3}{2}\right)$

$m _{ PD }=0$

$\frac{ b +3}{2}-1=0$

$b +3-2=0$

$b =-1$

$E \left(\frac{ b + a }{2}, \frac{5+ b }{2}\right)=\left(\frac{ af }{2}, 2\right)$

$m_{C B} \cdot m_{E P}=-1$

$\left(\frac{5-b}{b-a}\right)=\left(\frac{2-1}{\frac{a-1}{2}-1}\right)=-1$

$\left(\frac{6}{-1-a}\right)=\left(\frac{2}{a-3}\right)=-1$

$12=(1+a)(a-3)$

$12=a^{2}-3 a+a-3$

$a^{2}-2 a-15=0$

$(a-5)(a+3)=0$

$a=5$ or $a=-3$

Given $a b > 0$

$a (-1) > 0$

$- a > 0$

$a < 0$

$a=-3$ Accept

AP line $A (-3,3) P (1,1)$

$y-1=\left(\frac{3-1}{-3-1}\right)(x-1)$

$-2 y+2=x-1$

$x+2 y=3 \quad$ Appling $\ldots . .(1)$

Line $BC B(-1,5)$

$C (-3,-1)$

$( y -5)=\frac{6}{2}( x +1)$

$y-5=3 x+3$

$y=3 x+8$

Solving (1) \& (2)

$x+2(3 x+8)=3$

$7 x+16=3$

$7 x=-13$

$x=-\frac{13}{7}$

$y=3\left(-\frac{13}{7}\right)+8$

$=\frac{-39+56}{7}$

$y=\frac{17}{7}$

$x+y=\frac{-13+17}{7}=\frac{4}{7}$

$L_{2}: 8 x+6 y+11=0$

Equation of angle bisector of $L_{1}$ and $L_{2}$ of angle containing origin

$2(3 x-4 y+12)=8 x+6 y+11$

$2 x+14 y-13=04$

$\frac{3 \alpha-4 \beta+12}{5}=14$

$3 \alpha-4 \beta+7=0$

Solution of $2 x+14 y-13=0$ and $3 x-4 y+7=0$ gives the required point $P(\alpha, \beta), \alpha=\frac{-23}{25}, \beta=\frac{53}{50}$

$100(\alpha+\beta)=14$

And $\beta= y$-cordinate of $R$

$=\frac{2 \times 4+1 \times 0}{2+1}=\frac{8}{3} \ldots(1)$

Now $P'$ is image of $P$ in $y =0$ which will be $P ^{\prime}(2,-3)$

Equation of $P ^{\prime} Q$ is $( y +3)=\frac{4+3}{5-2}( x -2)$

i.e. $3 y +9=7 x -14$

$A \equiv\left(\frac{23}{7}, 0\right)$ by solving with $y =0$

$\therefore \alpha=\frac{23}{7}.........(2)$

By $(1), \,\,(2)$

$7 \alpha+3 \beta=23+8=31$

$a^{2}+11 a+3\left(m_{1}^{2}+\frac{1}{m_{1}^{2}}\right)=220$

Eq. of $AC$

$AC =(\cos \alpha-\sin \alpha)+(\sin \alpha+\cos \alpha) y =10$

$BD =(\sin \alpha-\cos \alpha) x +(\sin \alpha-\cos \alpha) y =0$

$(10(\cos \alpha-\sin \alpha), 10(\sin \alpha-\cos \alpha))$

$\text { Slope of } AC =\left(\frac{\sin \alpha-\cos \alpha}{\sin \alpha+\cos \alpha}\right)=\tan \theta= M$

Eq. of line making an angle $\pi_{4}$ with $AC$

$m _{1}, m _{2}=\frac{ m \pm tan }{1 \pm m \operatorname{ta}}$

$=\frac{ m +1}{1- m }$ or $\frac{ m -1}{1+ m }$

$\frac{\frac{\sin \alpha-\cos \alpha}{\sin \alpha+\cos \alpha}+1}{1-\left(\frac{\sin \alpha-\cos \alpha}{\sin \alpha+\cos \alpha}\right)}, \frac{\frac{\sin \alpha-\cos \alpha}{\sin \alpha+\cos \alpha}-1}{1+\frac{\sin \alpha-\cos \alpha}{\sin \alpha+\cos \alpha}}$

$m_{1}, m_{2}=\tan \alpha, \cot \alpha$

mid point of $AC \& BD$

$= M (5(\cos \alpha-\sin \alpha), 5(\cos \alpha+\sin \alpha))$

$B (10(\cos \alpha-\sin \alpha), 10(\cos \alpha+\sin \alpha))$

$a = AB =\sqrt{2} BM =\sqrt{2}(5 \sqrt{2})=10$

$a =10$

$\because a^{2}+11 a+3\left(m_{1}^{2}+\frac{1}{m_{1} 2}\right)=220$

$100+110+3\left(\tan ^{2} \alpha+\cot ^{2} \alpha\right)=220$

Hence $\tan ^{2} \alpha=3, \tan ^{2} \alpha=\frac{1}{3} \Rightarrow \alpha=\frac{\pi}{3}$ or $\frac{\pi}{6}$

Now $72\left(\sin ^{4} \alpha+\cos ^{4} \alpha\right)+a^{2}-3 a+13$

$=72\left(\frac{9}{16}+\frac{1}{16}\right)+100-30+13$

$=72\left(\frac{5}{8}\right)+83=45+83=128$

So, $x=4, y=4$

Hence, $(x-2)^{2}+(y-4)^{2}=4$

$x-$intercept : $-\frac{ c }{ m }$

$y-$intercept : $C$

$A.M$ of reciprocals of the intercepts :

$\frac{-\frac{ m }{ c }+\frac{1}{ c }}{2}=\frac{1}{4} \Rightarrow 2(1- m )= c$

line $: y=m x+2(1-m)=c$

$\Rightarrow \quad(y-2)-m(x-2)=0$

$\Rightarrow$ line always passes through $(2,2)$

$L_{2}: x+2 y=3$

$L _{3}: x + y =6$

on solving $L _{1}$ and $L _{2}$ :

$y = L$ and $x =1$

$L _{1}$ and $L _{3}$ :

$x=2$

$y=2$

$L _{2}$ and $L _{3}$ :

$x+y=3$

$2 x+y=6$

$x =3$

$y =0$

$AC =\sqrt{4+1}=\sqrt{5}$

$BC =\sqrt{4+1}=\sqrt{5}$

$AB =\sqrt{1+1}=\sqrt{2}$

so its an isosceles triangle

$\Rightarrow x \cos \alpha-\operatorname{ysin} \alpha=\frac{\mathrm{k}}{2} \cos 2 \alpha$

$\Rightarrow \mathrm{p}=\left|\frac{\mathrm{k}}{2} \cos \alpha\right| \Rightarrow 2 \mathrm{p}=|\mathrm{k} \cos 2 \alpha| \quad \ldots(i)$

second line is $x\, \sin \alpha+y \,\cos \alpha=\operatorname{k\,sin} 2 \alpha$

$\Rightarrow \mathrm{q}=|\mathrm{k\,sin} 2 \alpha|\ldots(ii)$

Hence $4 \mathrm{p}^{2}+\mathrm{q}^{2}=\mathrm{k}^{2}$ (From $(i)\, \&\, (ii))$

$\Delta=\frac{1}{2}\left|\begin{array}{lll}1 & 2 & 1 \\ 7 & 5 & 1 \\ 2 & 3 & 1\end{array}\right|$

$=\frac{1}{2}[1(5-3)-2(7-2)+1(21-10)]$

$=\frac{1}{2}[2-10+11]$

$\Delta \mathrm{DEF}=\frac{1}{2}(3)=\frac{3}{2}$

$\Delta \mathrm{ABC}=4 \Delta \mathrm{DEF}=4\left(\frac{3}{2}\right)=6$

point $D$ is equal of intersection of $4 x+5 y=0\,  \,11 x+7 y=9$

So, coordinates of point $D=\left(\frac{5}{3},-\frac{4}{3}\right)$

Also, point $B$ is point of intersection of $7 x+2 y=0\, \, 11 x+7 y=9$

So, coordinates of point $B=\left(-\frac{2}{3}, \frac{7}{3}\right)$

diagonals of parallelogram intersect at middle let middle point of $B,D$

$\Rightarrow\left(\frac{\frac{5}{3}-\frac{2}{3}}{2}, \frac{-4}{3}+\frac{7}{3}\right)=\left(\frac{1}{2}, \frac{1}{2}\right)$

equation of diagonal $AC$

$\Rightarrow(y-0)=\frac{\frac{1}{2}-0}{\frac{1}{2}-0}(x-0)$

$y=x$

diagonal $AC$ passes through $(2,2)$

$3 \mathrm{P}^{2}+1=4+(\mathrm{P}+1)^{2}$

$2 \mathrm{P}^{2}-2 P-4=0$

$\Rightarrow \mathrm{P}^{2}-\mathrm{P}-2=0$

$\mathrm{P}=2,-1$ (rejected)

$\cos \theta=\frac{\overline{V}_{1} \cdot \vec{V}_{2}}{\left|\overrightarrow{\mathrm{V}}_{1}\right| \cdot \overrightarrow{\mathrm{V}_{2}} \mid}$

$\cos \theta=\frac{4 \sqrt{3}+3}{\sqrt{13} \sqrt{13}}=\frac{4 \sqrt{3}+3}{13}$

$\tan \theta=\frac{\sqrt{112-24 \sqrt{3}}}{4 \sqrt{3}+\sqrt{3}}=\frac{6 \sqrt{3}-2}{4 \sqrt{3}+3}=\frac{\alpha \sqrt{3}-2}{4 \sqrt{3}+3}$

$\Rightarrow \alpha=6$

$3= m + c$

$\sqrt{2}=\left|\frac{m-3 \sqrt{2}}{1+3 \sqrt{2} m}\right|$

$=6 m+\sqrt{2}=m-3 \sqrt{2}$

$=\sin =-4 \sqrt{2} \rightarrow m =\frac{-4 \sqrt{2}}{5}$

$=6 m-\sqrt{2}=m-3 \sqrt{2}$

$=7 m -2 \sqrt{2} \rightarrow m =\frac{2 \sqrt{2}}{7}$

According to options take $m =\frac{-4 \sqrt{2}}{5}$

So $y=\frac{-4 \sqrt{2} x}{5}+\frac{3+4 \sqrt{2}}{5}$

$4 \sqrt{2} x+5 y-(15+4 \sqrt{2})=0$

$\Rightarrow 2 \mathrm{a}+\mathrm{b}=11 \ldots \ldots \ldots . . (i)$

$\because \mathrm{C}$ lies on $7 \mathrm{x}-4 \mathrm{y}=1$

$\Rightarrow 7 \mathrm{a}-4 \mathrm{~b}=1 \quad \ldots \ldots (ii)$

$\therefore$ by $(i)$ and $(ii) :$ $\mathrm{a}=3, \mathrm{~b}=5$

$\Rightarrow \mathrm{C}(3,5)$

$\therefore \quad \mathrm{m}_{\mathrm{AC}}=2 / 3$

Also, $m_{C D}=7 / 4$

$\Rightarrow \tan \frac{\theta}{2}=\left|\frac{\frac{2}{3}-\frac{4}{4}}{1+\frac{14}{12}}\right| \Rightarrow \tan \frac{\theta}{2}=\frac{1}{2}$

$\Rightarrow \tan \theta=\frac{2 \cdot \frac{1}{2}}{1-\frac{1}{4}}=\frac{4}{3}$

Perpendicular bisector of $\mathrm{AB}$ is

$(y-3)=\frac{t}{3}(x-t)$

So, $C=\left(0,3-\frac{t^{2}}{3}\right)$

Let $\mathrm{P}$ be $(\mathrm{h}, \mathrm{k})$

$\mathrm{h}=\frac{\mathrm{t}}{2} ; \mathrm{k}=\left(3-\frac{\mathrm{t}^{2}}{6}\right)$

$\Rightarrow \mathrm{k}=3-\frac{4 \mathrm{~h}^{2}}{6} \Rightarrow 2 \mathrm{x}^{2}+3 \mathrm{y}-9=0$

$4 \alpha-2 \beta=\pm 24+8$

$\Rightarrow 4 \alpha-2 \beta=+24+8 \Rightarrow 2 \alpha-\beta=16$

$2 \mathrm{x}-\mathrm{y}-16=0...(1)$

$\Rightarrow 4 \alpha-2 \beta=-24+8 \Rightarrow 2 \alpha-\beta=-8$

$2 \mathrm{x}-\mathrm{y}+8=0...(2)$

perpendicular distance of $(1)$ from $(0,0)$

$\left|\frac{0-0-16}{\sqrt{5}}\right|=\frac{16}{\sqrt{5}}$

perpendicular distance of $(2)$ from $(0,0)$ is

$\left|\frac{0-0+8}{\sqrt{5}}\right|=\frac{8}{\sqrt{5}}$

$\angle \mathrm{B}=90^{\circ}$

$\text { Circum-center }=\left(\frac{1}{2}, \frac{11}{2}\right)$

Mid point of $B C=\left(2, \frac{17}{2}\right)$

Line $:\left(y-\frac{11}{2}\right)=2\left(x-\frac{1}{2}\right) \Rightarrow y=2 x+\frac{9}{2}$

Passing through $\left(0, \frac{\alpha}{2}\right)$

$\frac{\alpha}{2}=\frac{9}{2} \Rightarrow \alpha=9$

$4 x-2 y+6=0$

$\left|\frac{\alpha-6}{255}\right|=\frac{1}{55}$

$|\alpha-6|=2 \Rightarrow \alpha=8,4$

$\operatorname{sum}=12$

again $6 x-3 y+\beta=0$

$6 x-3 y+9=0$

$\left|\frac{\beta-9}{3 \sqrt{5}}\right|=\frac{2}{\sqrt{5}}$

$|\beta-9|=6 \Rightarrow \beta=15,3$

$\operatorname{sum}=18$

sum of all values of $\alpha$ and $\beta$ is $=30$

Equation of $\perp^{ r }$ bisector is

$y+4=(k-1)(x-0)$

$\Rightarrow y +4= x ( k -1)$

$\Rightarrow \frac{7}{2}+4=\frac{ k +1}{2}( k -1)$

$\Rightarrow \frac{15}{2}=\frac{ k ^{2}-1}{2} \Rightarrow k ^{2}=16 \Rightarrow k =4,-4$

So, $($ Put $(1,2)$ in given line $)(\operatorname{Put}(\sin \theta, \cos \theta)$ in given line$)>0$

$\Rightarrow(1+2-1)(\sin \theta+\cos \theta-1)>0$

$\Rightarrow \sin \theta+\cos \theta>1\{\div \operatorname{by} \sqrt{2}\}$

$\Rightarrow \frac{1}{\sqrt{2}} \sin \theta+\frac{1}{\sqrt{2}} \cos \theta>\frac{1}{\sqrt{2}}$

$\Rightarrow \quad \sin \left(\theta+\frac{\pi}{4}\right)>\frac{1}{\sqrt{2}}$

$\Rightarrow \frac{\pi}{4}<\theta+\frac{\pi}{4}<\frac{3 \pi}{4}$

$\Rightarrow \quad 0<\theta<\frac{\pi}{2}$

Image of point (-1,-4)

$\frac{x+1}{1}=\frac{y+4}{3}=-2\left(\frac{-1-12-3}{10}\right)$

$\frac{x+1}{1}=\frac{y+4}{3}=\frac{16}{5}$

$( x , y ) \equiv\left(\frac{11}{5}, \frac{28}{5}\right)$

$3 \alpha=2 \beta$

$\mathrm{h}=\frac{2 \alpha+\beta}{2}$

$2 \mathrm{h}=\frac{7 \alpha}{2}$

$\mathrm{k}=\frac{\alpha+\beta}{2}$

$2 \mathrm{k}=\frac{5 \alpha}{2}$

$\frac{\mathrm{h}}{\mathrm{k}}=\frac{7}{5}$

$5 x=7 y$

Also, $\frac{1}{2} \times \sqrt{10} \times h=5$

$\Rightarrow h=\sqrt{10}$

$\Rightarrow \frac{|4 \lambda-2|}{\sqrt{10}}=\sqrt{10} \Rightarrow \lambda=3,-2$

line passing through intersection of $x+3 y-1=0$ and

$3 x-y+1=0,$ be given by

$(x+3 y-1)+\lambda(3 x-y+1)=0$

$\because$ It passes through $(2,2)$

$\Rightarrow \quad 7+5 \lambda=0 \Rightarrow \lambda=-\frac{7}{5}$

$\therefore \quad$ Required line is $8 x-11 y+6=0$

$(-9,-6)$ satisfies this equation

$\tan 60^{\circ}=\frac{2 \sqrt{3}- k }{2-1}$

$\sqrt{3}=2 \sqrt{3}- k$

$\therefore \quad k=\sqrt{3}$

so point $A (1, \sqrt{3})$

Now slope of line $AB$ is $m _{ AB }=\tan 120^{\circ}$

$m m _{ AB }=-\sqrt{3}$

Now equation of line $AB$ is

$y-\sqrt{3}=-\sqrt{3}(x-1)$

$\sqrt{3} x+y=2 \sqrt{3}$

Now satisfy options

Which is satisified by point $(7,17)$

$ \Rightarrow 3\left( 7 \right) + 2\left( {17} \right) + c = 0$

$ \Rightarrow c =  - 55$

$ \Rightarrow $ equation of line is $3x+2y-55=0$

$ \Rightarrow 3\left( {15} \right) + 2\left( \beta  \right) - 55 = 0$

$ \Rightarrow 2\beta  = 55 - 45 \Rightarrow \beta  = 5$

$ \Rightarrow \frac{{15 - 3t}}{5} = t\,\,$ or $\frac{{15 - 3t}}{5} =  - t$

$\therefore t = \frac{{15}}{8}\,$ or $t = \frac{{ - 15}}{2}$ 

So, $P\left( {\frac{{15}}{8}.\frac{{15}}{8}} \right) \in {I^{st}}$ quadrant 

or $P\left( {\frac{{ - 15}}{2}.\frac{{15}}{2}} \right) \in I{I^{st}}$ Quadrant

$\therefore {m_1}{m_2} =  - 1$

$ \Rightarrow \left( {\frac{{ - 1}}{{a - 1}}} \right)\left( {\frac{{ - 2}}{{{a^2}}}} \right) =  - 1$

$ \Rightarrow {a^3} - {a^2} + 2 = 0$

$ \Rightarrow \left( {a + 1} \right)\left( {{a^2} - 2a + 2} \right) = 0$

$\therefore a =  - 1$

So line are $\left. \begin{array}{l}
{L_1}:x - 2y + 1 = 0\\
{L_2}:2x + y - 1 = 0
\end{array} \right\}$

Solving these equation we get point of intersection 

$P\left( {\frac{1}{5},\frac{3}{6}} \right)$

Now distance of $P$ from origin 

$OP = \sqrt {\frac{1}{{25}} +  + \frac{9}{{25}}}  = \sqrt {\frac{2}{5}} $

$\left| {\frac{\lambda }{5}} \right| = \frac{3}{5}$

$ \Rightarrow \lambda  \pm 3$

So, required equation of the line is 

$4x - 3y + 3 = 0$ and $4x - 3y - 3 = 0$

$\left( 1 \right)\,4\left( { - \frac{1}{4}} \right) - 3\left( {\frac{2}{3}} \right) + 3 = 0$

$px + qy + \left( {\frac{{ - 3p - 2q}}{4}} \right) = 0$

$p\left( {x - \frac{3}{4}} \right) + q\left( {y - \frac{2}{4}} \right) = 0$

$x = \frac{3}{4}\,\,$ and $y = \frac{1}{2}$

Equation of $AC$ is $4x+5y-20=0$.

Equation of $BE$ is $2x+3y-5=0$

Equation of $CF$ is $5x-4y-1=0$

$ \Rightarrow $ Equation of  $BC$ is

$26x-122y=1675$

$ \Rightarrow \alpha  = 3h - 4$ and $\beta  = 3k - 2$

Because $\left( {\alpha ,\beta } \right)$ lies on $2x-3y+4=0$

$ \Rightarrow 2\left( {3h - 4} \right) - 3\left( {k - 2} \right) + 4 = 0$

$ \Rightarrow $ losus is $6x-9y+2=0$ whose slope is $\frac{2}{3}$

$ = \left( {\frac{{8\left( {60} \right) + 6\left( 0 \right) + 10\left( 0 \right)}}{{24}},\frac{{10\left( 0 \right) + 6\left( 8 \right) + 8\left( 0 \right)}}{{24}}} \right) = \left( {2,2} \right)$

$M=(4,6)$

$B \Rightarrow \left( {1,2} \right),D \to \left( {3,6} \right)$

Slope of  $AD = \frac{5}{3}$

Equation of $AD$ is $y - 2 = \frac{5}{3}\left( {x - 1} \right)$

$ \Rightarrow 5x - 3y + 1 = 0$

at $P$ will pass through $\left( {\frac{3}{2},0} \right)$.

Let Co-ordinates of $P$ be $s\left( {\frac{{{t^2}}}{4},\frac{t}{2}} \right)$

Hence equation of normal is $y + tx = \frac{t}{2} + \frac{{{t^2}}}{4}$

The line passes through $\left( {\frac{3}{2},0} \right)$

$\frac{{3{t^2}}}{2} = \frac{t}{2} + \frac{{{t^3}}}{4} \Rightarrow t = 2$         ($-2,0$ are rejected)

hence nearest point is $(1,1)$

distance $\sqrt {{{\left( {\frac{3}{2} - 1} \right)}^2} + {{\left( {1 - 0} \right)}^2}}  = \frac{{\sqrt 5 }}{2}$

$\left( { - 3,4} \right) = \left( {\frac{a}{2},\frac{b}{2}} \right)$

$a=-6, b=8$

equation of line is 

$4x-3y+24=0$

or ${L_2}$ is $2x - y = 5$

Their point of intersection is $(3,1)$

$ \Rightarrow \frac{k}{h} = \frac{1}{3}$

$y = 3 + r\,\sin \theta $

$ \Rightarrow 2 + r\,\cos \theta  + 3 + r\,\sin \theta  = 7$

$ \Rightarrow r\left( {\cos \theta  + \,\sin \theta } \right) = 2$

$ \Rightarrow \sin \theta  + \cos \theta  = \frac{2}{r} = \frac{2}{{ \pm 4}} =  \pm \frac{1}{2}$

$ \Rightarrow 1 + \sin \,2\theta  = \frac{1}{4}$

$ \Rightarrow \sin \,2\theta  =  - \frac{3}{4}$

$ \Rightarrow \frac{{2m}}{{1 + {m^2}}} =  - \frac{3}{4}$

$ \Rightarrow 3{m^2} + 8m + 3 = 0$

$ \Rightarrow m = \frac{{ - 4 \pm \sqrt 7 }}{{1 - 7}}$

$\frac{{1 - \sqrt 7 }}{{1 + \sqrt 7 }} = \frac{{{{\left( {1 - \sqrt 7 } \right)}^2}}}{{1 - 7}} = \frac{{8 - 2\sqrt 7 }}{{ - 6}} = \frac{{ - 4 + \sqrt 7 }}{3}$

$2y = \cos \,2 - \cos \,\left( {2x + 2} \right) - \,\left( {1 - \cos \left( {2x + 2} \right)} \right)$

$ = \cos \,2 - 1$

$2y =  - 2{\sin ^2}\frac{1}{2}$

$y =  - {\sin ^2}\frac{1}{2} \le 0$

given $OP$ makes ${60^o}$ with $x+y=0$

Let slope of $OP=m$

$ \Rightarrow \tan \,{60^o} = \left| {\frac{{m + 1}}{{1 - m}}} \right|$

$ \Rightarrow \frac{{m + 1}}{{1 - m}} = \sqrt 3 $ or $ - \sqrt 3 $

$m + 1 = \sqrt {3m}  - 3$ or $ \Rightarrow m + 1 = \sqrt {3 - } \sqrt {3m} $

$ \Rightarrow m\left( {\sqrt 3  - 1} \right) = \sqrt 3  + 1$  or $m\left( {1 + \sqrt 3 } \right) = \sqrt 3  - 1$

$ \Rightarrow m = \frac{{\sqrt 3  + 1}}{{\sqrt 3  - 1}}$ or $\,\,m = \frac{{\sqrt 3  - 1}}{{\sqrt 3  + 1}}$

$ \Rightarrow \tan \,\alpha  = \frac{{\sqrt 3  + 1}}{{\sqrt 3  - 1}}\,\,$ or $\tan \,\alpha  = \frac{{\sqrt 3  - 1}}{{\sqrt 3  + 1}}$

$ \Rightarrow $ equation of line $x\,\cos \,\alpha  + y\,\sin \,\alpha  = P$

$ \Rightarrow \left( {\sqrt 3  + 1} \right)x + \left( {\sqrt 3  - 1} \right)y = 8\sqrt 2 $ or $\left( {\sqrt 3  - 1} \right)x + \left( {\sqrt 3  + 1} \right)y = 8\sqrt 2 $

$\frac{x}{h} + \frac{y}{k} = 1\,\,\,\,\,\,\,......\left( 1 \right)$

Since, $(1)$ passes through the fixed point $(2,3)$ Then,

$\frac{2}{h} + \frac{3}{k} = 1\,$

Then, the locus of $R$ is $\frac{2}{x} + \frac{3}{y} = 1$ or $3x + 2y = xy$.

So the $x$ coordianate of $C$ is $4$. let $C \equiv \left( {4,y} \right)$

then the midpoint of $A(1,2)$ and $C(4,y)$ is $D$ 

which lies on the median through $B$.

$D \equiv \left( {\frac{{1 + 4}}{2},\frac{{2 + y}}{2}} \right)$

Now, $\frac{{1 + 4 + 2 + y}}{2} = 5 \Rightarrow y = 3$

So, $C \equiv \left( {4,3} \right)$.

The centroid of the triangle is the intersection of the mesians. Here the medians $x=4$ and $x+4$ and $x+y=5$ intersect

at $G(4,1)$.

The area of triangle $\Delta ABC = 3 \times \Delta AGC$

$ = 3 \times \frac{1}{2}\left[ {1\left( {1 - 3} \right) + 4\left( {3 - 2} \right) + 4\left( {2 - 1} \right)} \right] = 9$

Equations of the given lines are

$x-y+2=0$

$7x-y+3=0$

we know that the diagonals of the rhombus will be parallel to the angle bisectors of the two given lines; $y=x+2$ and $y=7x+3$

$\therefore $ equation of angle bisectors is given as:

$\frac{{x - y + 2}}{{\sqrt 2 }} =  \pm \frac{{7x - y + 3}}{{5\sqrt 2 }}$

$5x - 5y + 10 =  \pm \left( {7x - y + 3} \right)$

$\therefore $Parallel equation of the diagonals are $2x+4y-7=0$ and $12x-6y+13=0$

$\therefore $ slopes of diagonals are $\frac{{ - 1}}{2}$ and $2$.

Now, slope of the diagonal from $A(0,c)$ and passing through $P(1,2)$ is $(2-c)$

$\therefore 2 - c = 2 \Rightarrow c = 0$  (not possible)

$\therefore 2 - c = \frac{{ - 1}}{2} \Rightarrow c = \frac{5}{2}$

$\frac{{x - 0}}{3} = \frac{{y - 0}}{1} = r$

For foot of perpendicular

$r = \frac{{ - \left( {\left( {3 \times 0} \right) + \left( {1 \times 0} \right) - \lambda } \right)}}{{{3^2} + {1^2}}} = \frac{\lambda }{{10}}$

So, food of perpendicular $P = \left( {\frac{{3\lambda }}{{10}},\frac{\lambda }{{10}}} \right)$

Given the line meets $X$-asix at $A = \left( {\frac{\lambda }{3},0} \right)$

and meets $Y$-axis at $B = \left( {o,\lambda } \right)$

So,

$BP = \sqrt {{{\left( {\frac{{3\lambda }}{{10}}} \right)}^2} + {{\left( {\frac{\lambda }{{10}} - \lambda } \right)}^2}}  \Rightarrow BP = \sqrt {\frac{{9{\lambda ^2}}}{{100}} + \frac{{81{\lambda ^2}}}{{100}}} $

$ \Rightarrow BP = \sqrt {\frac{{90{\lambda ^2}}}{{100}}} $

Now,$PA = \sqrt {{{\left( {\frac{\lambda }{3} - \frac{{3\lambda }}{{10}}} \right)}^2} + {{\left( {0 + \frac{\lambda }{{10}}} \right)}^2}} $

 $PA = \sqrt {\frac{{{\lambda ^2}}}{{900}} + \frac{{{\lambda ^2}}}{{100}}}  \Rightarrow PA = \sqrt {\frac{{10{\lambda ^2}}}{{900}}} $

Therefore $BP:PA = 3:1$

$\frac{x-y+1}{\sqrt{2}}=\pm \frac{7 x-y-5}{5 \sqrt{2}}$

$\Rightarrow 5(\mathrm{x}-\mathrm{y}+1)=7 \mathrm{x}-\mathrm{y}-5$

and

$5(x-y+1)=-7 x+y+5$

$\therefore 2 \mathrm{x}+4 \mathrm{y}-10=0 \Rightarrow \mathrm{x}+2 \mathrm{y}-5=0$ and

$12 x-6 y=0 \Rightarrow 2 x-y=0$

Now equation of diagonals are

$(x+1)+2(y+2)=0 \Rightarrow x+2 y+5=0$           .......$(1)$

and

$2(x+1)-(y+2)=0 \Rightarrow 2 x-y=0$           ....$(2)$

Clearly $\left(\frac{1}{3},-\frac{8}{3}\right)$ lies on $( 1)$

${L_2}:3x + 4y - 12 = 0$

${L_1} + \lambda {L_2} = 0$

$\left( {4x + 3y - 12} \right) + \lambda \left( {3x + 4y - 12} \right) = 0$

$x\left( {4 + 3\lambda } \right) + y\left( {3 + 4\lambda } \right) - 12\left( {1 + \lambda } \right) = 0$

Point $A\left( {\frac{{12\left( {1 + \lambda } \right)}}{{4 + 3\lambda }},0} \right)$

Point $B\left( {0,\frac{{12\left( {1 + \lambda } \right)}}{{3 + 4\lambda }}} \right)$

mid point $ \Rightarrow h = \frac{{6\left( {1 + \lambda } \right)}}{{4 + 3\lambda }}\,\,\,\,\,\,\,....\left( i \right)$

$k = \frac{{6\left( {1 + \lambda } \right)}}{{3 + 4\lambda }}\,\,\,\,\,\,\,........\left( {ii} \right)$

Eliminate $\lambda $ from $(i)$ and $(ii)$, then

$6\left( {h + k} \right) = 7hk$

$6\left( {x + y} \right) = 7xy$

To find equation of $R$

slop of $L=0$ is $1$

$ \Rightarrow $slop of $QR=-1$

Let $QR$ is $y=mx+c$

           $y=-x+c$

           $x+y-c=o$

distance of $QR$ from $(2,1)$ is $2\sqrt 3 $

$2\sqrt 3  = \frac{{\left| {2 + 1 - c} \right|}}{{\sqrt 2 }}$

$2\sqrt 6  = \left| {3 - c} \right|$

$c - 3 =  \pm 2\sqrt 6 \,\,\,c = 3 \pm 2\sqrt 6 $

Line can be $x + y = 3 \pm 2\sqrt 6 $

$x + y = 3 - 2\sqrt 6 $

${P_1} = \frac{{10}}{5} = 2$

Length of $ \bot $ to $8x+6y+5=0$ from origin $(0,0)$

${P_2} = \frac{5}{{10}} = \frac{1}{2}$

Lines  are parallel to each other $ \Rightarrow $ ratio will be $4:1$ or $1:4$

$\therefore $ angle of incidence =angle of reflection

$\therefore \left| {\frac{{m - 7}}{{1 + 7m}}} \right| = \left| {\frac{{ - 2 - 7}}{{1 - 14}}} \right| = \frac{9}{{13}}$

$ \Rightarrow \frac{{m - 7}}{{1 + 7m}} = \frac{9}{{13}}\,\,\,\,$ or $\frac{{m - 7}}{{1 + 7m}} =  - \frac{9}{{13}}\,\,\,\,$

$ \Rightarrow 13m - 91 = 9 + 63m\,\,\,\,\,\,\,\,\,13m - 91 =  - 9 - 63m\,\,$

$ \Rightarrow 5m =  - 100 \Rightarrow 76m = 82$

$ \Rightarrow m =  - \frac{1}{2}\,\,\,\,\,\,\,\,\,\,m = \frac{{41}}{{38}}$

$ \Rightarrow y - 1 =  - \frac{1}{2}\left( {x - 0} \right)$

$ \Rightarrow y - 1 = \frac{{41}}{{38}}\left( {x - 0} \right)$

    i.e $x + 2y - 2 = 0$

$ \Rightarrow 38y - 38 - 41x = 0$

$ \Rightarrow 41x - 38y + 38 = 0$

Slope of $AB = \frac{1}{3}$

Slope of $BC = \frac{1}{3}$

so, lies on same line

(where $A$ is the point of intersection of the given lines)

$(x-2)(x-1)+(y-2)(y-3)=0$

$\Rightarrow\left(\frac{h+2}{2}-2\right)\left(\frac{h+2}{2}-1\right)+\left(\frac{k+3}{2}-2\right)\left(\frac{k+3}{2}-3\right)=0$

$\Rightarrow(h-2)(h)+(k-1)(k-3)=0$

$\Rightarrow \quad x^{2}-2 x+y^{2}-4 y+3=0$

$\Rightarrow \quad(x-1)^{2}+(y-2)^{2}=2$

$\frac{x}{2} + \frac{y}{4} = 1$

$2x + y = 4\,\,\,\,\,\,\,\,\,\,.....\left( 1 \right)$

For line 

$x - 2y =  - 4\,\,\,\,\,.......\left( 2 \right)$

solving equation  $(1)$ and $(2)$; we get point of intersection

$\left( {4/5,\frac{{12}}{5}} \right)$

$ \Rightarrow y =  - \sqrt 3 x + 1$

$ \Rightarrow $ (slope) ${m_2} =  - \sqrt 3 $

Let the other slope be ${m_1}$

$\therefore \tan {60^o} = \left| \begin{array}{l}
{m_1} - \left( { - \sqrt 3 } \right).\\
1 + \left( { - \sqrt 3 {m_1}} \right)
\end{array} \right.$

$ \Rightarrow {m_1} = 0,{m_2} = \sqrt 3 $

Since line $L$ is passing through $(3,-2)$

$\therefore y - \left( { - 2} \right) =  + \sqrt 3 \left( {x - 3} \right)$

$ \Rightarrow y + 2 = \sqrt 3 \left( {x - 3} \right)$

$y - \sqrt 3 x + 2 + 3\sqrt 3  = 0$

Slope of $\mathrm{PS}=\frac{-1}{\frac{9}{2}}=\frac{-2}{9}$

and line passes through $(1,-1)$

$\frac{y+1}{x-1}=\frac{-2}{9}$

$2 x+9 y+7=0$

$x + 3by + b = 0\,\,\,\,\,\,\,\,\,......\left( 2 \right)$

$x + 4ay + a = 0\,\,\,\,\,\,\,\,\,......\left( 3 \right)$

Subtracting equation $(3)$ from $(1)$

$ - 2y = 0$

$ay = 0 \Rightarrow y = 0$

Putting value of $y$ in equation $(1)$, we get

$x + 0 + a = 0$

$x =  - a$

Putting value of $x$ and $y$ in equation $(2)$, we get

$ - a + b = 0$

$a=b$

Thus, $(a,b)$ lies on astraight line

at $Y=0$,$X=2$ $[R (2,0)]$

as $PQ$ is parallel to $x,y$-coordinates of $Q$ is also $3$

putting value of $y$ in equation $(1)$, we get $Q(8,3)$

Centroid of $\Delta PQR = \left( {\frac{{8 + 5 + 2}}{3},\frac{{3 + 3}}{3}} \right)$

$=(5,2)$

Only $(2x - 5y = 0)$ satisfy the given coordinates.

$d = \frac{{a{x_1} + b{y_1} - c}}{{\sqrt {{a^2} + {b^2}} }}$

Now shortest distance of $P(1,2)$ from $3x+4y=9$ is

$PC = d = \frac{{3\left( 1 \right) + 4\left( 2 \right) - 9}}{{\sqrt {{3^2} + {4^2}} }} = \frac{2}{5}$

Given that $\Delta APB$ is equilateral triangle

Let $'a'$ be its side

then $PB = a,Cb = \frac{a}{2}$

Now,In $\Delta PCB,{\left( {PB} \right)^2} = {\left( {PC} \right)^2} + {\left( {CB} \right)^2}$

(By Pythagoras theorem )

${a^2} = {\left( {\frac{2}{5}} \right)^2} + \frac{{{a^2}}}{4}$

${a^2} - \frac{{{a^2}}}{4} = \frac{4}{{25}} \Rightarrow \frac{{3{a^2}}}{4} = \frac{4}{{25}}$

${a^2} = \frac{{16}}{{75}} \Rightarrow a = \sqrt {\frac{{16}}{{75}}}  = \frac{4}{{5\sqrt 3 }} \times \frac{{\sqrt 3 }}{{\sqrt 3 }} = \frac{{4\sqrt 3 }}{{15}}$

$\therefore $ Lengh of Equilateral triangle $(a) = \frac{{4\sqrt 3 }}{5}$

Centroid $ = \left( {\frac{{{{\left( {a + 1} \right)}^2}}}{2},\frac{{{{\left( {a - 1} \right)}^2}}}{2}} \right)$

We know the circumcenter $(O)$,

Centroid $(G)$ and orthocentre $(H)$ of a

triangle lie on the line joining the $O$ and $G$

Also, $\frac{{HG}}{{GO}} = \frac{2}{1}$

$ \Rightarrow $ Coordinate of orthocentre 

$ = \frac{{3{{\left( {a + 1} \right)}^2}}}{2},\frac{{3{{\left( {a - 1} \right)}^2}}}{2}$

Now, these coordinates satisfies eqn given in option $(d)$

Hence, required eqn of line is

${\left( {a - 1} \right)^2}x - {\left( {a + 1} \right)^2}y = 0$

Given that are of $\Delta AOB$  is $5$.

We know

$\left\{ {are,A = \frac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_2}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]} \right\}$

$ \Rightarrow 5 = \frac{1}{2}\left[ {c\left( {\frac{c}{5}} \right)} \right]$

$\because \left ( \begin{array}{l}
\left( {{x_1},{y_1}} \right) = \left( {10,0} \right)\left( {{x_3},{y_3}} \right) = \left( {0,\frac{c}{5}} \right)\\
\left( {{x_2},{y_2}} \right) = \left( {c,0} \right)
\end{array} \right)$

$ \Rightarrow c =  \pm \sqrt {50} $

$\therefore $ Equation of line $L$ is $x + 5y =  \pm \sqrt {50} $

Distance betwween $L$ and line $x+5y=0$ is 

$d = \left. {\frac{{ \pm \sqrt {50}  - 0}}{{\sqrt {{1^2} + {5^2}} }}} \right| = \frac{{\sqrt {50} }}{{\sqrt {26} }} = \frac{5}{{\sqrt {13} }}$

$(\alpha,-\alpha)$ lies on both lines $4 a x+2 a y+c=0$ and $5 b x+2 b y+d=0$

$\therefore 4 a \alpha-2 a \alpha+c=0 \Rightarrow \alpha=\frac{-c}{2 a} \ldots \ldots(i)$

$5 b \alpha-2 b \alpha+d=0 \Rightarrow \alpha=\frac{-d}{3 b} \ldots \ldots(i i)$

From Eqs. $(i)$ and $(ii),$

we get $\frac{-c}{2 a}=\frac{-d}{3 b}$

$\Rightarrow 3 b c=2 a d$

$\Rightarrow 2 a d-3 b c=0$

$ \Rightarrow 4 = \left( {\frac{{1 \times 0 + 2 \times a}}{{1 + 2}}} \right) = \frac{{2a}}{3}$

$ \Rightarrow a = 6 \Rightarrow $  coordinnate of $B$ is $B(6,0)$

$3 = \left( {\frac{{1 \times b + 2 \times 0}}{{1 + 2}}} \right) = \frac{b}{3}$

$ \Rightarrow b = 9 \Rightarrow $ and $C(0,9)$ 

Slope of line passing through $(6,0),(0,9)$

slop, $m = \frac{9}{{ - 6}} =  - \frac{3}{2}$

Aquation of line $y - 0 =  - \frac{3}{2}\left( {x - 6} \right)$

$2y =  - 3x + 18$

$3x + 2y = 18$

Equation of $\mathrm{AB}^{\prime}$ is

$ y-0=\frac{-1-0}{0-\sqrt{3}}(x-\sqrt{3}) $

$ \Rightarrow -\sqrt{3} y=-x+\sqrt{3}$

$ \Rightarrow  x-\sqrt{3} y=\sqrt{3} $

$ \Rightarrow  \sqrt{3} y=x-\sqrt{3} $

Slope of $PQ=1$

Slope of the line $L=-1$

Mid-point $(3,4)$ line on the line $L$.

equation of line $L$,

$y - 4 =  - 1\left( {x - 3} \right) \Rightarrow x + y - 7 = 0\,\,\,\,\,.....\left( i \right)$

Let image of point $R(0,0)$ be $S\left( {{x_1},{y_1}} \right)$

Mid-point of $RS = \left( {\frac{{{x_1}}}{2},\frac{{{y_1}}}{2}} \right)$

Mid-point $\left( {\frac{{{x_1}}}{2},\frac{{{y_1}}}{2}} \right)$ line on the line $(i)$

$\therefore {x_1} + {y_1} = 14\,\,\,\,\,\,\,\,.....\left( {ii} \right)$

Slope of $RS = \frac{{{y_1}}}{{{x_1}}}$

Since $RS \bot $ line $L$

$\therefore \frac{{{y_1}}}{{{x_1}}} \times \left( { - 1} \right) =  - 1$

$\therefore {x_1} = {y_1}\,\,\,\,\,\,\,\,\,\,\,.......\left( {iii} \right)$

From $(ii)$ and $(iii)$

${x_1} = {y_1} = 7\,\,$

Hence the image of $R=(7,7)$

$AC = BC \Rightarrow t = \sqrt {4{a^2} + {{(a - t)}^2}} \Rightarrow $$t = \frac{{5a}}{2}$

So the coordinates of third vertex $C$ are $\left( {2a,\frac{{5a}}{2}} \right)$

Therefore area of the triangle

$ = \pm \frac{1}{2}\left| {\,\begin{array}{*{20}{c}}{2a}&{\frac{{5a}}{2}}&1\\{2a}&0&1\\0&a&1\end{array}\,} \right| = \left| {\,\begin{array}{*{20}{c}}a&{\frac{{5a}}{2}}&1\\0&{ - \frac{{5a}}{2}}&0\\0&a&1\end{array}\,} \right| = \frac{{5{a^2}}}{2}sq.$units.

$ax+2y=q$ and $ax+y=r$

from aright angled triangle

$\therefore $ product of sloper of any two lines $=-1$

Suppose $ax+2y=q$ and $x-3y=p$ are $ \bot $ to each other.

$\therefore \frac{{ - a}}{2} \times \frac{1}{3} =  - 1 \Rightarrow a = 6$

Now, consider option one by one $a=6$ satisfies noly option $(a)$

$\therefore $ Required answer is ${a^2} - 9a + 18 = 0$

Centroid, $E = \left( {\frac{{{x_1} - 5}}{3},\frac{{{x_1} - 3}}{3}} \right)$

Since centroid lies on the line 

$3x + 4y + 2 = 0$

$\therefore 3\left( {\frac{{{x_1} - 5}}{3}} \right) + 4\left( {\frac{{{x_1} - 3}}{3}} \right) + 2 = 0$

$ \Rightarrow 3{x_1} + 4{y_1} + 3 = 0$

Hence vertex $\left( {{x_1},{y_1}} \right)$ lies on the line 

$3x + 4y + 3 = 0$

$\therefore $  $Q=(x,0)$

$\tan \,\theta  = \frac{{0 - 7}}{{x - 6}},\tan \left( {{{180}^o} - \theta } \right) = \frac{{0 - 3}}{{x - 1}}$

Now, $\tan \left( {{{180}^o} - \theta } \right) =  - \tan \,\theta $

$\therefore \frac{{ - 3}}{{x - 1}} = \frac{{ - 7}}{{x - 6}} \Rightarrow x = \frac{5}{2}$