Straight Lines and Pair of Straight Lines

Let the slopes of lines $L_1, L_2$ and $L_3$ be $m_1, m_2$ and $m_3$.

Here, $m_1=-2$ and $m_3=-\frac{1}{2}$

Now, angle between $L_1$ and $L_3$ will be equal to angle between $L_2$ and $L_3$.

$ \begin{array}{ll} \therefore & \tan ^{-1}\left|\frac{m_3-m_1}{1+m_1 m_3}\right| \\\\ & =\tan ^{-1}\left|\frac{m_3-m_2}{1+m_2 m_3}\right| \\\\ \Rightarrow & \left|\frac{-\frac{1}{2}+2}{1+(-2)\left(-\frac{1}{2}\right)}\right|=\left|\frac{-\frac{1}{2}-m_2}{1-\frac{1}{2} m_2}\right| \end{array} $

$ \begin{aligned} \Rightarrow \quad\left(\frac{-\frac{1}{2}-m_2}{1-\frac{1}{2} m_2}\right)^2 & =\left(\frac{3}{4}\right)^2 \\\\ \Rightarrow \quad m_2 & =-2 \frac{2}{11} \end{aligned} $

Discard $m_2=-2$ because if $m_2=-2$, then $L_1$ and $L_2$ will be parallel to each other.

$ \therefore \quad m_2=\frac{2}{11} $

The line $L_2$ is passing through the point ( $5, \mathrm{f}$ ).

$ \begin{array}{ll} \therefore & L_2:(y-1)=\frac{2}{11}(x-5) \\\\ \Rightarrow & 2 x-11 y+1=0 \end{array} $

Hence, $a=2 b=-11$ and $c=1$

So, $\frac{b^2}{|a c|}=\frac{(-11)^2}{|2 \times 1|}=\frac{121}{2}$

Given pair of lines

$ 2 x^2+h x y+6 y^2=0 $

Let the equations of lines be $y=m x$ and

$ y=3 m x . $

Now, pair of equation of these lines will

$ \begin{aligned} & \text { be }(y-m x)(y-3 m x)=0 \\\\ & \Rightarrow y^2-m x y-3 m x y+3 m^2 x^2=0 \\\\ & \Rightarrow 3 m^2 x^2-4 m x y+y^2=0 \\\\ & \therefore \frac{3 m^2}{2}=\frac{-4 m}{h}=\frac{1}{6} \Rightarrow \frac{3 m^2}{2}=\frac{1}{6} \\\\ & \Rightarrow m^2=\frac{1}{9} \text { and } \frac{-4 m}{h}=\frac{1}{6} \\\\ & \Rightarrow h=-24 m \\\\ & \Rightarrow \quad h^2=(24)^2 m^2=576\left(\frac{1}{9}\right) \\\\ & \Rightarrow \quad h^2=64 \\\\ & \Rightarrow h= \pm 8 \end{aligned} $

When origin is shifted to $(2, b)$ by translation of axes, the coordinates of the point $(a, 4)$ have changed to $(6,8)$

$\therefore X=x-2 \Rightarrow Y=y-b $

For point $(a, 4)$ transforming to $(6,8)$

$ \begin{aligned} & 6=a-2 \Rightarrow a=8 \\\\ & 8=4-b \Rightarrow b=-4 \end{aligned} $

So, the point $(a, b)$ is $(8,-4)$.

Now, when origin is shifted to $(a, b)$

$ \Rightarrow \quad X=x-8 \Rightarrow Y=y+4 $

and transformed equation of $x^2+4 x y+y^2=0$ is

$ \begin{aligned} & (X+8)^2+4(X+8)(Y-4)+(Y-4)^2=0 \\\\ & X^2+Y^2+4 X Y+24 Y-48=0 \end{aligned} $

On comparing with

$ X^2+2 H X Y+Y^2+2 G X+2 F Y+C=0 $

we get, $G=0, H=2, F=12, C=-48$

$ \therefore 2 H(G+F)=2 \times 2(0+12)=48=-C $

Since, slope of line $L$ passing through $(-2,-3)$ is not defined $\Rightarrow \quad L_1: x=-2$ and $m_1= \pm \infty$ and given line is $a x-2 y+3=0(a>0)$

$ m_2=\frac{a}{2} $

Given, $\tan 45^{\circ}=\left|\frac{m_2-m_1}{1+m_1 m_2}\right|$

$\tan 45^{\circ}=\left|\frac{\frac{a}{2}-\infty}{1+\frac{a}{2} \cdot \infty}\right|$ or $\left|\frac{\frac{a}{2}+\infty}{1-\frac{a}{2} \cdot \infty}\right|$

$\Rightarrow a=2$ (considering the $45^{\circ}$ angle with the vertical line)

Let $\theta$ be the angle made by the line $x+a y-4=0$ with positive $X$-axis in the anti-clockwise direction.

$ \begin{array}{rlrl} & \therefore & \tan \theta & =\frac{-1}{a}=\frac{-1}{2} \\\\ \Rightarrow & \theta & =\tan ^{-1}\left(-\frac{1}{2}\right)=\pi-\tan ^{-1}\left(\frac{1}{2}\right) \end{array} $

We have,

$ \begin{aligned} & x-3 y+3=0\,\,\,\,\,\,.......(i) \\\\ & k x+y+k=0 \,\,\,.......(ii)\\\\ & 2 x+y-8=0 \,\,\, ........(iii) \end{aligned} $

On solving Eqs. (i) and (iii), we get

$ x=3, y=2 $

From Eq. (ii), we get

$ 3 k+2+k=0 \Rightarrow K=-\frac{1}{2} $

( $\because$ Eqs. (i), (ii) and (iii) are concurrent) and $a=3, b=2$ [point of concurrency is $(3,2)]$

So, $L \equiv 3 x-2 y-1=0$

given, distance from origin to $L$ is $P$

$ P=\frac{|0-0-1|}{\sqrt{3^2+(-2)^2}}=\frac{1}{\sqrt{13}} $

Hence, perpendicular distance from the point $(2,3)$ to $L$ is

$ =\frac{|6-6-1|}{\sqrt{3^2+(-2)^2}}=\frac{1}{\sqrt{13}}=P $

Slope of diagonal $=\frac{-2-3}{1-4}=\frac{5}{3}$

Let slope of a side of square be $m$.

$\because$ Angle between a diagonal and a side in square $=45^{\circ}$

$\therefore \tan 45^{\circ}=\left|\frac{\frac{5}{3}-m}{1+\frac{5}{3} m}\right| \Rightarrow 1=\left|\frac{\frac{5}{3}-m}{1+\frac{5}{3} m}\right|$

$\Rightarrow \quad m=\frac{1}{4},-4$

Then, equation of one of the side of square is

$ \begin{aligned} y-(-2) & =\frac{1}{4}(x-1) \Rightarrow 4(y+2)=x-1 \\\\ x-4 y-9 & =0 \end{aligned} $

We have, $12 x^2-20 x y+7 y^2=0$

$ (6 x-7 y)(2 x-y)=0 $

$\therefore$ Sides of triangle are

$ \begin{aligned} & 6 x-7 y=0 \quad \ldots \text { (i) } \\\\ & 2 x-y=0 \quad \ldots \text { (ii) } \\\\ & x+y=-1 \quad \ldots \text { (iii) } \end{aligned} $

From Eqs. (i), (ii) and (iii), we get the coordinates of vertices are

$ \begin{aligned} & \begin{aligned} (0,0),\left(-\frac{1}{3},-\frac{2}{3}\right),\left(-\frac{7}{13},-\frac{6}{13}\right) \end{aligned} \\\\ & \begin{aligned} \text { So, area of triangle } & \left.=\left|\frac{1}{2}\right| \begin{array}{ccc} 0 & 0 & 1 \\\\ -\frac{1}{3} & -\frac{2}{3} & 1 \\\\ -\frac{7}{13} & -\frac{6}{13} & 1 \end{array} \right\rvert\, \\\\ & =\left|\frac{1}{2}\left(\frac{6}{39}-\frac{14}{39}\right)\right| \\\\ & =\left|\frac{1}{2} \times\left(-\frac{8}{39}\right)\right|=\frac{4}{39} \end{aligned} \end{aligned} $

$ \text { Equation of line } $

$ \begin{aligned} &\frac{x}{a}+\frac{y}{b}=1\\ &\text { Let the co-ordinate of } C \equiv(h, k)\\ &\Rightarrow \quad \frac{x}{h}+\frac{y}{k}=1 \end{aligned} $

$ \begin{array}{r} (-3,4) \Rightarrow \frac{-3}{h}+\frac{4}{k}=1 \\ 4 x-3 y-x y=0 \end{array} $

$2 x^2+y^2-4 x+4 y=0$

or $2(x-1)^2+(y+2)^2=6$

Now, let origin is shifted to $0^{\prime}(h, k)$, so the new equation will be

$ 2(x-h-1)^2+(y-k+2)^2=6 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Which is given as

$ \begin{array}{r} 2 x^2+y^2-8 x+8 y+18=0 \\or\,\, 2(x-2)^2+(y+4)^2=6 \,\,...(ii)\end{array} $

On comparing Eqs. (i) and (ii), we get

$ \begin{aligned} h+1 & =2 \text { and } k-2=-4 \\ h & =1 \text { and } k=-2 \end{aligned} $

Hence, the equation of line $x+2 y+2=0$ will be

$ \begin{aligned} & (x-1)+2(y+2)+2=0 \\ & \Rightarrow \quad x+2 y+5=0 \end{aligned} $

Given, lines $x+y-1=0$,

$k x+2 y+1=0$ and $4 x+2 k y+7=0$ are concurrent, then $\Delta=0$

$ \begin{aligned} & \because \quad\left|\begin{array}{ccc} 1 & 1 & -1 \\ k & 2 & 1 \\ 4 & 2 k & 7 \end{array}\right|=0 \\ & \Rightarrow 1(14-2 k)-1(7 k-4)-1\left(2 k^2-8\right)=0 \\ & \Rightarrow \quad-2 k^2-9 k+26=0 \\ & \Rightarrow \quad 2 k^2+9 k-26=0 \\ & \Rightarrow \quad k=\frac{-13}{2}, 2 \end{aligned} $

But at $k=2$, lines $x+y-1=0 2 x+2 y+1=0$ and $4 x+4 y+7=0$ are parallel.

$ \therefore \quad k=\frac{-13}{2} $

\begin{aligned}

& \text { Given, } 2 x+(3-2 k) y+2 k+1=0 \

& \text { and } k x+(k-1) y-4=0 \

& \qquad m_1=\frac{-2}{3-2 k}

\end{aligned}

$ \begin{array}{rlrl} \text { and } & m_2 =\frac{-k}{k-1} \\ \Rightarrow & m_1 m_2 =-1 \\ \Rightarrow & \frac{2 k}{(3-2 k)(k-1)} =-1 \\ \Rightarrow & 2 k =2 k^2-5 k+3 \\ \Rightarrow & 2 k^2-7 k+3 =0 \\ \Rightarrow & k =3, \frac{1}{2} \\ \Rightarrow & \alpha =3 \text { and } \beta=\frac{1}{2} \\ & \therefore \alpha^2+2 \beta =9+1=10 \end{array} $

$ \begin{aligned} & \text { } k=\frac{a+b}{a b} \Rightarrow k=\frac{1}{a}+\frac{1}{b} \\ & \Rightarrow \quad \frac{1}{k a}+\frac{1}{k b}=1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Now, line $\frac{x}{a}+\frac{y}{b}=1$

To have the equation verified, we need

$ \begin{aligned} x & =\frac{1}{k} \text { and } y=\frac{1}{k} \\ \therefore \text { Point } & =\left(\frac{1}{k}, \frac{1}{k}\right) \end{aligned} $

$3 x^2-y^2-2 x+4 y=0$

Let the chord

$ a x+b y=1 $

Now, making homogeneous to Eq. (i) with the help of Eq. (ii),

$ \begin{array}{r} 3 x^2-y^2-2 x(a x+b y)+4 y(a x+b y)=0 \\ \Rightarrow(3-2 a) x^2+(4 b-1) y^2-2 b x y \ +4 a x y=0 \end{array} $

Chord subtracts right angle at origin

$ \Rightarrow 3-2 a+4 b-1=0 \Rightarrow 2 a-4 b=2 $

or  $ a-2 b=1 $

On comparing with $a x+b y=1$, we get fixed point $(1,-2)$

Given, parallel lines are

$ \begin{aligned} & 3 x-2 y+5=0 \text { and } 3 x-2 y+5+2 \sqrt{13}=0 \\ & \text { So, } d=\frac{|5+2 \sqrt{13}-5|}{\sqrt{(3)^2+(-2)^2}}=\frac{2 \sqrt{13}}{\sqrt{13}}=2 \end{aligned} $

Then, $\frac{4 d}{\sqrt{13}}=\frac{8}{\sqrt{13}}$ and $\frac{3 d}{\sqrt{13}}=\frac{6}{\sqrt{13}}$

According to the question,

$ \frac{\left|k_1-5\right|}{\sqrt{(3)^2+(-2)^2}}=\frac{8}{\sqrt{13}} $

$ \begin{array}{lrl} \Rightarrow & \left|k_1-5\right|=8 & \\ \Rightarrow & k_1-5= \pm 8 & \\ \Rightarrow & k_1=13,-3 & \\ \Rightarrow & k_1=13 & \left(\because k_1>0\right) \end{array} $

Similarly, $\frac{\left|k_2-5\right|}{\sqrt{13}}=\frac{6}{\sqrt{13}}$

$ \begin{aligned} \Rightarrow & & k_2-5 & = \pm 6 \\ \Rightarrow & & k_2 & =11,-1 \\ \Rightarrow & & k_2 & =11 \,\,\,\,\,\,\,\,\,\,\,\left(\because k_2>0\right)\end{aligned} $

Now, $L_1=(3 x-2 y+13)=0$ and $L_2=(3 x-2 y+11)=0$

Therefore, the combine equation of $L_1=0$ and $L_2=0$ is

$ \begin{aligned} (3 x-2 y+13)(3 x-2 y+11) & =0 \\ \Rightarrow(3 x-2 y)^2+24(3 x-2 y)+143 & =0 \end{aligned} $

$ \begin{aligned} & \frac{h-3}{2}=\frac{k+4}{-3}=\frac{-2(2 \times 3-3 \times-4-5)}{4+9} \\ & \Rightarrow \frac{h-3}{2}=\frac{k+4}{-3}=\frac{-2(6+12-5)}{13} \\ & \Rightarrow \frac{h-3}{2}=\frac{k+4}{-3}=-2 \\ & \Rightarrow h-3=-4 \text { and } k+4=6 \\ & \Rightarrow h=-1 \text { and } k=2 \end{aligned} $

Also, $(l, m)$ is the foot of the perpendicular from $(h, k)$

$ \begin{aligned} & \therefore \frac{l-h}{3}=\frac{m-k}{2}=\frac{-(3 \times h+2 \times k+12)}{9+4} \\ & \Rightarrow \frac{l+1}{3}=\frac{m-2}{2}=-\frac{(-3+4+12)}{13} \\ & \Rightarrow \frac{l+1}{3}=\frac{m-2}{2}=-1 \\ & \Rightarrow l+1=-3 \text { and } m-2=-2 \\ & \Rightarrow l=-4 \text { and } m=0 \\ & \therefore l h+m k+1 \\ & \quad=(-4)(-1)+(0)(2)+1=4+0+1=5 \end{aligned} $

Equation of line, parallel to the line $y=\sqrt{3} x$ is $y=\sqrt{3} x+k$, it passes through $(2,3)$.

$ \begin{aligned} & \text { So, } 3=2 \sqrt{3}+k \Rightarrow k=3-2 \sqrt{3} \\ & \text { Thent, } y=\sqrt{3} x+3-2 \sqrt{3} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \text { and } 2 x+4 y-27=0 \\ & \Rightarrow \quad 4 y=-2 x+27 \\ & \Rightarrow \quad y=\frac{-2 x+27}{4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{aligned} $

By using Eqs. (i) and (ii), we get

$ \begin{aligned} & \sqrt{3} x+3-2 \sqrt{3}=\frac{-2 x+27}{4} \\ \Rightarrow & 4 \sqrt{3} x+12-8 \sqrt{3}=-2 x+27 \\ \Rightarrow & x(4 \sqrt{3}+2)=15+8 \sqrt{3} \\ \Rightarrow & x=\frac{15+8 \sqrt{3}}{4 \sqrt{3}+2} \end{aligned} $

So, $y=\frac{\sqrt{3}(15+8 \sqrt{3})}{(4 \sqrt{3}+2)}+3-2 \sqrt{3}$

[From Eq. (i)]

$ \begin{aligned} & =\frac{15 \sqrt{3}+24+(3-2 \sqrt{3})(4 \sqrt{3}+2)}{(4 \sqrt{3}+2)} \\ & =\frac{15 \sqrt{3}+24+12 \sqrt{3}+6-24-4 \sqrt{3}}{(4 \sqrt{3}+2)} \\ & =\frac{23 \sqrt{3}+6}{4 \sqrt{3}+2} \end{aligned} $

So, $P=\left(\frac{15+8 \sqrt{3}}{4 \sqrt{3}+2}, \frac{23 \sqrt{3}+6}{4 \sqrt{3}+2}\right)$

Then,

$ \begin{aligned} P Q & =\sqrt{\left(2-\frac{15+8 \sqrt{3}}{4 \sqrt{3}+2}\right)^2+\left(3-\frac{23 \sqrt{3}+6}{4 \sqrt{3}+2}\right)^2} \\ & =\sqrt{\left(\frac{4-15}{4 \sqrt{3}+2}\right)^2+\left(\frac{-11 \sqrt{3}}{4 \sqrt{3}+2}\right)^2} \\ & =\sqrt{\frac{121+363}{(4 \sqrt{3}+2)^2}}=\frac{22}{(4 \sqrt{3}+2} \\ & =\frac{11}{2 \sqrt{3}+1}=\frac{11(2 \sqrt{3}-1)}{(2 \sqrt{3}+1)(2 \sqrt{3}-1)} \\ & =(2 \sqrt{3}-1) \end{aligned} $

$ \begin{aligned} & \text { Area of } \triangle A O B=\frac{1}{2}(O A) \cdot(O B) \\ & \Rightarrow \frac{1}{2}\left(\frac{k}{a}\right) \cdot\left(\frac{k}{2}\right)=3 \\ & \Rightarrow \quad k^2=12 a \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \text { Slope of } a x+2 y=k \text { is } \\ & \qquad \begin{aligned} m_1 & =-a / 2 \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { and slope of } 2 x-3 y+7=0 \text { is }\\ &\begin{array}{lrl} \because & m_2 & =2 / 3 \\ \Rightarrow & m_1 \cdot m_2 & =-1 \\ \Rightarrow & \frac{-2 a}{6} & =-1 \\ \Rightarrow & a & =3 \end{array} \end{aligned} $

From Eq. (i), we get

$ \begin{aligned} & & k^2 & =12 \times 3 \\ \Rightarrow & & k & = \pm 6 \Rightarrow k=+6,-6 \end{aligned} $

So, product of all possible values of

$ k=6 \times(-6)=-36 $

Given sides of triangle are

$ \begin{array}{r} x+y+10=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \\ x-y-2=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ 2 x+y-7=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{array} $

Vertex of these two sides,

$ x+y+10=0 \text { and } x-y-2=0 $

On solving, we get

[using Eqs. (i) and (ii)]

$ x=-4, y=-6 \Rightarrow \text { vertex }=(-4,-6) $

For altitude from $(-4,-6)$ on

$ \begin{aligned} & 2 x+y-7=0 \\ & \text { Slope }=\frac{-1}{(-2)}=\frac{1}{2} \\ & \therefore y=m x+c \Rightarrow y=\frac{x}{2}+c \\ & \Rightarrow-6=-2+c \Rightarrow c=-4 \\ & \Rightarrow y=\frac{x}{2}-4 \\ & \Rightarrow 2 y=x-8 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\end{aligned} $

Similarly,

Vertex of these two sides,

$ x-y-2=0 \text { and } 2 x+y-7=0 $

On solving $x=3, y=1 \Rightarrow$ Vertex $=(3,1)$

Altitude on $x+y+10=0$

$ \Rightarrow y=-x-10 $

Slope of altitude $=1$

$ \begin{gathered} y=1 \cdot x+c \Rightarrow 1=3+c \Rightarrow c=-2 \\ y=x-2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(v) \end{gathered} $

From Eqs. (iv) and (v), we get

$ x=-4, y=-6 $

Hence, $(-4,-6)$ is orthocenter of the triangle formed by the lines $x+y+10=0, x-y-2=0$ and $2 x+y-7=0$.

$l \in R,(2 l-3) x^2+2 l x y-y^2=0$.

As, we know that equation $a x^2+2 h x y+b y^2=0$ pair of real and distinct lines, if $h^2>a b$

From Eq. (i), we get

$ \begin{aligned} h=l, a & =2 l-3, b=-1 \\ h^2 & >a b \\ l^2 & >(2 l-3)(-1) \\ l^2 & >-2 l+3 \end{aligned} $

$ \begin{aligned} & \qquad l^2+2 l-3>0 \Rightarrow l^2+3 l-l-3>0 \\ & (l+3)(l-1)>0 \Rightarrow l \in(-\infty,-3) \cup(1, \infty) \\ & \text { or } l \in R-[-3,1] \end{aligned} $

On solving equations $4 x+3 y+k=0$ and $3 x+4 y+k=0$, we get

$ x=-\frac{k}{7} \text { and } y=-\frac{k}{7} $

Given, $P(a, 2)$ is the point of intersection of $4 x+3 y+k=0$ and $3 x+4 y+k=0$

$ \begin{aligned} \therefore & a=-\frac{k}{7} \text { and }-\frac{k}{7}=2 \\ \Rightarrow & k=-14 \end{aligned} $

i.e., $\quad a=2$

Point $P(2,2)$ and $Q(1, b)$ lies on either side of the line $2 x-3 y+1=0$

$ \begin{array}{lr} \therefore(2 \times 2-3 \times 2+1) & (2 \times 1-3 \times b+1)<0 \\ \Rightarrow & -(2-3 b+1)<0 \\ \Rightarrow & 3-3 b>0 \\ \Rightarrow & b<1 \\ \therefore (-\infty, 1) \end{array} $

Given, $x-2 y+3=0$

$ \begin{aligned}and\,\, k x-y+2 & =0 \\ x-2 y+3 & =0 \\ y & =\frac{x}{2}+\frac{3}{2} \end{aligned} $

[comparing with $y=m x+c$ ]

$ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,m_1=1 / 2 $

$ \begin{aligned}and\,\, k x-y+2 & =0 \\ y & =k x+2 \end{aligned} $

[comparing with $y=m x+c$ ]

$ \begin{aligned} m_2=k & \\ \text { Now, } \tan \theta & =\left|\frac{m_1-m_2}{1+m_1 m_2}\right| \\ \tan 45^{\circ} & =\left|\frac{\frac{1}{2}-k}{1+\frac{1}{2} \times k}\right| \quad\left[\text { given } \theta=45^{\circ}\right] \\ \pm 1 & =\frac{\frac{1}{2}-k}{1+\frac{1 k}{2}} \\ \Rightarrow \quad \pm 1 & =\frac{1-2 k}{2+k} \end{aligned} $

Either, $\quad 1=\frac{1-2 k}{2+k} \Rightarrow 2+k=1-2 k$

$ \begin{aligned} & \Rightarrow \quad 3 k=-1 \\ & \Rightarrow \quad k=-\frac{1}{3} \text { or }-1=\frac{1-2 k}{2+k} \\ & \Rightarrow \quad-2-k=1-2 k \Rightarrow k=3 \end{aligned} $

Let $k_1=3$ and $k_2=-\frac{1}{3}$

Now, $k_1-2=3-2=1=-3 \times\left(-\frac{1}{3}\right)=-3 k_2$

Given, lines $4 x+3 y-k=0$,

$ 2 x+y+3=0 $

and $3 x+2 y+k=0$ are concurrent.

We know that three lines

$ \begin{aligned} & a_1 x+b_1 y+c_1=0 \\ & a_2 x+b_2 y+c_2=0 \text { and } a_3 x+b_3 y+c_3=0 \end{aligned} $

are said to be concurrent, if

$ \begin{aligned} \left|\begin{array}{lll} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array}\right| \\ \therefore \quad\left|\begin{array}{ccc} 4 & 3 & -k \\ 2 & 1 & 3 \\ 3 & 2 & k \end{array}\right| & =0 \end{aligned} $

$ \begin{aligned} & 4(k-6)-3(2 k-9)-k(4-3)=0 \\ & 4 k-24-6 k+27-4 k+3 k=0 \\ & \Rightarrow \quad-3 k+3=0 \\ & \Rightarrow \quad \begin{aligned} & k=1 \end{aligned} \end{aligned} $

On putting the value of $k$ in the given equation, we get

$ \begin{aligned} & 4 x+3 y-1=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \\ & 2 x+y+3=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \\ & 3 x+2 y+1=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii) \end{aligned} $

On solving Eqs. (i) and (ii), we get

$ x=-5 \text { and } y=7 $

Now, perpendicular distance of $(-5,7)$.

From $3 x+4 y+2=0$, is $\left|\frac{-5 \times 3+4(7)+2}{5}\right|$

$ =\left|\frac{15}{5}\right|=3 \text { units } $

Given, $A(1,3), B(2,5), C(h, k)$, $B C \perp A C$ and $\angle C A B=\angle C B A$

So,

Now, $A C=B C$

$ \begin{aligned} & \Rightarrow \sqrt{(h-1)^2+(k-3)^2}=\sqrt{(h-2)^2+(k-5)^2} \\ & \Rightarrow \quad 2 h+4 k-19=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

and $B C \perp A C$

$\Rightarrow \quad($ slope of $B C) \times($ slope of $A C)=-1$

$ \Rightarrow \quad \frac{k-5}{h-2} \times \frac{k-3}{h-1}=-1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

On solving Eqs. (i) and (ii), we get

$ h=\frac{1}{2}, k=\frac{9}{2} \text { and } h=\frac{5}{2}, k=\frac{7}{2} $

Thus, $h=\frac{1}{2}$ or $\frac{5}{2}$

$ \begin{aligned} &\text { Equation of pair of line } O A \text { and } O B\\ &\begin{array}{r} \left.x^2+2 x y+5 y^2+2 x(2 x-3 y)+3 y\right)(2 x-3 y) -(2 x-3 y)^2=0 \end{array} \end{aligned} $

(Homoginise Eq. (i) with the help of Eq. (ii)),

$ \begin{aligned} x^2+14 x y-13 y^2=0 \\ a=1, h=7 \text { and } b=-13 \\ \tan \theta=\frac{2 \sqrt{h^2-a b}}{|a+b|}=2 \frac{\sqrt{49+13}}{|1-13|}=\frac{\sqrt{62}}{6} \\ \therefore \cos \theta=\frac{6}{\sqrt{62+36}}=\frac{6}{\sqrt{98}}=\frac{6}{7 \sqrt{2}}=\frac{3 \sqrt{2}}{7} \end{aligned} $

Length of perpendicular from origin to the line

$ \begin{gathered} 3 x-4 y+5=0 \text { is } \\ \frac{5}{\sqrt{25}}=1 \end{gathered} $

If $\alpha$ is the angle made by perpendicular, then equation of line in normal form

$ \begin{array}{r} x \cos \alpha+y \sin \alpha=P \\ \text { or } \quad x \cos \alpha+y \sin \alpha=1 \,\,...(i)\end{array} $

On comparing Eq. (i) with

$ \begin{aligned} 3 x-4 y+5 & =0 \\ \text { or } \quad x\left(-\frac{3}{5}\right)+y\left(\frac{4}{5}\right) & =1 \end{aligned} $

We have,

$ \begin{gathered} \cos \alpha=-\frac{3}{5}, \sin \alpha=\frac{4}{5} \\ \tan \alpha=-\frac{4}{3} \end{gathered} $

Equation of line passing through $(1,-1)$ and slope $\left(-\frac{4}{3}\right)$ is

$ \begin{aligned} y+1 & =-\frac{4}{3}(x-1) \\ 3 y+3 & =-4 x+4 \\ 4 x+3 y & =1 \end{aligned} $

On comparing with $a x+b y=1$, we have

$ \begin{aligned} & a=4, b=3 \\ & a+a b+b=4+12+3=19 \end{aligned} $

Equation of $L_1$ perpendicular to

$ \begin{aligned} & 3 x-4 y+k=0 \\ & \text { is } 4 x+3 y+\lambda=0 \end{aligned} $

It passes through $P(4,-3)$

$ \begin{array}{rrl} \therefore & 4 \cdot 4+3(-3)+\lambda=0 \\ \Rightarrow & \lambda=-7 \\ \therefore & L_1: 4 x+3 y-7=0 ...(i)\end{array} $

and given line $\quad 5 x-3 y-2=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Solving Eqs. (i) and (ii), we get

$ x=1, y=1 $

Point of intersection is $(1,1)$

Distance between $(1,1)$ and $(4,-3)$ is

$ \sqrt{(1-4)^2+(1+3)^2}=\sqrt{9+16}=5 $

$L_1$ is passing through intersection of $2 x+3 y-5=0$ and $4 x-5 y+7=0$, then equation of $L_1$ is given by

$ \begin{aligned} & (2 x+3 y-5)+\lambda(4 x-5 y+7)=0 \\ & \Rightarrow(2+4 \lambda) x+(3-5 \lambda) y+(7 \lambda-5)=0 \end{aligned} $

This line divides the line segment joining the points $(2,3)$ and $(1,-1)$ in the ratio $2: 1$. Then, by section formula

$ \begin{gathered} x=\frac{2 \times 1+1 \times 2}{2+1}=\frac{4}{3} \\ y=\frac{2 \times(-1)+1 \times 3}{2+1}=\frac{1}{3} \\ (x, y)=\left(\frac{4}{3}, \frac{1}{3}\right) \text { lies on Eq. (i) } \\ \therefore \quad(2+4 \lambda) \frac{4}{3}+(3-5 \lambda) \frac{1}{3}+(7 \lambda-5)=0 \\ \Rightarrow \quad 8+16 \lambda+3-5 \lambda+21 \lambda-15=0 \\ \Rightarrow \quad 32 \lambda-4=0 \Rightarrow \lambda=\frac{1}{8} \end{gathered} $

From Eq. (i), we get

$ \begin{aligned} & \left(2+4 \times \frac{1}{8}\right) x+\left(3-5 \times \frac{1}{8}\right) y+\frac{7 \times 1}{8}-5=0 \\ & 20 x+19 y+(-33)=0 \\ & \Rightarrow \quad \frac{20}{33}+\frac{19}{33} y=1 \end{aligned} $

On comparing with $a x+b y=1$

$ \begin{aligned} & a=\frac{20}{33} \text { and } b =\frac{19}{33} \\ \therefore & 33(a-b) =33\left(\frac{20}{33}-\frac{19}{33}\right)=1 \end{aligned} $

Let $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$ be the coordinates of vertex $B$ and $C$ respectively.

Then, coordinate of $D$

$ \left(\frac{x_1+1}{2}, \frac{y_1+2}{2}\right) $

Also, $x-3 y-5=0$ is perpendicular to $A B$

$ \begin{array}{ll} \Rightarrow & \frac{1}{3}\left(\frac{y_1-2}{x_1-1}\right)=-1 \\ \Rightarrow & 3 x_1+y_1-5=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{array} $

and $D$ lies on the line $x-3 y-5=0$

$ \begin{array}{ll} \therefore & \frac{x_1+1}{2}-3\left(\frac{y_1+2}{2}\right)-5=0 \\ \Rightarrow & x_1-3 y_1-15=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} $

On solving Eqs. (i) and (ii), we get

$ \begin{aligned} & x_1=3 \text { and } y_1=-4 \\ & B(3,-4) \end{aligned} $

Now, coordinate of $E\left(\frac{x_2+3}{2}, \frac{y_2-4}{2}\right)$.

$E$ lies on line $x+5 y-9=0$

$ \begin{aligned} \frac{x_2+3}{2}+5\left(\frac{y_2-4}{2}\right)-9 & =0 \\ \Rightarrow \quad x_2+5 y_2-35 & =0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} $

Also, $x+5 y-9=0$ is perpendicular to $B C$

$ \begin{aligned} & \left(-\frac{1}{5}\right)\left(\frac{y_2+4}{x_2-3}\right)=-1 \\ & 5 x_2-y_2-19=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\end{aligned} $

Solving Eqs. (iii) and (iv), we get

$ \begin{aligned} & x_2=5, y_2=6 \\ & C(5,6) \end{aligned} $

Length of $A C=\sqrt{(5-1)^2+(6-2)^2}$

$ =\sqrt{4^2+4^2}=4 \sqrt{2} $

$ A(4,3,5), B(1,-2,1) \text { and } C(3,2,1) $

$A D$ is angle bisector of $A$

$ \Rightarrow \quad \frac{B D}{D C}=\frac{A B}{A C}=\frac{\sqrt{3^2+5^2+4^2}}{\sqrt{1^2+1^2+4^2}}=\frac{5}{3} $

$D$ divides $B C$ in the ratio $5: 3$

$ \begin{gathered} D\left(\frac{15+3}{8}, \frac{10-6}{8}, \frac{5+3}{8}\right)=D\left(\frac{18}{8}, \frac{4}{8}, 1\right) \\ C D=\sqrt{\left(\frac{18}{8}-3\right)^2+\left(\frac{4}{8}-2\right)^2+(1-1)^2} \\ \quad=\sqrt{\frac{36}{64}+\frac{144}{64}}=\frac{\sqrt{180}}{8}=\frac{6 \sqrt{5}}{8}=\frac{3 \sqrt{5}}{4} \end{gathered} $

Let us suppose that the axis are rotated in the anti-clockwise direction through an angle $\alpha$. The equation of the line $L$ with respect to the old axis is given by $\frac{x}{a}+\frac{y}{b}=1$.

Now, to find the equation of $L$ with respect to the new axes, we replace ' $x$ by $x \cos \alpha-y \sin \alpha$ and $y$ by $x \sin \alpha+y \cos \alpha$ so that the equation of ' $L$ ' with respect to the new axis is

$ \frac{1}{a}(x \cos \alpha-y \sin \alpha) $

$ +\frac{1}{b}(x \sin \alpha+y \cos \alpha)=1 $

Since, the line ' $L$ ' intercepts $p$ and $q$ on the new axes. So, we have on putting ( $P, 0$ ) and then ( $0, q$ )

So, $\frac{1}{p}=\frac{1}{a} \cos \alpha+\frac{1}{b} \sin \alpha$ and $\frac{1}{q}=-\frac{1}{a} \sin \alpha+\frac{1}{b} \cos \alpha$

Now, eliminating ' $\alpha$ ', we get

$ \frac{1}{p^2}+\frac{1}{q^2}=\frac{1}{a^2}+\frac{1}{b^2} $

Given point $ P(1,2) $, we need to find the coordinates of points on lines $ L_1 $ and $ L_2 $ which intersect the line $ x+y=4 $ at a distance of $\frac{\sqrt{6}}{3}$ units from $ P $.

Using the parametric formula for points at a given distance, the coordinates are:

$ \left(1+\sqrt{\frac{2}{3}} \cos \theta, 2+\sqrt{\frac{2}{3}} \sin \theta\right) $

This point must satisfy the equation of the line $ x+y=4 $:

$ \begin{align*} \Rightarrow & \left(1+\sqrt{\frac{2}{3}} \cos \theta\right) + \left(2+\sqrt{\frac{2}{3}} \sin \theta\right) = 4 \\ \Rightarrow & 3 + \sqrt{\frac{2}{3}} (\cos \theta + \sin \theta) = 4 \\ \Rightarrow & \sqrt{\frac{2}{3}} (\cos \theta + \sin \theta) = 1 \end{align*} $

Squaring both sides:

$ \begin{align*} & \frac{2}{3} (1 + \sin 2\theta) = 1 \\ \Rightarrow & 1 + \sin 2\theta = \frac{3}{2} \\ \Rightarrow & \sin 2\theta = \frac{1}{2} \end{align*} $

The solutions to $\sin 2\theta = \frac{1}{2}$ are:

$ \theta = \frac{1}{2} \sin^{-1}\left(\frac{1}{2}\right) $

Which gives the angles:

$ \theta = \frac{\pi}{12} \quad \text{and} \quad \frac{5\pi}{12} $

Thus, the angles made by lines $ L_1 $ and $ L_2 $ with the positive $ X $-axis are $\frac{\pi}{12}$ and $\frac{5\pi}{12}$.

Let ${ }^{\prime} L^{\prime}=2 x+3 y=6$

Any line through the origin making an angle of $45^{\circ}$ with the given line $2 x+3 y=6$ is of the form $y=m x$, where

$ \begin{aligned} & \tan \left( \pm 45^{\circ}\right)=\frac{m-\left(-\frac{2}{3}\right)}{1+m\left(-\frac{2}{3}\right)} \\ & \Rightarrow \pm 1=\frac{3 m+2}{(3-2 m)} \Rightarrow(3 m+2)= \pm(3-2 m) \end{aligned} $

Either, $3 m+2=3-2 m$

$ \Rightarrow \quad m=\frac{1}{5} $

$ \text { Or, } \quad 3 m+2=-(3-2 m) $

$ \Rightarrow \quad m=-5 $

Hence, the other two sides are $y=\frac{1}{5} x$ and $y=-5 x$

i.e. $x-5 y=0$ and $5 x+y=0$

Let ' $P$ ' be the perpendicular from $A(0,0)$ to the base $2 x+3 y=6$

So, $P=\left|\frac{2 \cdot 0+3 \cdot 0-6}{\sqrt{4+9}}\right|=\frac{6}{\sqrt{13}}$

Now, the required area is

$ \begin{aligned} & =\frac{1}{2} \times B C \times A D \\ & =\frac{1}{2} \times 2 P \times P=P^2=\frac{36}{13} \text { sq unit } \end{aligned} $

The given line is, $\sqrt{3} x+y=1$

So, slope of the line is $m_1=-\sqrt{3}$

Let, the slope of the line $L$ which passing through the point $(3,2)$ is $m_2$.

Now,

$ \begin{aligned} & \tan 60^{\circ}=\left|\frac{m_2+\sqrt{3}}{1-\sqrt{3} m_2}\right|= \pm \sqrt{3} \\ & \Rightarrow\left(m_2+\sqrt{3}\right)= \pm \sqrt{3}\left(1-\sqrt{3} m_2\right) \\ & \Rightarrow m_2=0 \text { and } m_2=\sqrt{3} \end{aligned} $

Now, equation of the line $L$ passing through $(3,2)$ is

$ \begin{aligned} & (y-2)=\sqrt{3}(x-3) \\ \Rightarrow & y-2=\sqrt{3} x-3 \sqrt{3} \\ \Rightarrow & \sqrt{3} x-y+(2-3 \sqrt{3})=0 \end{aligned} $

$ \begin{aligned} &\text { Given lines are }\\ &\begin{aligned} & & \sqrt{3} x+y-6 & =0 \\ \Rightarrow & & y & =-\sqrt{3} x+6 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \text { and } & \sqrt{3} x+y+9 & =0 \\ \Rightarrow & & y & =-\sqrt{3} x-9 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} \end{aligned} $

$\because$ Slope of both lines are same.

$\therefore$ These line are parallel.

So, perpendicular distance between both lines.

$ =\frac{\left|c_1-c_2\right|}{\sqrt{1+m^2}}=\frac{|6-(-9)|}{\sqrt{1+(-\sqrt{3})^2}}=\frac{15}{2} $

Then, equilateral triangle will be

$ A D=\frac{15}{2} \text { and } A B=B C=A C=a $

Then, $D C=\frac{B C}{2}=\frac{a}{2}$

In $\triangle A D C, \tan 60^{\circ}=\frac{A D}{D C}=\frac{(15 / 2)}{(a / 2)}$

$ \Rightarrow \quad a=\frac{15}{\sqrt{3}}=5 \sqrt{3} $

Therefore, Area ( $\triangle A B C$ )

$ =\frac{\sqrt{3}}{4} a^2=\frac{\sqrt{3}}{4}(5 \sqrt{3})^2=\frac{75 \sqrt{3}}{4}(\text { units })^2 $

Equation of the line

$ x+y+2=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

and the equation of the curve is

$ x^2+x y+y^2+x+3 y+1=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Now, from Eq. (i), we get $x+y+2=0$

$ \begin{aligned} & \Rightarrow x+y=-2 \\ & \Rightarrow \quad 1=\frac{-x-y}{2} \end{aligned} $

From Eq. (ii), we get

$ \begin{aligned} & \begin{array}{r} x^2+x y+y^2+x+3 y+1=0 \\ \Rightarrow x^2+x y+y^2+x(1)+3 y(1)+(1)^2=0 \\ \begin{aligned} \Rightarrow x^2+x y+y^2+x\left(\frac{-x-y}{2}\right)+3 y\left(\frac{-x-y}{2}\right) \\ +\left(\frac{-x-y}{2}\right)^2=0 \end{aligned} \\ \begin{array}{r} \Rightarrow x^2+x y+y^2+\left(\frac{-x^2-x y}{2}\right)+\left(\frac{-3 x y-3 y^2}{2}\right) \\ +\left(\frac{x^2+y^2+2 x y}{4}\right)=0 \end{array} \\ \begin{array}{r} \Rightarrow 4 x^2+4 x y+4 y^2-2 x^2-2 x y-6 x y \\ -6 y^2+x^2+y^2+2 x y=0 \end{array} \\ \begin{array}{r} \Rightarrow 3 x^2-y^2-2 x y=0 \\ \Rightarrow 3 x^2-2 x y-y^2=0 \end{array} \end{array} \end{aligned} $

Now, comparing this equation with $a x^2+2 h x y+b y^2=0$

Thus, $a=3, h=-1, b=-1$

Now, $\theta$ be the acute angle (given)

So, $\cos \theta=\frac{|a+b|}{\sqrt{(a-b)^2+4 h^2}}$

$ =\frac{|3-1|}{\sqrt{(4)^2+4(-1)^2}}=\frac{2}{\sqrt{20}}=\frac{1}{\sqrt{5}} $

Let axes be rotated through an angle $\theta$, then old coordinates are

$ \begin{aligned} & x=x \cos \theta-y \sin \theta \\ & y=x \sin \theta+y \cos \theta \\ & \begin{array}{l} \therefore \sqrt{3}(x \cos \theta-y \sin \theta)^2 \\ \quad+(\sqrt{3}-1)(x \cos \theta-y \sin \theta) \end{array} \\ & \quad \begin{array}{l} (x \sin \theta+y \cos \theta)-(x \sin \theta+y \cos \theta)^2 \end{array} \\ & \text { Coefficient of } x y=0 \\ & \Rightarrow-2 \sqrt{3} \cos \theta \cdot \sin \theta+(\sqrt{3}-1) \\ & \quad\left(\cos ^2 \theta-\sin ^2 \theta\right)-2 \sin \theta \cdot \cos \theta=0 \\ & \Rightarrow-\sqrt{3} \sin 2 \theta+(\sqrt{3}-1) \cos 2 \theta-\sin 2 \theta=0 \end{aligned} $

$\begin{aligned} & \Rightarrow(\sqrt{3}-1) \cos 2 \theta-(\sqrt{3}+1) \sin 2 \theta=0 \\ & \Rightarrow \tan 2 \theta=\frac{\sqrt{3}-1}{\sqrt{3}+1}=2-\sqrt{3} \\ & \Rightarrow 2 \theta=15^{\circ} \Rightarrow \theta=7.5^{\circ}\end{aligned}$

To find the coordinates of the points that are 5 units away from $A(3, 2)$ on a line with a slope of $ \frac{3}{4} $, we use the angle θ where $ \tan \theta = \frac{3}{4} $.

First, we find the trigonometric values:

$\cos \theta = \frac{4}{5}$

$\sin \theta = \frac{3}{5}$

Using the distance formula along the direction of the line, we set up the equations as follows:

$ \frac{x - x_1}{\cos \theta} = \frac{y - y_1}{\sin \theta} = \pm r $

Substitute the known values:

$\frac{x - 3}{\frac{4}{5}} = \pm 5$

$\frac{y - 2}{\frac{3}{5}} = \pm 5$

Solve for $x$ and $y$:

$x - 3 = \pm 4$, so $x = 3 \pm 4$.

$y - 2 = \pm 3$, so $y = 2 \pm 3$.

This results in the coordinates:

$A_1 = (3 + 4, 2 + 3) = (7, 5)$

$A_2 = (3 - 4, 2 - 3) = (-1, -1)$

So, the points on the line that are 5 units away from $A(3, 2)$ are $(7, 5)$ and $(-1, -1)$.

$ \text { Equation of } A B $

$ y+1=\frac{-3+1}{5-2}(x-2) $

$ \begin{aligned} & \Rightarrow \quad y+1=-\frac{2}{3}(x-2) \\ & \Rightarrow \quad 3 y+3=-2 x+4 \\ & \Rightarrow \quad 2 x+3 y=1 \\ & \because(a, 4) \text { and }(-2, b) \text { lies on } A B \\ & \Rightarrow 2 a+12=1 \text { and }-4+3 b=1 \\ & \qquad a=-\frac{11}{2}, b=\frac{5}{3} \\ & \text { (a,b) i.e. }\left(-\frac{11}{2}, \frac{5}{3}\right) \text { lies on } 2 x+6 y+1=0 \end{aligned} $

$ \begin{aligned} & \text { Let } C \equiv(0, h) \\ & \text { So, } D \equiv(0, h+4) \end{aligned} $

Equation of $A C$ in slope intercept form,

$ \begin{array}{ll} & \frac{x}{-4}+\frac{y}{h}=1 \\ \Rightarrow & \frac{y}{h}=\frac{x+4}{4} \\ \Rightarrow & h=\frac{4 y}{x+4} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{array} $

Similarly, equation of $B D$,

$ \begin{aligned} & & \frac{x}{4}+\frac{y}{h+4} & =1 \\ \Rightarrow & & \frac{4-x}{4} & =\frac{y}{h+4} \\ \Rightarrow & & h+4 & =\frac{4 y}{4-x} \\ & \Rightarrow & h & =\frac{4 y}{4-x}-4 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

$ \begin{aligned} &\text { On comparing Eqs. (i) and (ii), we get }\\ &\frac{4 y}{x+4}=\frac{4 y}{4-x}-4 \end{aligned} $

$ \begin{aligned} & \Rightarrow 4 y(4+x)=4 y(4-x)+4(4+x)(4-x) \\ & \Rightarrow x^2+2 x y-16=0 \end{aligned} $

Given,

$ 4 x^2+12 x y+9 y^2+6 x+9 y+2=0 $

Since, axes are rotating an angle of $30^{\circ}$ in anticlockwise direction.

$ \begin{aligned} \therefore \quad x & =X \cos 30^{\circ}-Y \sin 30^{\circ} \\ & =\frac{\sqrt{3} X-Y}{2} \end{aligned} $

$ \begin{aligned}and\,\, y & =\frac{x \sin 30^{\circ}+y \cos 30^{\circ}}{2} \\ & =\frac{X+\sqrt{3} Y}{2} \end{aligned} $

Now, Required Equation

$ \begin{aligned} & 4\left(\frac{\sqrt{3} X-Y}{2}\right)^2+12\left(\frac{\sqrt{3} X-Y}{2}\right) \\ & \left(\frac{X+\sqrt{3} Y}{2}\right)+9\left(\frac{X+\sqrt{3} Y}{2}\right)^2 +6\left(\frac{\sqrt{3} X-Y}{2}\right)+9\left(\frac{X+\sqrt{3} Y}{2}\right)+2=0 \end{aligned} $

$ \begin{aligned} & \Rightarrow 3 X^2+Y^2-2 \sqrt{3} X Y+3 \sqrt{3} X^2+9 X Y \\ & -3 X Y-3 \sqrt{3} Y^2+\frac{9}{4} X^2+\frac{27}{4} Y^2 \\ & +\frac{9 \sqrt{3} X Y}{2}+3 \sqrt{3} X-3 Y \\ & +\frac{9 X}{2}+\frac{9 \sqrt{3} Y}{2}+2=0 \\ & \Rightarrow 12 X^2+4 Y^2-8 \sqrt{3} X Y+12 \sqrt{3} X^2 \\ & +36 X Y-12 X Y-12 \sqrt{3} Y^2+9 X^2 \\ & +27 Y^2+18 \sqrt{3} X Y+12 \sqrt{3} X-12 Y \\ & +18 X+18 \sqrt{3} Y+8=0 \\ & \Rightarrow X^2(21+12 \sqrt{3})+Y^2(31-12 \sqrt{3} \\ & +X Y(10 \sqrt{3}+36)+X(12 \sqrt{3}+18) \\ & +Y(18 \sqrt{3}-12)+8=0 \end{aligned} $

$ \begin{aligned} \Rightarrow \quad a & =21+12 \sqrt{3}, b=31-12 \sqrt{3} . \\ h & =5 \sqrt{3}+18, g=6 \sqrt{3}+9 \\ f & =3 \sqrt{3}-6 \end{aligned} $

$ \therefore \quad \frac{g}{f}=\frac{6 \sqrt{3}+9}{9 \sqrt{3}-6}=\frac{2 \sqrt{3}+3}{3 \sqrt{3}-2} $

$ \begin{aligned} &\text { Given, third vertex be }\left(x_1, y_1\right) \text {. }\\ &\therefore \quad A C=A B \end{aligned} $

$ \begin{array}{ll} \Rightarrow & \quad x_1^2+\left(y_1-a\right)^2=y_1^2 \\ \Rightarrow & \quad(2 a)^2+y_1^2+a^2-2 a y_1=y_1^2 \\ & \quad\left[\because x_1=2 a\right] \\ \Rightarrow & \quad 5 a^2+y_1^2-2 a y_1=y_1^2 \\ \Rightarrow & \quad 5 a^2-2 a y_1=0 \\ \Rightarrow & y_1=\frac{5 a}{2} \text { and } x_1=2 a \\ \therefore& \quad x_1+y_1=2 a+\frac{5 a}{2}=\frac{9 a}{2} \end{array} $

$ \begin{aligned} & L_1=x-2 y+3=0 \\ & L_2=2 x+y+1=0 \\ & L_3=3 x+y+c=0 \end{aligned} $

∴ All three lines are concurrent.

$ \therefore \quad\left|\begin{array}{ccc} 1 & -2 & 3 \\ 2 & 1 & 1 \\ 3 & 1 & c \end{array}\right|=0 $

$ \begin{array}{lrl} \Rightarrow & 1(c-1)+2(2 c-3)+3(2-3) & =0 \\ \Rightarrow & c-1+4 c-6-3 & =0 \\ \Rightarrow & 5 c =10\\ \Rightarrow & c=2 \end{array} $

$ \begin{aligned} & L_1 \Rightarrow y=\frac{1}{2} x+\frac{3}{2} \\ & L_3 \Rightarrow y=-3 x-c \Rightarrow y=-3 x-2 \\ & m_1=\frac{1}{2}, m_2=-3 \\ & \tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right| \\ & \tan \theta=\left|\frac{\frac{1}{2}-(-3)}{1+\frac{1}{2}(-3)}\right|=\left|\frac{\frac{7}{2}}{\frac{-1}{2}}\right|=7 \\ & \tan \theta=2+5=c+5 \end{aligned} $

$(a, b)$ is equidistance from

$ \begin{aligned} & 2 x+3 y+4=0 \text { and } 3 x-2 y+4=0 \\ & \text { So, } \quad \frac{|2 a+3 b+4|}{\sqrt{2^2+3^2}}=\frac{|3 a-2 b+4|}{\sqrt{2^2+3^2}} \\ & \Rightarrow \quad 2 a+3 b+4= \pm(3 a-2 b+4) \end{aligned} $

For ( + )ve sign.

$ \begin{aligned} & & 2 a+3 b+4 & =3 a-2 b+4 \\ \Rightarrow & & -a+5 b & =0 \Rightarrow a=5 b \\ \Rightarrow & & x-5 y & =0 \end{aligned} $

For ( - ) ve sign,

$ \begin{aligned} & & 2 a+3 b+4 & =-(3 a-2 b+4) \\ \Rightarrow & & 5 a+b+8 & =0 \\ \Rightarrow & & 5 x+y+8 & =0 \end{aligned} $

The points $\left(-\frac{4}{3}, \frac{1}{3}\right)$ is satisfy all three lines Hence, the point concurrence of the lines.

$ \begin{aligned} &\\ &\begin{aligned} & \text { For } a x^2+2 h x y+b y^2=0 \\ & \qquad \tan \theta=\frac{2 \sqrt{h^2-a b}}{a+b} \\ & \text { For } 12 x^2+2 h x y+7 y^2=0 \\ & \qquad \tan \theta=\frac{2 \sqrt{h^2-(12 \times 7)}}{12+7} \\ & \Rightarrow \quad \frac{8}{19}=\frac{2 \sqrt{h^2-84}}{19} \\ & {\left[\because \tan \theta=\frac{8}{19} \text { given }\right]} \\ & \Rightarrow \quad 4=\sqrt{h^2-84} \Rightarrow 16=h^2-84 \\ & \Rightarrow \quad h^2=100 \Rightarrow h= \pm 10 \end{aligned} \end{aligned} $

$ \begin{aligned} \left(\alpha^2+12|\alpha|\right) x^2 & +6 x y +(18-21|\alpha|) y^2=0 \end{aligned} $

∵ Lines are at right angled.

Coefficient of $x^2+$ Coefficient of $y^2=0$

$ \begin{array}{lr} \Rightarrow & \alpha^2+12|\alpha|+18-21|\alpha|=0 \\ \Rightarrow & \alpha^2-9|\alpha|+18=0 \\ \Rightarrow & |\alpha|^2-6|\alpha|-3|\alpha|+18=0 \\ \Rightarrow & |\alpha|(|\alpha|-6)-3| | \alpha \mid-6)=0 \\ \Rightarrow & \quad(|\alpha|-6)(|\alpha|-3)=0 \\ \Rightarrow & |\alpha|=6,|\alpha|=3 \\ \Rightarrow & \alpha= \pm 6, \alpha= \pm 3 \end{array} $

Given line, $x+2 y-c=0$

$ \Rightarrow \frac{x+2 y}{c}=1 $

and curve, $x^2+y^2-3 x-6 y+3=0$

∴ By homogenisation,

$ \begin{array}{r} x^2+y^2-3 x\left(\frac{x+2 y}{c}\right)-6 y\left(\frac{x+2 y}{c}\right) +3\left(\frac{x+2 y}{c}\right)^2=0 \end{array} $

$ \begin{array}{rlr} \Rightarrow & c^2 x^2+c^2 y^2-3 c x(x+2 y)-6 c y(x+2 y) \\ & +3\left(x^2+4 y^2+4 x y\right)=0 \\ \Rightarrow & c^2 x^2+c^2 y^2-3 c x^2-6 c x y-6 c x y-12 c y^2 \\ & +3 x^2+12 y^2+12 x y=0 \\ \Rightarrow & x^2\left(c^2-3 c+3\right)+y^2\left(c^2-12 c+12\right) \\ & +x y(-12 c+12)=0 \\ \because & \angle P O Q=\pi / 2 \\ \therefore & a+b=0 \\ \Rightarrow & c^2-3 c+3+c^2-12 c+12=0 \\ \Rightarrow & 2 c^2-15 c+15=0 \\ \Rightarrow & 2 c^2-15 c=-15 \end{array} $

According to the question,

Let $A(a, b), B(0,0), C(c, 0)$

Since, $P$ divides $A B$ in ratio $1: 2$.

So, $\quad P\left(\frac{2 a}{3}, \frac{2 b}{3}\right)$

and $Q$ divides line $B C$ in ratio $1: 2$.

So, $\quad Q\left(\frac{c}{3}, 0\right)$

Let $D$ divides $A Q$ line in ratio $\lambda_2: 1$.

Then, $y$ coordinate of $D: \frac{b}{\lambda_2+1}$

and also let $D$ divides line $P C$ into $\lambda_1: 1$ ratio

they $y$ coordinate of $D: \frac{2 b}{3\left(\lambda_1+1\right)}$

From Eqs. (i) and (ii), we get

$ \begin{array}{rlrl}& \frac{b}{\lambda_2+1} =\frac{2 b}{3\left(\lambda_1+1\right)} \\ \Rightarrow & 3 \lambda_1+3 =2 \lambda_2+2 \\ \Rightarrow & 3 \lambda_1-2 \lambda_2+1 =0 \end{array} $

also $x$-coordinate will be equal

$ \begin{aligned} & \frac{\frac{2 a}{3}+\lambda_1 c}{\lambda_1+1} =\frac{a+\lambda_2 \frac{c}{3}}{\lambda_2+1} \\ \Rightarrow & c =\frac{2}{3}\left(\lambda_2 \frac{3}{3}+a\right) \frac{2 a}{3}+\lambda_1 \end{aligned} $

$ \begin{aligned} & \Rightarrow \lambda_1=\frac{2 \lambda_2}{9} \text { on putting in Eq. (iii), } \\ & \frac{2 \lambda_2}{3}-2 \lambda_2+1=0 \\ & \Rightarrow \quad \frac{-4 \lambda_2}{3}=-1 \Rightarrow \lambda_2=\frac{3}{4}, \lambda_1=\frac{1}{6} \\ & \text { Now, } \frac{\operatorname{ar} \triangle A B C}{\operatorname{ar} \triangle B C D}=\frac{\frac{1}{2} \times B C \times b}{\frac{1}{2} \times B C \times \frac{b}{\lambda_2+1}} \\ & \quad \quad=\lambda_2+1=\frac{3}{4}+1=\frac{7}{4} \\ & \begin{aligned} \Rightarrow \quad & \frac{k}{\operatorname{ar} \triangle B C D}=\frac{7}{4} \\ \Rightarrow \quad \text { ar } \triangle B C D & =\frac{4}{7} k \end{aligned} \end{aligned} $

Given, area of quadrilateral $=10$ sq. units

$ \Rightarrow \quad \frac{1}{2}\left|\begin{array}{ccccc} x & 2 & 5 & 1 & x \\ y & 3 & -2 & -1 & y \end{array}\right|=10 $

$ \begin{aligned} & \Rightarrow|3 x-2 y-4-15-5+2+y+x|=20 \\ & \Rightarrow \quad|4 x-y-22|=20 \\ & \Rightarrow \quad 4 x-y-22= \pm 20 \\ & \text { Now, } 4 x-y-22=20 \\ & \Rightarrow \quad 4 x-y-42=0 \\ & \text { and } \quad 4 x-y-22=-20 \\ & \Rightarrow \quad 4 x-y-2=0 \\ & \text { Locus: }(4 x-y-42)(4 x-y-2)=0 \end{aligned} $

$x$-intercept of line $=5$

So, $a=5(5,0)$

$y$-intercept of line $=7$

So, $b=7(0,7)$

Equation of line : $\frac{x}{a}+\frac{y}{b}=1$

$ \Rightarrow \quad \frac{x}{5}+\frac{y}{7}=1 $

Equation of line w.r.t. old axis

$ :7 x+5 y=35 $

Now replace $x \rightarrow x \cos \theta-y \sin \theta$,

$ y \rightarrow x \sin \theta+y \cos \theta $

Equation of line $L$ w.r.t. new axis

$ \begin{array}{r} 7(x \cos \theta-y \sin \theta)+5(x \sin \theta+y \cos \theta) =35 \\ (7 \cos \theta+5 \sin \theta) x+(5 \cos \theta-7 \sin \theta) y =35 \end{array} $

For $x$-intercept, on putting $x=0$,

$ x \text {-intercept }=\frac{35}{7 \cos \theta+5 \sin \theta} $

For $y$-intercept, on putting $x=0$,

$ y \text {-intercept }=\frac{35}{5 \cos \theta-7 \sin \theta} $

Given, $x$-intercept $=y$-intercept

$ \begin{aligned} & \therefore & 7 & \cos \theta+5 \\ \Rightarrow & & \tan \theta & =5 \cos \theta-7 \sin \theta \\ & \therefore & & \tan \theta \\ & & |\tan \theta| & =\frac{1}{6} \end{aligned} $