Maths M.C.Q (1 Marks)

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

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