Straight Lines and Pair of Straight Lines

We are given the line equation $2x + by + 5 = 0$ and the quadratic equation $ax^2 - 96bxy + ky^2 = 0$. We need to find the value of $a + 3k$ when these form an equilateral triangle.

The angle between the lines derived from the quadratic equation can be represented as:

$ 2\left|\frac{\sqrt{h^2 - ak}}{a + k}\right| = \tan 60^\circ $

Given that $\tan 60^\circ = \sqrt{3}$, substituting into the equation we get:

$ \sqrt{3} = \frac{2\sqrt{(48b)^2 - ak}}{a + k} $

Squaring both sides:

$ 3(a + k)^2 = 4(48b)^2 - 4ak $

Expanding and rearranging terms, we have:

$ 3a^2 + 3k^2 + 6ak = 4(48)^2b^2 - 4ak $

Adding $4ak$ to both sides:

$ 3a^2 + 3k^2 + 10ak = 4 \times 48^2 b^2 $

Hence, factoring gives:

$ (a + 3k)(3a + k) = 4 \times 48 \times 48 b^2 $

Therefore, the possible values for $a + 3k$ could be $192$ or $48b^2$. Comparing with the options, it is clear:

$ a + 3k = 192 $

Let the variable point be $ P(h, k) $.

The distance from $ P $ to the point $(4, 3)$ can be expressed as:

$ \sqrt{(h-4)^2 + (k-3)^2} $

The perpendicular distance from $ P $ to the line $ x + 2y - 1 = 0 $ is given by:

$ \left|\frac{h + 2k - 1}{\sqrt{1^2 + 2^2}}\right| = \left|\frac{h + 2k - 1}{\sqrt{5}}\right| $

According to the problem, these two distances are equal:

$ \sqrt{(h-4)^2 + (k-3)^2} = \left|\frac{h + 2k - 1}{\sqrt{5}}\right| $

Squaring both sides, we get:

$ (h-4)^2 + (k-3)^2 = \frac{(h + 2k - 1)^2}{5} $

Expanding both sides:

$ h^2 - 8h + 16 + k^2 - 6k + 9 = \frac{h^2 + 4hk + 4k^2 - 2h - 4k + 1}{5} $

Multiply through by 5 to eliminate the fraction:

$ 5(h^2 - 8h + 16 + k^2 - 6k + 9) = h^2 + 4hk + 4k^2 - 2h - 4k + 1 $

Simplify and collect like terms:

$ 5h^2 + 5k^2 - 40h - 30k + 125 = h^2 + 4hk + 4k^2 - 2h - 4k + 1 $

Rearranging gives:

$ 4h^2 + k^2 - 4hk - 38h - 26k + 124 = 0 $

Finally, replace $ (h, k) $ with $ (x, y) $:

$ 4x^2 + y^2 - 4xy - 38x - 26y + 124 = 0 $

Put $ x=X $ and $ y=Y+k $

Then,

$ \begin{array}{l} \Rightarrow a X^{2}-2 X(Y+k)+b(Y+k)-2 X \\\\ +4(Y+k)+1=0 \\\\ \Rightarrow a X^{2}-2 X Y-2 k X+b Y^{2}+b k^{2} \\\\ +26 k Y-2 X+4 Y+4 k+1=0 \\\\ \text { Now, } \\\\ \because \text { coefficient of } x=0 \\\\ \Rightarrow \quad-2 k-2=0 \\\\ \therefore \quad k=-1 \\\\ \because \text { Coefficient of } y=0 \\\\ \Rightarrow \quad 2 b k+4=0 \Rightarrow-2 b+4=0 \\\\ \therefore \quad b=2 \end{array} $

Now, put

$ \begin{array}{l} \quad X=x \cos \theta-y \sin \theta \\\\ \text { and } \quad Y=x \sin \theta+y \cos \theta \\\\ \Rightarrow a(x \cos \theta-y \sin \theta)^{2}-2(x \cos \theta-y \sin \theta) \\\\ (x \sin \theta+y \cos \theta-1) \\\\ +2(x \sin \theta+y \cos \theta-1)^{2}-2(x \cos \theta-y \sin \theta) \\\\ +4(x \sin \theta+y \cos \theta-1)+1=0 \end{array} $

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

Now, coefficient of $ x y=0 $

$ \begin{array}{r} \Rightarrow-a(2 \cos \theta \sin \theta)-2\left(\cos ^{2} \theta-\sin ^{2} \theta\right) \\\\ +2(2 \sin \theta \cos \theta)=0 \\\\ \Rightarrow-a \sin 2 \theta-2 \cos 2 \theta+2 \sin 2 \theta=0 \quad \ldots \text { (iii) } \end{array} $

Let $ \frac{1}{2} \tan ^{-1}(2)=\theta $

$ \Rightarrow \quad \tan 2 \theta=2 $

Now, from Eq. (iii), we get

$ \begin{array}{l} \Rightarrow \frac{-a \times 2}{\sqrt{5}}-\frac{2 \times 1}{\sqrt{5}}+\frac{2 \times 2}{\sqrt{5}}=0 \\\\ \quad\left[\because \sin 2 \theta=\frac{2}{\sqrt{5}} \text { and } \cos \theta=\frac{1}{\sqrt{5}}\right] \\\\ \Rightarrow a=1 \end{array} $

Hence, $ a+b=1+2=3 $

For $ x $-intercept, put $ y=0 $

$ \begin{array}{rlrl} & & a & =2 \\\\ \therefore & P & =\left|\frac{0+0-2}{\sqrt{2}}\right|=\sqrt{2}\end{array} $

Now; $ \tan \beta=\frac{p}{a}=\frac{1}{\sqrt{2}} $

So, $ \quad 2 \tan \beta+P^{2} \Rightarrow 2 \tan 45^{\circ}+(\sqrt{2})^{2} $

$ \Rightarrow \quad 2+2=4 $

Using section formula

$ C\left(\frac{-2 a+b}{a+b}, \frac{a+2 b}{a+b}\right) $

Now, substitute point $ C $ on line equation $ (2 x+y-3=0) $.

$ \Rightarrow \quad 2\left(\frac{-2 a+b}{a+b}\right)+\frac{a+2 b}{a+b}=3 $

On solving, we get

$ b=6 a $

On substituting $ b=6 a $ in point $ C $, we get

$ \begin{aligned} & \left(\frac{4}{7}, \frac{13}{7}\right) \\\\ \because & P\left(\frac{b}{3 a},-3\right)=P(2,-3) \end{aligned} $

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

Now, coordinate of $ C $ is

$ C\left(\frac{-3 p+2 q}{p+q}, \frac{-2 p-3 q}{p+q}\right) $

On comparing, we get

$ \begin{array}{r|r} \frac{-3 p+2 q}{p+q}=\frac{4}{7} & \frac{-2 p-3 q}{p+q}=\frac{13}{7} \\\\ \frac{p}{q}=\frac{2}{5} & \frac{p}{q}=\frac{-34}{27} \end{array} $

Now, $ \frac{p}{q}+\frac{q}{p}=\frac{2}{5}+\frac{5}{2}=\frac{4+25}{10}=\frac{29}{10} $

Let $ Q \equiv(a, b) $ and $ R=(c, d) $

Then,

$ \Rightarrow \frac{a-2}{1}=\frac{b-3}{-1}=\frac{-2(2-3+2)}{2} $

$ \Rightarrow a-2=3-b=-(1) $

$ \Rightarrow a-2=-1 ; 3-b=-1 \Rightarrow a=1, b=4 $ $ \because \frac{c-2}{2}=\frac{d-3}{1}=\frac{-2(4+3-2)}{5} $

$ \begin{array}{l} \Rightarrow c-2=-4, d-3=-2 \\\\ \Rightarrow \quad c=-2, d=1 \end{array} $

Option (c)

$ \begin{array}{l} P(Q)=P(1,4)=1+4+2=7 \text { (positive) } \\\\ P(R)=P(-2,1)=-2+1+2=1 \text { (positive) } \end{array} $

Hence, point lie on same sides.

To find point of intersection we $ d_{0} $ partial differentiation

On differentiating w.r.t. $ x $, we get

$ 4 x+a y+b=0 $

On differentiating w.r.t. $ y $, we get

$ a x+6 y+c=0 $

$ a x+6 y+c=0 $

On putting ( $ 2,-1 $ ) in Eqs. (i) and (ii), we get

$ \begin{array}{l} 8-a+b=0 \Rightarrow a-b=8 \\\\ 2 a-6+c=0 \Rightarrow 2 a+c=6 \end{array} $

Also,

$ (2,-1) $ satisfies the given equation

$ \begin{aligned} 8-2 a+3+2 b-c-3 & =0 \\\\ -2 a+2 b-c & =-8 \\\\ 2 a-2 b+c & =8 \end{aligned} $

Put $ 2 a+c=6 $ from Eqs. (iv) to (v)

$ \begin{array}{l} \Rightarrow \quad 6-2 b=8 \Rightarrow-2 b=2 \\\\ b=-1 \Rightarrow \quad a=7 \\\\ \therefore \quad c=-8 \end{array} $

Hence, $ 3 a+2 b+c=21-2-8=11 $.

Let $P=(h, k)$

Now, according to the question,

$ \begin{aligned} & \frac{\sqrt{(h-1)^2+(k-1)^2}}{\left|\frac{h-k+2}{\sqrt{1^2+(-1)^2}}\right|}=\frac{1}{\sqrt{2}} \\\\ & \Rightarrow \frac{2\left[(h-1)^2+(k-1)^2\right]}{(h-k+2)^2}=\frac{1}{2} \end{aligned} $

[squaring both sides]

$ \begin{array}{ll} \Rightarrow & 4\left[\left(h^2-2 h+1\right)+\left(k^2-2 k+1\right)\right] \\\\ & =h^2+k^2+4-2 h k-4 k+4 h \\\\ \Rightarrow & 3 h^2+2 h k+3 k^2-12 h-4 k+4=0 \end{array} $

$\therefore$ The equation of the locus of $P$ is

$ 3 x^2+2 x y+3 y^2-12 x-4 y+4=0 $

Interchanging $x$ with $x+\frac{3}{2}$ and $y$ with $y+(-2)$ i.e. $y-2$ in the given equation to get the required equation.

$\therefore$ Required equation :

$ \begin{aligned} & 2\left(x+\frac{3}{2}\right)^2+4\left(x+\frac{3}{2}\right)(y-2) \\\\ & \quad+(y-2)^2+2\left(x+\frac{3}{2}\right)-2(y-2)+1=0 \\\\ & \Rightarrow \frac{2(2 x+3)^2}{4}+\frac{4(2 x+3)}{2}(y-2) \\\\ & \quad+(y-2)^2+\frac{2(2 x+3)}{2}-2(y-2)+1=0 \\\\ & \Rightarrow \frac{1}{2}\left[4 x^2+12 x+9+(8 x+12)\right. \\\\ & \left.(y-2)+2(y-2)^2+4 x+6-4(y-2)+2\right] \\\\ & =0 \\\\ & \Rightarrow 4 x^2+12 x+9+8 x y+12 y \\\\ & \quad-16 x-24+2 y^2-8 y+8+4 x+6 \\\\ & \Rightarrow 4 y+8+2=0 \\\\ & \Rightarrow 4 x^2+8 x y+2 y^2+9=0 \end{aligned} $

Given,

$ L \equiv x \cos \alpha+y \sin \alpha-p=0 $

$L$ is perpendicular to the line

$ x+y+1=0 $

$p$ is positive, $\alpha \in\left(270^{\circ}, 360^{\circ}\right)$ and

Distance of point $(\sqrt{2}, \sqrt{2})$ to the line $L$ is 5 units.

Slope of $L=\frac{-\cos \alpha}{\sin \alpha}=-\cot \alpha$

Slope of line $x+y+1=0$ is -1

$ \begin{array}{cc} \therefore & (-\cot \alpha)(-1) \end{array}=-1 \Rightarrow \cot \alpha=-1 $

Now, $L \equiv x \cos 315^{\circ}+y \sin 315^{\circ}-p=0$

$ \begin{aligned} & \Rightarrow \quad x\left(\frac{1}{\sqrt{2}}\right)+y\left(-\frac{1}{\sqrt{2}}\right)-p=0 \\\\ & \Rightarrow \quad x-y-p \sqrt{2}=0 \end{aligned} $

According to the question,

$ \begin{array}{rlrlrl} & & \frac{|\sqrt{2}-\sqrt{2}-p \sqrt{2}|}{\sqrt{1^2+(-1)^2}} & =5 & & \\\\ \Rightarrow & & p \sqrt{2} \mid & =5 \sqrt{2} \\\\ \Rightarrow & & p & =5 & {[\because p>0]} \end{array} $

Given, $(-2,-1),(2,5)$ and $(m, n)$ are three vertices of a triangle whose orthocenter is $\left(2, \frac{5}{3}\right)$.

Here, line $A H \perp$ line $B C$ and line $C H \perp$ line $A B$

$ \begin{gathered} m_{A B}=\frac{5-(-1)}{2-(-2)}=\frac{6}{4}=\frac{3}{2} \\\\ m_{A H}=\frac{\frac{5}{3}-(-1)}{2-(-2)}=\frac{8}{12}=\frac{2}{3} \\\\ m_{B C}=\frac{n-5}{m-2} \\\\ \Rightarrow \quad m_{C H}=\frac{\frac{5}{3}-n}{2-m}=\frac{5-3 n}{6-3 m} \\\\ \Rightarrow \quad m_{A H} \times m_{B C}=-1 \\\\ \frac{2}{3} \times \frac{n-5}{m-2}=-1 \\\\ \Rightarrow \quad m_{C H} \times m_{A B}=-1 \quad \ldots \text { (i) } \\\\ \frac{5-3 n}{6-3 m} \times \frac{3}{2}=-1 \end{gathered} $

On solving Eqs. (i) and (ii), we get $m=6$ and $n=-1$

$ \therefore m+n=6+(-1)=5 $

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} $

Let equation of $A B$ be $x \cos \alpha+y \sin \alpha=p$

$ \begin{aligned} & \Rightarrow \quad \frac{x \cos \alpha}{p}+\frac{y \sin \alpha}{p}=1 \\ & \Rightarrow \quad \frac{x}{\frac{p}{\cos \alpha}}+\frac{y}{\frac{p}{\sin \alpha}}=1 \end{aligned} $

So, coordinate of $A$ and $B$ are

$ \left(\frac{p}{\cos \alpha}, 0\right) \text { and }\left(0, \frac{p}{\sin \alpha}\right) $

$\therefore$ Mid-point of $A$ and $B$ are

$ \begin{aligned} & \left(\frac{p}{2 \cos \alpha}, \frac{p}{2 \sin \alpha}\right) \\ & \therefore \quad x=\frac{p}{2 \cos \alpha} \Rightarrow \cos \alpha=\frac{p}{2 x} \\ & y=\frac{p}{2 \sin \alpha} \\ & \Rightarrow \sin \alpha=\frac{p}{2 y} \end{aligned} $

$ \begin{aligned} & \text { Now, } \cos ^2 \alpha+\sin ^2 \alpha=1 \\ & \Rightarrow \quad \frac{p^2}{4 x^2}+\frac{p^2}{4 y^2}=1 \\ & \Rightarrow \quad \frac{1}{x^2}+\frac{1}{y^2}=\frac{4}{p^2} \end{aligned} $

When the origin is shifted to $(2,3)$ and the axes are rotated counterclockwise by an angle $\theta$, the transformation for the coordinates is given by:

$ x = X \cos \theta - Y \sin \theta + 2 $

$ y = X \sin \theta + Y \cos \theta + 3 $

By substituting these expressions for $x$ and $y$ into the provided equation:

$ 3x^2 + 2xy + 3y^2 - 18x - 22y + 50 = 0 $

we perform the transformation and simplify it:

$ \begin{aligned} &\Rightarrow 3(X \cos \theta - Y \sin \theta + 2)^2 \\ &- 2(X \cos \theta - Y \sin \theta + 2) \\ &(X \sin \theta + Y \cos \theta + 3) \\ &+ 3(X \sin \theta + Y \cos \theta + 3)^2 \\ &- 18(X \cos \theta - Y \sin \theta + 2) \\ &- 22(X \sin \theta + Y \cos \theta + 3) + 50 = 0 \end{aligned} $

After simplification, we match it to the given transformed equation:

$ 4X^2 + 2Y^2 - 1 = 0 $

This requires the coefficients from the original equation after transformation to satisfy the following equation:

$ 3\cos^2 \theta + 2\sin \theta \cos \theta + 3\sin^2 \theta = 4 $

Solving this, we simplify to:

$ \sin 2\theta + 3 = 4 $

which implies:

$ \sin 2\theta = 1 $

Thus:

$ 2\theta = \frac{\pi}{2} \Rightarrow \theta = \frac{\pi}{4} $

To find the length of the segment $ PQ $, we begin by determining the equation of the line passing through the point $ P(3, 4) $ and making an angle $\frac{\pi}{6}$ with the positive $ X $-axis.

The direction ratios of the line, given the angle $\frac{\pi}{6}$ or $30^\circ$, are $\cos 30^\circ = \frac{\sqrt{3}}{2}$ and $\sin 30^\circ = \frac{1}{2}$. Using these, we construct the parametric equation of the line:

$ \frac{x - 3}{\frac{\sqrt{3}}{2}} = \frac{y - 4}{\frac{1}{2}} = d $

Here, $d$ is a parameter representing the distance from the point $ (3, 4) $ along the line.

From this, the coordinates of any point $ (x, y) $ on the line can be expressed as:

$ x = 3 + \frac{\sqrt{3}}{2}d, \quad y = 4 + \frac{d}{2} $

This point also lies on the line given by the equation $ 12x + 5y + 10 = 0 $. Substitute $ x $ and $ y $ into this equation:

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

Simplifying this expression:

$ 36 + 6\sqrt{3}d + 20 + \frac{5}{2}d + 10 = 0 $

Further simplification gives:

$ 66 + \left(6\sqrt{3} + \frac{5}{2}\right) d = 0 $

Solving for $d$, we find:

$ d = \frac{-66}{6\sqrt{3} + \frac{5}{2}} $

Rationalizing the denominator and simplifying provides:

$ d = \frac{132}{12\sqrt{3} + 5} $

Thus, the length of segment $ PQ $ is $\frac{132}{12 \sqrt{3} + 5}$.

Let $O M$ and $O N$ be the perpendicular bisectors of $A B$ and $A C$, respectively.

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

Coordinates of $M$ and $N$ are $\left(\frac{1+x_1}{2}, \frac{-2+y_1}{2}\right)$ and $\left(\frac{1+x_2}{2}, \frac{-2+y_2}{2}\right)$, respectively.

$M$ lies on line $x-y+5=0$

$ \begin{aligned} & \Rightarrow \quad \frac{1+x_1+2-y_1}{2}+5=0 \\ & \Rightarrow \quad x_1-y_1=-13 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

$N$ lies on line $O N$,

$ \begin{aligned} & \Rightarrow \quad \frac{1+x_2}{2}+2\left(\frac{-2+y_2}{2}\right)=0 \\ & \quad 1+x_2-4+2 y_2=0 \\ & \Rightarrow \quad x_2+2 y_2=3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{aligned} $

Slope of line $O M$ is 1 and slope of line $A B$ is $\frac{y_1+2}{x_1-1}$.

As $O M$ is perpendicular to $A B$,

$ \begin{aligned} & \left(\frac{y_1+2}{x_1-1}\right) \times(1)=-1 \\ \Rightarrow & y_1+2=1-x_1 \\ \Rightarrow & x_1+y_1=-1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{aligned} $

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

$ x_1=-7 \text { and } y_2=6 $

$\therefore$ Coordinates of $B$ is $(-7,6)$

Slope of line $O N=-\frac{1}{2}$

Slope of line $A C=\frac{y_2+2}{x_2-1}$

As $O N$ is perpendicular to $A C$,

$ \begin{aligned} & \left(\frac{y_2+2}{x_2-1}\right) \times\left(-\frac{1}{2}\right)=-1 \\ & y_2+2=2 x_2-2 \\ & 2 x_2-y_2=4 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\end{aligned} $

On solving Eqs. (ii) and (iv), we get $y_2=\frac{2}{5}$ and $x_2=\frac{11}{5}$

$\therefore$ Coordinate of $C$ is $\left(\frac{11}{5}, \frac{2}{5}\right)$.

Equation of line $A C$ is

$ \begin{aligned} & y-6=\left(\frac{\frac{2}{5}-6}{\frac{11}{5}+7}\right)(x+7) \\ & y-6=\left(-\frac{28}{46}\right)(x+7) \\ & 23 y-138=-14 x-98 \\ & 14 x+23 y-40=0 \end{aligned} $

Here, $O A B$ is the required triangle as shown in figure and $O M$ is the height of triangle

$ O M=\frac{|0+0-6|}{\sqrt{4+9}}=\frac{6}{\sqrt{13}} $

As $O A B$ is an isosceles $\triangle$, so $A M=M B$ and $\angle A O M=\angle B O M=45^{\circ}$

$\triangle O M B$ is, also an isosceles right angle

$ \begin{array}{ll} \Rightarrow & O M=M B=A M=\frac{6}{\sqrt{13}} \\ \Rightarrow & A B=\frac{12}{\sqrt{13}} \end{array} $

Area of $\triangle O A B=\frac{1}{2} \times \frac{6}{\sqrt{13}} \times \frac{12}{\sqrt{13}}=\frac{36}{13}$

Substitute $y=-x-2$

in the equation of curve to find the point of intersection.

$ \begin{aligned} & x^2+x^2+4+4 x-x^2-2 x+x \\ & -3 x-6+1=0 \\ & x^2-1=0 \Rightarrow x= \pm 1 \end{aligned} $

If $x=1$, then $y=-3$

and if $x=-1$, then $y=-1$

$\therefore$ The point of intersection are $(1,-3)$ and $(-1,-1)$

Let coordinates of $A$ and $B$ be $(1,-3)$ and $(-1,-1)$ respectively.

Equation of pair of line is

$ \begin{array}{r} (y+3 x)(y-x)=0 \\ y^2+3 x y-x y-3 x^2=0 \\ y^2+2 x y-3 x^2=0 \end{array} $

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

$\therefore$ Required combined equation is

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

The locus of a variable point which form a triangle of fixed area with two fixed point will be a pair of parallel line.

Let $\theta$ be the inclination of line through $A(-5,-4)$

Therefore, equation of this line is

$ \begin{aligned} & \frac{x-x_1}{\cos \theta}=\frac{y-y_1}{\sin \theta}=r \\ & \frac{x+5}{\cos \theta}=\frac{y+4}{\sin \theta}=r \,\,\,\,\,\,\,\,\,\,\,\,\,(say)\end{aligned} $

So, $x=(r \cos \theta-5)$ and $y=(r \sin \theta-4)$

So, coordinate of $p$ are

$ (-5+r \cos \theta,-4+r \sin \theta) $

If $P Q=r_1$ and $P R=r_2$, then

$ \begin{aligned} & Q\left(r_1 \cos \theta-5, r_1 \sin \theta-4\right) \\ & R\left(r_2 \cos \theta-5, r_2 \sin \theta-4\right) \end{aligned} $

Given, Qlies on $x-y-5=0$

$ \begin{array}{ll} \Rightarrow & r_1 \cos \theta-5-r_1 \sin \theta+4-5=0 \\ \Rightarrow & r_1(\cos \theta-\sin \theta)=6 \\ \Rightarrow & \cos \theta-\sin \theta=\frac{6}{P Q}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, ...(i)\end{array} $

Given, $R$ lies on $x+3 y+2=0$

$ \begin{aligned} \Rightarrow & r_2 \cos \theta-5+3 r_2 \sin \theta-12+2=0 \\ \Rightarrow & r_2(\cos \theta+3 \sin \theta)=15 \\ & \cos \theta+3 \sin \theta=\frac{15}{P R}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, ...(ii) \end{aligned} $

Now, $\frac{18}{P Q}+\frac{15}{P R}=2$

$ \Rightarrow \quad 3 \cos \theta-3 \sin \theta+\cos \theta+3 \sin \theta=2 $

$ \begin{aligned} &\Rightarrow \quad \cos \theta=\frac{1}{2}\\ &\text { So, slope of line } L \text { is } \tan \theta= \pm \sqrt{3} \text {. } \end{aligned} $

Reflaction of point $A(2,3)$ in $x-axis$

$ =\operatorname{Point}(B)=B(2,-3) $

Reflaction of $B$ along the line $x+y=0$

$ =\operatorname{Point}(C)=x=-y $

swap the value and change the sign

$ =\operatorname{Point}(3,-2) $

Reflaction of $C$ along $x-y=0$

$ x=y $

swap the value of point is

$ D(-2,3) $

Equation of line $A B$ is

$ \begin{aligned} & y-3=\frac{-3-3}{2-2}(x-2) \Rightarrow x=2 \\ & A B: x-2=0 \end{aligned} $

Equation of line $C D$ is

$ \begin{array}{ll} & y+2=\frac{3+2}{-2-3}(x-3) \\ \Rightarrow \quad & y+2=-x+3 \\ & C D: x+y=1 \end{array} $

Intersection point of $A B$ and $C D$ are

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

So, intersection point $(2,-1)$

Given pair of equation of linesis

$ x y-x-y+1=0 $

Factorize this equation

$ (x-1)(y-1)=0 $

Equation of line

From the option, only $x-y=5$ wil makes an angle of $45^{\circ}$ with each of the pair of line.

Given pair of lines is

$ 8 x^2+a x y+y^2=0 $

We know that $m$ is the slope of

$ a x^2+2 h x y+b y^2=0 $

it slopes $m_1$ and $m_2$

Then, $m_1+m_2=-\frac{2 h}{b}$ and $m_1 m_2=\frac{a}{b}$

$ m_1+m_2=-\frac{a}{1}\,\,\,\,\,\,\,\,\,\,\,\,\, [from Eq.(i)]$

$and\,\,\,\, m_1 m_2=8 / 1 $

and given, if $m_1=m$

$ m_2=3 m $

$ \text { } \begin{aligned} So,\,\,\,4 m & =-a \\ 3 m^2 & =8 \Rightarrow m= \pm \sqrt{8 / 3} \\ \Rightarrow \quad a & =+4 \times \sqrt{\frac{8}{3}}=8 \sqrt{\frac{2}{3}} \end{aligned} $

To find the equation of the locus of points equidistant from the points $(2,3)$ and $(4,5)$, we start with the distance formula:

The distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

$ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $

For a point $ P(x, y) $ equidistant from both $(2,3)$ and $(4,5)$, the distances are given by:

For point $(2,3)$:

$ d_1 = \sqrt{(x - 2)^2 + (y - 3)^2} $

For point $(4,5)$:

$ d_2 = \sqrt{(x - 4)^2 + (y - 5)^2} $

According to the problem, these distances are equal:

$ d_1 = d_2 $

Squaring both sides to remove the square roots gives:

$ d_1^2 = d_2^2 $

Simplifying, we have:

$ (x - 2)^2 + (y - 3)^2 = (x - 4)^2 + (y - 5)^2 $

Expanding both sides:

$ x^2 - 4x + 4 + y^2 - 6y + 9 = x^2 - 8x + 16 + y^2 - 10y + 25 $

Simplifying further:

$ 4x + 4y = 28 $

Divide the entire equation by 4:

$ x + y = 7 $

Therefore, the equation of the locus is $x + y = 7$.

$ \begin{aligned} &\text { } x+y=2\\ &\text { Distance }=\left|\frac{a x_1+b y_1+c}{\sqrt{a^2+b^2}}\right| \end{aligned} $

$ \begin{aligned} &A D=\left|\frac{1(2)+1(-1)-2}{\sqrt{1^2+1^2}}\right|=\frac{1}{\sqrt{2}}\\ &\text { In' } \triangle A B D, \sin 60^{\circ}=\frac{A D}{A B}\\ &A B=\frac{A D}{\sin 60^{\circ}}=\frac{1 / \sqrt{2}}{\sqrt{3} / 2}=\sqrt{\frac{2}{3}} \end{aligned} $

$ \begin{array}{r} x+y+1=0\,\,\,\,\,\,\,\,\,\, ...(i)\\ x-y-1=0\,\,\,\,\,\,\,\,\,\, ...(ii) \\ 3 x+4 y+5=0 \,\,\,\,\,\,\,\,\,\, ...(iii)\end{array} $

$ \text { Slope of line }=-\frac{x \text { coordinate }}{y \text { coordinate }} $

For Eq. (i) $m_1=\frac{-1}{1}=-1$

for Eq. (ii) $m_2=\frac{-1}{-1}=1$

$ \begin{aligned}Now,\,\,\,\, & m_1 \cdot m_2=-1 \times 1=-1 \\ & m_1 \cdot m_2=-1 \end{aligned} $

So, we can say Eq. (i) perpendicular to Eq. (ii) therefore triangle formed by these three lines will be a right angle triangle.

Let, $\triangle A B C$ be the new triangle.

We know that for right angle triangle orthocentre will be the point of right angle.

i.e. $B$ is the required point.

For coordinates of $\boldsymbol{B}$

On solving Eq. (i) and Eq. (ii)

$ \begin{aligned} & x+y+1=0 \\ & x-y-1=0 \end{aligned} $

$ \begin{aligned} \Rightarrow \quad 2 x & =0 \Rightarrow x=0 \\ y+1 & =0\,\,\,\, [from (i)]\\ y & =-1 \end{aligned} $

So, $B=(0,-1)$

The problem involves finding the angle between a pair of lines represented by the equation $2x^2 + 3xy + Ky^2 = 0$ when one of the line's slopes is $2$.

First, rewrite the equation in terms of the slope $m$ by substituting $y = mx$:

$ 2x^2 + 3x(mx) + K(mx)^2 = 0 $

$ x^2(2 + 3m + Km^2) = 0 $

Since $x \neq 0$, we must have:

$ Km^2 + 3m + 2 = 0 $

Given that the slope $m_1 = 2$ for one of the lines, we can find $K$ using:

$ m_1 + m_2 = -\frac{3}{K} $

$ 2 + m_2 = -\frac{3}{K} $

Since:

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

$ 2K + 1 = -3 $

$ 2K = -4 $

$ K = -2 $

Next, use the property $m_1 \cdot m_2 = \frac{2}{K}$:

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

$ m_2 = \frac{1}{K} $

Substitute $K = -2$ into the original equation:

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

This implies:

$ a = 2, \quad h = \frac{3}{2}, \quad b = -2 $

Then, calculate the angle $\theta$ between the lines using the formula:

$ \tan \theta = \frac{2 \sqrt{h^2 - ab}}{a + b} $

$ \tan \theta = \frac{\sqrt{\left(\frac{3}{2}\right)^2 - 2 \times (-2)}}{2 - 2} $

$ = \frac{\sqrt{\frac{9}{4} + 4}}{0} \rightarrow \infty $

Thus, the angle $\theta$ is:

$ \tan \theta = \tan \frac{\pi}{2} $

Therefore, the angle between the lines is $\frac{\pi}{2}$.

To find the length of the $x$-intercept made by the pair of lines given by the equation $2x^2 + xy - 6y^2 - 2x + 17y - 12 = 0$, follow these steps:

First, set $y = 0$ to determine the points where the lines intersect the $x$-axis:

$ 2x^2 - 2x - 12 = 0 $

This is a quadratic equation in $x$. To solve it, use the quadratic formula:

$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $

where $a = 2$, $b = -2$, and $c = -12$. Plug these values into the formula:

$ x = \frac{2 \pm \sqrt{(-2)^2 - 4 \times 2 \times (-12)}}{2 \times 2} $

Simplify inside the square root and solve:

$ x = \frac{2 \pm \sqrt{4 + 96}}{4} $

$ x = \frac{2 \pm \sqrt{100}}{4} $

$ x = \frac{2 \pm 10}{4} $

This gives the solutions:

$ x_1 = \frac{12}{4} = 3, \quad x_2 = \frac{-8}{4} = -2 $

Therefore, the length of the $x$-intercept is the distance between $x_1$ and $x_2$:

$ x_1 - x_2 = 3 - (-2) = 3 + 2 = 5 $

So, the length of the $x$-intercept is 5.

We have, the equation

$ 3 x^2+2 \sqrt{3} x y+y^2=0 $

A conic equation of the type of

$ A x^2+B x y+C y^2+D x+E y+F=0 \text { is } $

rotated by an angle $\theta$, to form a new cartesion plane with coordinate $\left(x^{\prime}, y^{\prime}\right)$. If $\theta$ is the angle of rotation.

Then, the relation between coordinates $(x, y)$ and $\left(x^{\prime}, y^{\prime}\right)$ can be expressed as

$ \begin{gathered} \Rightarrow x=x^{\prime} \cos \theta-y^{\prime} \sin \theta \text { and } \\ y=x^{\prime} \sin \theta+y^{\prime} \cos \theta \end{gathered} $

or

$ \begin{aligned} \Rightarrow y^{\prime} & =x \sin \theta+y \cos \theta \text { and } \\ x^{\prime} & =x \cos \theta+y \sin \theta \end{aligned} $

For this we need to have $\theta$ given by

$ \cot 2 \theta=\frac{A-C}{B} $

In the given case as equation is

$ \begin{aligned} & 3 x^2+2 \sqrt{3} x y+y^2=0 \text { we have, } \\ & \Rightarrow A=3, B=2 \sqrt{3}, C=1 \end{aligned} $

Hence, $\cot 2 \theta=\frac{3-1}{2 \sqrt{3}}=\frac{2}{2 \sqrt{3}}=\frac{1}{\sqrt{3}}$

$ \text { i.e. } \theta=\frac{\pi}{6} $

Hence, rotation is given by

$ \begin{aligned} \Rightarrow x & =x^{\prime} \cos \left(\frac{\pi}{6}\right)-y^{\prime} \sin \left(\frac{\pi}{6}\right) \\ \Rightarrow x & =\frac{x^{\prime} \sqrt{3}}{2}-\frac{y^{\prime}}{2} \\ y & =x^{\prime} \sin \left(\frac{\pi}{6}\right)+y^{\prime} \cos \left(\frac{\pi}{6}\right) \\ \Rightarrow y & =\frac{x^{\prime}}{2}+\frac{y^{\prime} \sqrt{3}}{2} \end{aligned} $

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

On putting the value of $x$ and $y$ in Eq. (i),

we get

$ \begin{aligned} & \Rightarrow\left(\frac{x^{\prime} \sqrt{3}}{2}-\frac{y^{\prime}}{2}\right)^2+\left(\frac{x^{\prime}}{2}+\frac{y^{\prime} \sqrt{3}}{2}\right)^2+2\left(\frac{x^{\prime} \sqrt{3}}{2}-\frac{y^{\prime}}{2}\right)\left(\frac{x^{\prime}}{2}+\frac{y^{\prime} \sqrt{3}}{2}\right)=2\\ & \Rightarrow\left(x^{\prime} \sqrt{3}-y^{\prime}\right)^2+\left(x^{\prime}+y^{\prime} \sqrt{3}\right)^2 +2\left(x^{\prime} \sqrt{3}-y^{\prime}\right)\left(x^{\prime}+y^{\prime} \sqrt{3}\right)=8 \\ & \Rightarrow 3 x^{\prime 2}+y^{\prime 2}-2 \sqrt{3} x^{\prime} y^{\prime}+x^{\prime 2}+y^{\prime 2} 3 +2 \sqrt{3} x^{\prime} y^{\prime}+2\left(\sqrt{3} x^{\prime 2}+3 x^{\prime} y^{\prime}-x^{\prime} y^{\prime}\right) \left.-\sqrt{3} y^{\prime 2}\right)=8 \\ & \Rightarrow 4 x^{\prime 2}+4 y^{\prime 2}+2 \sqrt{3} x^{\prime 2}+4 x^{\prime} y^{\prime}-2 \sqrt{3} y^{\prime 2}=8 \end{aligned} $

$ \begin{aligned} & \left.\Rightarrow 2 x^{\prime 2}(2+\sqrt{3})+2 y^{\prime 2}(2-\sqrt{3})+4 x^{\prime} y^{\prime}\right)=8 \\ & \Rightarrow x^{\prime 2}(2+\sqrt{3})+y^{\prime 2}(2-\sqrt{3})+2 x^{\prime} y^{\prime} \leqslant 4 \end{aligned} $

transformation of Eq. (i).

Let the point $P$ be $(h, k)$ which lies on line

$ x+y+5=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

If P lie on line, then if will satisfy the equation, then

$ h+k+5=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Distance betwen Eq. (ii) and $2 x+3 y+3=0$ is given by

$ \begin{aligned} & \Rightarrow \quad d=\left|\frac{2 h+3 k+3}{\sqrt{2^2+3^2}}\right|=\sqrt{13} \\ & \Rightarrow|2 h+3 k+3|=13 \\ & \Rightarrow|2(-k-5)+3 k+3|=13 \quad \text { [from } \mathrm{Eq}((ii)] \\ & \Rightarrow|-2 k-10+3 k+3|=13 \Rightarrow|k-7|=13 \\ & \Rightarrow k=20 \text { and } k=-6 \\ & \\ & h=-k-5 \\ & \Rightarrow h=-25 \text { and } h=1 \end{aligned} $

Hence, the coordinate are $(1,-6)$ and (-25,20)

We have equation of line for $\lambda, \mu \in I$

$ \begin{aligned} & (x-2 y-1)+\lambda(3 x+2 y-11)=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & (3 x+4 y-11)+\mu(-x+2 y-3)=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

The intersection of first Eq. (i)families d line can be obtained by solving Eq. (i)

$ \begin{array}{rlrl} \Rightarrow & x-2 y-1 =0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\\ \Rightarrow & x =2 y+1 \\ & 3 x+2 y-11 =0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(vi)\end{array} $

$ \begin{aligned} \Rightarrow & & 3(2 y+1)+2 y-11 & =0 \\ \Rightarrow & & 6 y+3+2 y-11 & =0 \\ \Rightarrow & & 8 y-8 & =0 \\ \Rightarrow & & y & =1 \\ \Rightarrow & & x & =3 \text { (from Eq.(iil) } \end{aligned} $

Intersection point of Eq. (i) is $(3,1)$ Similarly, for Eq. (ii), we get

$ \begin{array}{ll} \Rightarrow & 3 x+4 y-11=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(v)\\ \Rightarrow & -x+2 y-3=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(vi)\end{array} $

On solving Eq. (v) and (vi), we get the intersection of lines of Eq. (i) is ( 1,2 ). Equation of line common to both.

$ \begin{array}{cc} \Rightarrow \frac{y-1}{x-3}=\frac{2-1}{1-3}=\frac{1}{-2} \\ \Rightarrow -2 y+2=x-3 \\ \Rightarrow x+2 y-5=0 \\ \Rightarrow & x+2 y-5=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(vii)\end{array} $

On comparing Eq. (vii) with the given equation of line common to both the families of lines is $a x+b y-5=0$, we got

$ \Rightarrow a=1 \Rightarrow b=2 $

$ \begin{aligned} & \text { Then, } 2 a+b=2 \times 1+2=4 \Rightarrow 2 a+b=4 \end{aligned} $

$ \begin{aligned} &\text { Equation of lines }\\ &\begin{array}{ll} L_1 \equiv 3 x^2-5 x y+P y^2=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \text { and } L_2 \equiv 6 x^2-x y-5 y^2=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} \end{aligned} $

From Eq. (ii), we get

$ \Rightarrow 6 x^2-x y-5 y^2=0 $

On substituting $t=\frac{x}{y}$, we get

$ \Rightarrow 6 t^2-t-5=0 $

On solving, we get

$ \begin{aligned} & \Rightarrow \quad t=\frac{1 \pm \sqrt{1+120}}{12} \Rightarrow t=\frac{1 \pm 11}{12} \\ & \Rightarrow \quad t=1, \frac{-5}{6} \\ & \Rightarrow y=x \text { or } y=\frac{-6}{5} x \end{aligned} $

$\therefore$ From Eq. (i), $3 x^2-5 x y+P y^2=0$

On Substituting $t=\frac{x}{y}$, we get

$ \Rightarrow 3 t^2-5 t+P=0 $

If $t=1$, then $P=2$

If $t=\frac{-5}{6}$, then $P=5\left(\frac{-5}{6}\right)-3\left(\frac{25}{36}\right)$

$ \Rightarrow P=-\frac{225}{36}=-\frac{25}{4} $

So, $P=2$ or $P=\frac{-25}{4}$ are two possible values.

$\therefore$ Sum of all possible values

$ =2+\frac{(-25)}{4}=\frac{-17}{4} $

Given, distance of $P(x, y)$ from A (4,0) is twice the distance of $P$ from $B(-4,0)$

$ \begin{aligned} & \text { The distance of } P(x, y) \text { from } A \text { is } \\ & =\sqrt{(x-4)^2+(y-0)^2}=\sqrt{(x-4)^2+y^2} \end{aligned} $

The distance of $P(x, y)$ from $B$ is

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

On squaring both sides, we get

$ \begin{aligned} & (x-4)^2+y^2=4\left[(x+4)^2+y^2\right] \\ & x^2+16-8 x+y^2=4\left(x^2+16+8 x+y^2\right] \\ & -3 x^2-3 y^2-48-40 x=0 \\ & 3 x^2+3 y^2+48+40 x=0 \end{aligned} $

Now, we find the distance between C and $D$ where locus intersects the line

$ 3 y-3 x=20 $

On substitute $3 y=3 x+20$

$ \begin{aligned} & y=\frac{1}{3}(3 x+20) \\ & 3 x^2+3 \frac{1}{9}[3 x+20]^2+48+40 x=0 \\ & 9 x^2+9 x^2+400+120 x+(48+40 x)^{x^3}=0 \\ & 18 x^2+400+120 x+144+120 x=0 \\ & 18 x^2+240 x+544=0 \\ & 9 x^2+120 x+272=0 \end{aligned} $

On using quadratic formula

$ x_1=\frac{-20-8 \sqrt{2}}{3}, x_2=\frac{-20+8 \sqrt{2}}{3} $

Now, by using $x_1$ and $x_2$ we will find $y_1$ and $y_2$.

$ \begin{aligned} & y_1=\frac{1}{3}[-20-8 \sqrt{2}+20]=-\frac{8 \sqrt{2}}{3} \\ & y_2=\frac{1}{3}[-20+8 \sqrt{2}+20]=\frac{8 \sqrt{2}}{3} \\ & \text { So, } C \equiv\left(x_1, y_1\right) \equiv\left(\frac{-20-8 \sqrt{2}}{3} \frac{-8 \sqrt{2}}{3}\right) \\ & \text { and } D \equiv\left(x_2, y_2\right) \equiv\left(\frac{-20+8 \sqrt{2}}{3}, \frac{8 \sqrt{2}}{3}\right) \end{aligned} $

$ \begin{aligned} &\text { Distaance between } C \text { and } D\\ &D_{C D}=\sqrt{\begin{array}{r} \left.\frac{-20-8 \sqrt{2}}{3}-\frac{-20+8 \sqrt{2}}{3}\right)^2 +\left(-\frac{8 \sqrt{2}}{3}-\frac{8 \sqrt{2}}{3}\right)^2 \end{array}} \end{aligned} $

$ =\sqrt{2\left(\frac{6 \sqrt{2}}{3}\right)^2}=\frac{16 \times 2}{3}=\frac{32}{3} $

Given, when we shift the origin from $(0,0)$ to $(h, k)$, the coordinates $(x, y)$ transform to $\left(x^{\prime}, y^{\prime}\right)$ where $x^{\prime}=x-h$ and $y^{\prime}=y-k$.

Given the equation,

$ x^2+2 x+2 y-7=0 $

On substitute $x=x^{\prime}+h, y=y^{\prime}+k$, into the original equation

$ \begin{aligned} & \left(x^{\prime}+h\right)^2+2\left(x^{\prime}+h\right)+2\left(y^{\prime}+k\right)-7=0 \\ & x^{\prime 2}+2 h x^{\prime}+h^2+2 x^{\prime}+2 h+2 h+2 y^{\prime}+2 k \\ & -7=0 \end{aligned} $

Since, the transformed equation does not contain $x^{\prime}$ and constant terms, we must have equation that does not contain $x^{\prime}$ and constant terms.

$ \begin{aligned} & \left(x^{\prime}\right)^2+2 h x^{\prime}+h^2+2 x^{\prime}+2 h \\ & +2 y^{\prime}+2 k-7=0 \\ & \Rightarrow \quad(2 h+2) x^{\prime}=0 \Rightarrow h=-1 \\ & \text { and } h^2+2 h+2 k-7=0 \\ & \Rightarrow(-1)^2=2(-1)+2 k-7=0 \\ & \Rightarrow \quad 1-2+2 k-7=0 \\ & \Rightarrow \quad-1+2 k-7=0 \\ & \Rightarrow \quad 2 k-8=0 \\ & \Rightarrow 1 \quad k=4 \\ & \text { Solve for, } 2 h+k=2 \times(-1)+4 \\ & =-2+4=2 \end{aligned} $

To determine the value of $ h^2 + k^2 $, consider the problem of finding $ h $ and $ k $ such that the line $(\alpha+1) x + \alpha y + \alpha = 1$ passes through the fixed point $(h, k)$ for all values of $\alpha$.

Start by substituting $ x = h $ and $ y = k $ into the line equation:

$ (\alpha + 1)h + \alpha k + \alpha = 1 $

Simplify this equation:

$ \alpha h + h + \alpha k + \alpha = 1 $

Further simplify to group terms containing $\alpha$:

$ \alpha(h + k + 1) + h = 1 $

For this equation to hold true for all $\alpha$, the coefficient of $\alpha$ must be zero, and the constant term should equal 1. Therefore:

$ h + k + 1 = 0 $

This implies:

$ h + k = -1 $

Since the equation must hold for all values of $\alpha$, set the constant term:

$ h = 1 $

Substitute $ h = 1 $ into $ h + k = -1 $ to find $ k $:

$ 1 + k = -1 $

Solving for $ k $ gives:

$ k = -2 $

Finally, calculate $ h^2 + k^2 $:

$ \begin{aligned} h^2 + k^2 & = 1^2 + (-2)^2 \\ & = 1 + 4 \\ & = 5 \end{aligned} $

Thus, the value of $ h^2 + k^2 $ is 5.

Given, lines are

$ 3 x+y+15=0 $

and $\quad 3 x^2+12 x y-13 y^2=0$

We can write as

$ (3 x-y)(x+13 y)=0 $

Thus, two lines

$ 3 x-y=0 $

Now, we have three lines

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

$(3 x-y=0) \Rightarrow y=3 x$

$(x+13 y=0)$

Intersection of $3 x-y=0$ and $x+13 y=0$

$ y=3 x $

put in Eq. (iii), we get

$ \begin{array}{r} x+13 \times 3 x=0 \\ 40 x=0 \\ x=0 \\and\,\,\,\,\,\,\,\,\,\,\,\,\,\, y=0 \end{array} $

On putting in Eq. (i), we get

$ \begin{aligned} 3 x+3 x+15 & =0 \\ 6 x & =-15 \\ x & =\frac{-5}{2}, y=\frac{-15}{2} \end{aligned} $

Intersection of $(3 x+y+15=0)$ and $(x+13 y=0)$

$ y=\frac{-x}{13}, 3 x-\frac{x}{13}+15=0 $

$ 39 x-x+13 \times 15=0 $

$ 38 x=-195 $

$ x=-\frac{195}{38} $

$ y=-\left(\frac{-195}{38}\right) \frac{1}{\times 13} $

$ y=\frac{15}{38} $

$ \begin{aligned} &\text { Area }\\ &\begin{aligned} & \left.=\frac{1}{2} \right\rvert\,\left(x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right) \mid\right. \\ & =\frac{1}{2} \left\lvert\, 0\left(\frac{-15}{2}-\frac{5}{38}\right)+\left(\frac{-5}{2}\right)\left(\frac{5}{38}-0\right)+\left(\frac{-195}{38}\right)\left(0+\frac{15}{2}\right) \right\rvert\, \\ & =\frac{1}{2}\left[\frac{-5}{2} \times \frac{5}{38}-\frac{195}{38} \times \frac{15}{2}\right] \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & =\frac{1}{2}\left[\left|\frac{-75}{76}-\frac{2925}{76}\right|\right] \\ & =\frac{1}{2}\left[\frac{75+2925}{76}\right]=\frac{1}{2} \times \frac{3000}{76} \\ & =\frac{1500}{76} \approx 19.74 \text { sq units. } \end{aligned}\\ &\text { Check from option (a) } \frac{15 \sqrt{3}}{2} \approx 19.74 \text {. } \end{aligned} $

Curve, $2 x^2-y^2+3 x+2 y=0$

Let the chord be $y=m x+c$

So, find $O A \times O B$ where $A$ and $B$ are points of intersection of chord and circle, we use homogenisation.

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

$\therefore O A$ and $O B$ are right angles

$ \begin{aligned} m_1 m_2=\frac{a}{b} & =-1 \\ \frac{4 c-3 m}{2-2 c} & =-1 \\ 4 c-3 m & =-2+2 c \\ 4 c-2 c & =-2+3 m \\ 2 c & =3 m-2 \\ c & =\frac{3}{2} m-1 \end{aligned} $

Hence, chord is

$ \begin{aligned} & y=m x+\frac{3}{2} m-1 \\ & y=m\left(x+\frac{3}{2}\right)-1 \end{aligned} $

The equation always satisfy $\frac{-3}{2},-1$ or

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

So, $(-3,-2)$ are the fixed point.

Given, equation is

$ S=2 x^2+4 x y-3 y^2+4 x+4 y-14=0 \quad \ldots \text { (i) }$

On partial differentiation of Eq. (i) w.r.t $x$ and $y$, we get

$ \begin{aligned} & \frac{\delta S}{\delta x}=4 x+4 y+4=0 \quad \ldots \text { (ii) } \\ & \frac{\delta S}{\delta y}=4 x-6 y+4=0 \quad \ldots \text { (iii) } \end{aligned} $

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

$ x=-1 \text { and } y=0 $

Thus, the point $(x, y)=(-1,0)$

Let the new coordinates of the point $(x, y)$ be

$ x=X-1, y=Y+0 $

$\therefore$ The transformed equation of

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

$(X-1)^2+(Y+0)^2-3(X-1)(Y+0)+4(Y+0)+3=0$

$ \begin{gathered} \Rightarrow X^2+1-2 X+Y^2-3 X Y+3 Y +4 Y+3=0 \\ \Rightarrow X^2+Y^2-3 X Y-2 X+7 Y+4=0 \end{gathered} $

Replace $X$ and $Y$ by $x$ and $y$, we get

$ x^2+y^2-3 x y-2 x+7 y+4=0 $

Given lines are

$ \begin{array}{r} x+y+2=0 \\ 2 x+y+8=0 \\ x-y-2=0 \end{array} $

The graph of these lines is as follows

These lines form a triangle as shown in the figure. These equations will form right angle triangle. So, circumcentre will be at mid-point of $A C$.

$\therefore$ Circumcentre is $(-4,0)$ i.e. point $D $(-4,0)$.

To find $\alpha + \beta$, we start with the line equation $2x - 3y + 5 = 0$, which can be rewritten in slope-intercept form as:

$ y = \frac{2}{3}x + \frac{5}{3} $

The slope of this line is $\frac{2}{3}$. A line segment is perpendicular to this line if their slopes multiply to $-1$. Therefore, if the line segment's slope is $m_1$, then:

$ \frac{2}{3} \cdot m_1 = -1 $

We define the slope of the line segment connecting $(1, -2)$ and $(\alpha, \beta)$ as:

$ m_1 = \frac{\beta + 2}{\alpha - 1} $

Substituting in the perpendicular condition:

$ \frac{2}{3} \cdot \frac{\beta + 2}{\alpha - 1} = -1 $

Solving this, we multiply both sides by $(\alpha - 1)$:

$ 2(\beta + 2) = -3(\alpha - 1) $

Expanding and simplifying:

$ 2\beta + 4 = -3\alpha + 3 $

$ 3\alpha + 2\beta + 1 = 0 \quad \ldots \text{(i)} $

Next, find the midpoint of the segment connecting $(1, -2)$ and $(\alpha, \beta)$:

$ \left(\frac{1 + \alpha}{2}, \frac{-2 + \beta}{2}\right) $

Substituting this midpoint into the given line equation:

$ 2\left(\frac{1 + \alpha}{2}\right) - 3\left(\frac{-2 + \beta}{2}\right) + 5 = 0 $

Expanding the expression:

$ 1 + \alpha + 3 + \frac{3\beta}{2} + 5 = 0 $

Simplifying:

$ 2\alpha - 3\beta + 18 = 0 \quad \ldots \text{(ii)} $

Now, solve the system of linear equations formed by equations (i) and (ii):

From (i):

$ 3\alpha + 2\beta + 1 = 0 $

From (ii):

$ 2\alpha - 3\beta + 18 = 0 $

To solve, let's use substitution or elimination. Subtract (i) from (ii):

$ (2\alpha - 3\beta + 18) - (3\alpha + 2\beta + 1) = 0 $

Simplifying:

$ - \alpha - 5\beta + 17 = 0 $

$ \alpha + 5\beta = 17 \quad \ldots \text{(iii)} $

Now solve equations (i) and (iii):

$3\alpha + 2\beta + 1 = 0$

$\alpha + 5\beta = 17$

Multiply equation 2 by 3, then subtract equations:

$ 3\alpha + 15\beta = 51 $

Subtract from equation 1:

$ (3\alpha + 2\beta + 1) - (3\alpha + 15\beta) = 0 $

Simplifying:

$ -13\beta = -52 $

$ \beta = 4 $

Substitute $\beta = 4$ back into equation $\alpha + 5\beta = 17$:

$ \alpha + 5(4) = 17 $

$ \alpha + 20 = 17 $

$ \alpha = -3 $

Thus, $\alpha + \beta = -3 + 4 = 1$.

To find the area of the triangle formed by the lines given, we start with the lines:

$ -15 x^2 + 4 xy + 4 y^2 = 0 $

and

$ x = \alpha $

These equations indicate that we need the points of intersection resulting from these lines.

Finding Intersection Points

From the first equation, factor to find:

$ -15 \alpha^2 + 4 \alpha y + 4 y^2 = 0 $

Re-arrange to express it in a simpler form:

$ 4(y^2 + \alpha y) = 15 \alpha^2 $

Factor further:

$ 4y(y+\alpha) = 15 \alpha^2 $

This implies the solutions for $ y $ are:

$ y = \frac{15}{4} \alpha^2 \quad \text{or} \quad y = 15 \alpha^2 - \alpha $

Vertices of the Triangle

The vertices of the triangle are:

$ A(0,0) $

$ B\left(\alpha, \frac{15}{4} \alpha^2 \right) $

$ C\left(\alpha, 15 \alpha^2 - \alpha \right) $

Calculating the Area

To find the area of the triangle $ ABC $, use the determinant formula:

$ \text{Area} = \frac{1}{2}\left|\begin{array}{ccc} 0 & 0 & 1 \\ \alpha & \frac{15}{4} \alpha^2 & 1 \\ \alpha & 15 \alpha^2 - \alpha & 1 \end{array}\right| $

Simplify the determinant by taking $\alpha$ common:

$ = \frac{1}{2} \alpha^2\left|\begin{array}{ccc} 0 & 0 & 1 \\ 1 & \frac{15}{4} \alpha & 1 \\ 1 & 15 \alpha - 1 & 1 \end{array}\right| $

Evaluate the determinant:

$ = \frac{1}{2} \alpha^2 \left(15 \alpha - 1 - \frac{15}{4} \alpha \right) $

$ = \frac{1}{2} \alpha^2 \left(\frac{60 \alpha - 4 - 15 \alpha}{4}\right) $

$ = \frac{1}{8} \alpha^2 (45 \alpha - 4) $

Solving for $|\alpha|$

Given that the area is 200 square units:

$ \frac{1}{8} \alpha^2 (45 \alpha - 4) = 200 $

$ \alpha^2 (45 \alpha - 4) = 1600 $

Solving for $|\alpha|$, we find:

$ \alpha^2 = 1600 $

$ |\alpha| = 40 $

To find the equation of a straight line passing through the point of intersection of the lines given by the equation $x^2 + 4xy + 3y^2 - 4x - 10y + 3 = 0$ and the point $ (2, 2) $, follow these steps:

Identify the line equation coefficients:

$ a = 1 $, $ b = 3 $, $ h = 2 $, $ g = -2 $, $ f = -5 $, $ c = 3 $.

Calculate the point of intersection:

The coordinates for the point of intersection of the lines are given by:

$ \left(\frac{hf - bg}{ab - h^2}, \frac{gh - af}{ab - h^2}\right) $

Substituting the values:

$ \left(\frac{2 \times -5 - 3 \times -2}{1 \times 3 - 2^2}, \frac{-2 \times 2 - 1 \times -5}{1 \times 3 - 2^2}\right) $

$ = \left(\frac{-10 + 6}{3 - 4}, \frac{-4 + 5}{3 - 4}\right) $

$ = (4, -1) $

Find the slope of the line between $ (4, -1) $ and $ (2, 2) $:

The slope $ m $ is:

$ m = \frac{2 - (-1)}{2 - 4} = \frac{3}{-2} $

Equation of the line using point-slope form:

Use the point-slope form $ y - y_1 = m(x - x_1) $:

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

Simplifying:

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

$ -2y + 4 = 3x - 6 $

$ 3x + 2y - 10 = 0 $

Thus, the equation of the line is $ 3x + 2y - 10 = 0 $.

To eliminate the $y$-term from the given equation $x^2-y^2+2y-1=0$, we can shift the origin to the point $(0, K)$. Here's how the transformation is done:

Let the transformation replace $y$ with $Y + K$ and $x$ with $X$. Thus:

$ y = Y + K \quad \text{and} \quad x = X $

Substitute these into the original equation:

$ X^2 - (Y + K)^2 + 2(Y + K) - 1 = 0 $

Simplify the expression:

$ X^2 - (Y + K)^2 + 2(Y + K) - 1 = X^2 - Y^2 - 2YK - K^2 + 2Y + 2K - 1 = 0 $

Rearrange the terms to group them properly:

$ X^2 - Y^2 - Y(2K - 2) - K^2 + 2K - 1 = 0 $

To remove the $Y$ term, set the coefficient of $Y$ to zero:

$ 2K - 2 = 0 \quad \Rightarrow \quad K = 1 $

Substitute $K = 1$ back into the equation:

$ X^2 - Y^2 - 1 + 2 - 1 = 0 \quad \Rightarrow \quad X^2 - Y^2 = 0 $

Therefore, the transformed equation after the translation is $x^2 - y^2 = 0$.

To find the value of $ a + 2b + 1 $ for the line $ L $ described by the equation $\frac{x}{a} + \frac{y}{b} = 1$, follow the steps below:

Line Equation and Intersection Points:

Let the coordinates of $ A $ be $(h, k)$ and for $ B $ be $(m, n)$.

The midpoint of $ AB $ is given to be $(1, 2)$. Thus:

$ \frac{h + m}{2} = 1 \quad \text{and} \quad \frac{k + n}{2} = 2 $

Solving gives:

$ m = 2 - h \quad \text{and} \quad n = 4 - k $

Determine the Coordinates:

$ A $ lies on the line $ 3x - 2y - 1 = 0 $:

$ 3h - 2k - 1 = 0 \quad \Rightarrow \quad 3h - 2k = 1 \quad \text{(Equation 1)} $

$ B $ lies on the line $ x + 2y + 1 = 0 $:

$ (2-h) + 2(4-k) + 1 = 0 \quad \Rightarrow \quad 2 - h + 8 - 2k + 1 = 0 $

Simplifying:

$ h + 2k = 11 \quad \text{(Equation 2)} $

Solve for $ h $ and $ k $:

By solving Equations 1 and 2 simultaneously:

$ \begin{align*} 3h - 2k &= 1 \\ h + 2k &= 11 \end{align*} $

Adding these equations:

$ 4h = 12 \quad \Rightarrow \quad h = 3 $

Substituting into Equation 2:

$ 3 + 2k = 11 \quad \Rightarrow \quad 2k = 8 \quad \Rightarrow \quad k = 4 $

Hence, the coordinates of $ A $ are $(3, 4)$ and $ B $ becomes $(-1, 0)$.

Find $ a $ and $ b $:

Both points $(3, 4)$ and $(-1, 0)$ satisfy the line $ L $:

$ \frac{3}{a} + \frac{4}{b} = 1 \quad \text{and} \quad -\frac{1}{a} = 1 $

Solving for $ a $:

$ -\frac{1}{a} = 1 \quad \Rightarrow \quad a = -1 $

Substitute $ a = -1 $ into the equation with $(3, 4)$:

$ -3 + \frac{4}{b} = 1 \quad \Rightarrow \quad \frac{4}{b} = 4 \quad \Rightarrow \quad b = 1 $

Calculate $ a + 2b + 1 $:

$ a + 2b + 1 = -1 + 2 \times 1 + 1 = 2 $

This confirms that the value of $ a + 2b + 1 $ is 2.