Straight Lines and Pair of Straight Lines

Equation of $A B: y-2=-1(x)$

Equation of $B C: y-2=1(x)$

Equation of $A C$ : $y-0=0(x-2)$

Let the variable point be $(x, y)$.

$ \begin{gathered} a=\frac{|x+y-2|}{\sqrt{2}}, \\ b=\frac{|x-y+2|}{\sqrt{2}} \\ c=|y| \\ 2 b=a+c \\ 2\left|\frac{x-y+2}{\sqrt{2}}\right|=\left|\frac{x+y-2}{\sqrt{2}}\right|+|y| \end{gathered} $

Hence lows is

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

$ \begin{aligned} & \text { } 4 a^2+9 b^2-12 a b-c^2=0 \\ & \Rightarrow(2 a-3 b)^2-c^2=0 \\ & \Rightarrow(2 a-3 b-c)(2 a-3 b+c)=0 \\ & \Rightarrow 2 a-3 b-c=0 \text { or } 2 a-3 b+c=0 \end{aligned} $

$\Rightarrow a x+b y+c$ is passing through $(-2, 3)$ and also passing through ( $2,-3$ )

Let slope of line $A B$ be -1 and slope of $A D$ be 7

$ \begin{aligned} & \tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right|=\left|\frac{7-(-1)}{1+7 \times-1}\right| \\ & \tan \theta=\frac{8}{6}=\frac{4}{3} \end{aligned} $

Let slope of $A C$ be $m$.

$ \begin{aligned} & \Rightarrow\left|\frac{-1-m}{1-m}\right|=\left|\frac{m-7}{1+7 m}\right| \\ & \Rightarrow \frac{-1-m}{1-m}= \pm\left(\frac{m-7}{1+7 m}\right) \end{aligned} $

$ \begin{aligned} & \Rightarrow(-1-m)(1+7 m)=(1-m)(m-7) \\ & \Rightarrow-1-7 m-m-7 m^2=m-7-m^2+7 m \\ & \Rightarrow-7 m^2-8 m-1=-m^2+8 m-7 \\ & \Rightarrow 6 m^2+16 m-6=0 \\ & \Rightarrow 3 m^2+8 m-3=0 \\ & \Rightarrow 3 m^2+9 m-m-3=0 \\ & \Rightarrow 3 m(m+3)-1(m+3)=0 \\ & \Rightarrow m=-3 \text { or } m=\frac{1}{3} \end{aligned} $

Equation of $A C: y-3=-3(x-1)$

$ \begin{aligned} & y-3=-3 x+3 \\ & \Rightarrow 3 x+y-6=0 \end{aligned} $

Solving $3 x+y-6=0$ and $15 x-5 y=6$

we get $\alpha$ and $\beta$

$ \begin{aligned} & 3 x+y=6 \\ & \Rightarrow 3 x-y=\frac{6}{5} \Rightarrow 2 y=6-\frac{6}{5} \\ & \Rightarrow y=\frac{12}{5}=\beta \Rightarrow 3 x=\frac{6-12}{5} \\ & \Rightarrow 3 x=\frac{18}{5} \Rightarrow x=\frac{6}{5}=\alpha \\ & \Rightarrow \alpha+\beta=\frac{18}{5} \end{aligned} $

The first given equation is $3x^2 + 2hxy - 3y^2 = 0$. This equation represents two straight lines passing through the origin because there are no constant terms.

This equation can be written as the product of two linear factors: $3x^2 + 2hxy - 3y^2 = (3x - y)(x + 3y) = 0$, which means $h=4$.

The second equation is $3x^2 + 2hxy - 3y^2 + 2x - 4y + c = 0$. This also represents two straight lines, but they are not passing through the origin.

Since both equations together represent the four sides of a square, the sides must be parallel and equal in pairs. Therefore, the second equation must be the same as the first one but shifted by some constants.

The second equation can be written as $(3x - y + c_1)(x + 3y + c_2) = 0$ for some $c_1$ and $c_2$.

Now, expanding: $(3x - y + c_1)(x + 3y + c_2)$ gives $3x^2 + 2xy - 3y^2 + (3c_2 + c_1)x + (3c_1 - c_2)y + (c_1c_2)$.

Comparing this with $3x^2 + 2hxy - 3y^2 + 2x - 4y + c = 0$, we match coefficients:

By comparing the $x$ terms: $3c_2 + c_1 = 2$.

By comparing the $y$ terms: $3c_1 - c_2 = -4$.

Solving these two equations:
$3c_2 + c_1 = 2$ … (i)
$3c_1 - c_2 = -4$ … (ii)

Multiply (i) by 3: $9c_2 + 3c_1 = 6$.
Add (ii): $3c_1 - c_2 = -4$
$9c_2 + 3c_1 + 3c_1 - c_2 = 6 -4$
$8c_2 + 6c_1 = 2$

Insert the solution from the original explanation: $c_1 = -1$ and $c_2 = 1$.

The constant term is $c = c_1c_2 = -1 \times 1 = -1$.

From earlier, we found $h = 4$, so $\frac{h}{c} = \frac{4}{-1} = -4$.

$ \begin{aligned} & \text { }(2,3) \xrightarrow[\text { shifted to }(3,2)]{\text { Origin }}(a, b) \\ & \Rightarrow \quad 3+a=2 \text { and } b+2=3 \\ & \Rightarrow \quad a=-1 \text { and } b=1 \end{aligned} $

$ \begin{aligned} &\begin{array}{lll} \hline & -1 & 1 \\ \hline c & \cos \theta & \sin \theta \\ \hline d & -\sin \theta & \cos \theta \\ \hline \end{array}\\ &\begin{gathered} c=-\cos \theta+\sin \theta \\ d=+\sin \theta+\cos \theta \\ -c+d=2 \cos \theta=2 \times \frac{1}{\sqrt{2}}=\sqrt{2} \end{gathered} \end{aligned} $

Solving the lines equation, we get point $A, B$ and $C$.

$ \begin{aligned} & \left.\begin{array}{l} x+y+4=0 \\ x-2 y-4=0 \end{array}\right\} \Rightarrow \begin{array}{l} C_x=\frac{-4}{3} \\ C_y=\frac{-8}{3} \end{array} \\ & \left.\begin{array}{l} x+y+4=0 \\ 3 x+4 y-2=0 \end{array}\right\} \begin{array}{l} B_x=-18 \\ B_y=14 \end{array} \\ & \left.\begin{array}{l} 3 x+4 y-2=0 \\ x-2 y-4=0 \end{array}\right\} \begin{array}{l} A_x=2 \\ A_y=-1 \end{array} \end{aligned} $

$A B \neq A C \neq B C \Rightarrow \triangle A B C$ is a scalene triangle.

$ \begin{aligned} &\text { Let equation of line } L=0 \text { be }\\ &y=m x+c \end{aligned} $

$ \begin{aligned} & 4=12 m+c \\ & \Rightarrow y=m x+4-12 m \end{aligned} $

⇒ Length of $X$-intercept of $L=\left|\frac{12 m-4}{m}\right|$

Length of $Y$-intercept of $L=|12 m-4|$

$ \begin{aligned} & \frac{1}{2} \times \frac{(12 m-4)^2}{|m|}=12 \\ & \Rightarrow \quad 8(3 m-1)^2=12 m(m>0) \\ & \Rightarrow \quad 2(3 m-1)^2=3 m \\ & \Rightarrow \quad 18 m^2-12 m+2=3 m \\ & \Rightarrow \quad 18 m^2-15 m+2=0 \\ & \Rightarrow \quad 18 m^2-3 m-12 m+2=0 \\ & \Rightarrow \quad 3 m(6 m-1)-2(6 m-1)=0 \\ & \quad m=\frac{1}{6} \text { or } m=\frac{2}{3} . \end{aligned} $

$ \begin{aligned} &\text { As intercept of } L \text { on } X \text {-axis is negative }\\ &\begin{aligned} \Rightarrow & m=\frac{1}{6}\left(m<\frac{1}{3}\right) \\ \Rightarrow & P=\frac{12 m-4}{m} \times\{(4-12 m)\}^2 \\ \Rightarrow & P=(12 m-4)^3 \times \frac{1}{m} \\ & =\left(12 \times \frac{1}{6}-4\right)^3 \times 6 \\ & =-8 \times 6=-48 \end{aligned} \end{aligned} $

Quadrilateral is a parallelogram.

∴ Area of parallelogram

$ \begin{aligned} & =\frac{\left|c_1-c_2\right| \times\left|d_1-d_2\right|}{\left|a_1 b_2-a_2 b_1\right|} \\ & =\frac{\left|\frac{9}{2}-3\right| \times\left|3-\frac{11}{3}\right|}{|1 \times-2-1 \times 2|} \\ & =\frac{3 \times 2}{2 \times 3 \times 4}=\frac{1}{4} \end{aligned} $

Let

$ \begin{aligned} & S=2 x^2+5 x y-3 y^2+2 g x+2 f y+C \\ & \Rightarrow \quad \frac{\partial S}{\partial x}=4 x+5 y+2 g \\ & \Rightarrow \quad \frac{\partial S}{\partial y}=5 x-6 y+2 f \end{aligned} $

$[\because(-1,-1)$ is the point in intersection]

$ \begin{array}{ll} \Rightarrow & 4 \times(-1)-5 \times 1+2 g=0 \\ \Rightarrow & 2 g=9 \\ \text { and } & 5 \times(-1)-6 \times(-1)+2 f=0 \\ & 2 f=-1 \\ & g+f=4 \end{array} $

Also, $(-1,-1)$ will satisfy

$ \begin{aligned} & 2 x^2+5 x y-3 y^2+2 g x+2 f y+c=0 \\ & \Rightarrow \quad 2+5-3-2 g-2 f+c=0 \\ & \Rightarrow \quad 4-8+c=0 \\ & \,\,\,\,\,\,\,\,\, \quad c=4 \\ & \Rightarrow \quad g+f=c \end{aligned} $

$ \begin{aligned} &h=\frac{3 a}{5}, k=\frac{2 b}{5}\\ &\text { ∴ Equation of } A B\\ &\frac{3}{a}+\frac{2}{b}-1=0 \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \frac{3}{\frac{5 h}{3}}+\frac{2}{5 k}=1 \quad\left[\because h=\frac{3 a}{5}, k=\frac{2 b}{5}\right] \\ & \frac{9}{h}+\frac{4}{k}=5 \end{aligned}\\ &\text { Locus, }\\ &\begin{aligned} & \frac{9}{x}+\frac{4}{y}=5 \\ & 4 x+9 y=5 x y \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { } \begin{aligned} & X= x-h, Y=y-k \\ & X= x+1, Y=y-2 \\ & x=X-1, y=Y+2 \\ & \Rightarrow \quad 2(X-1)^2-(X-1)(Y+2)+(Y+2)^2-3(X-1)+4(Y+2) -5=0\end{aligned} \end{aligned} $

$ \begin{gathered} \Rightarrow \quad 2 X^2-4 X+2-X Y-2 X+Y+2 \\ +Y^2+4 y+4 \\ -3 X+3+4 Y+8-5=0 \end{gathered} $

$ \begin{array}{lr} 2 h=-1 & a=2 \\ 2 g=-4-2-3=-9 & b=1 \\ 2 f=1+4+4=9 & c=2+2+4+3+8-5 \\ & =14 \end{array} $

$ \begin{array}{ll} \therefore & 2(f+g+h)=-1 \\ \text { And } & c-5(a+b)=14-15=-1 \end{array} $

Using parametric form of line

$ \begin{aligned} & \frac{x-x_1}{\cos \theta}=\frac{y-y_1}{\sin \theta}= \pm r \\ \Rightarrow \quad & \frac{x+2}{\cos 60^{\circ}}=\frac{y-4}{\sin 60^{\circ}}= \pm 6 \\ \Rightarrow & x= \pm 6 \cos 60^{\circ}-2, y= \pm 6 \sin 60^{\circ}+4 \\ & x= \pm 6\left(\frac{1}{2}\right)-2 y= \pm 3 \sqrt{3}+4 \\ = & \pm 3-2 \end{aligned} $

∵ Point lies in 3rd 24

$ \begin{aligned} & x=-5, y=-3 \sqrt{3}+4 \\ & p=-5, q=4-3 \sqrt{3} \end{aligned} $

$ \begin{aligned} & \therefore \sqrt{p^2+q^2-8 q} \\ & \quad=\sqrt{25+16+27-24 \sqrt{3}-32+24 \sqrt{3}} \\ & \quad=\sqrt{36}=6 \end{aligned} $

Foot of perpendicular of point $(2,-3)$ with respect to line $4 x-3 y+8=0$

$ \begin{aligned} & \frac{a-2}{4}=\frac{b+3}{-3}=\frac{-1(4(2)-3(-3)+8)}{4^2+3^2} \\ \Rightarrow & \frac{a-2}{4}=\frac{b+3}{-3}=-1 \\ \Rightarrow & x=-2, b=0 \\ \therefore & \quad a^3-b^3=k^3 \Rightarrow k=-2 \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } Q(\alpha, \beta) \text { and } R(\gamma, \delta), P(1,2) \\ & \therefore \frac{\alpha-1}{1}=\frac{\beta-2}{1}=\frac{-2(1+2+1)}{1+1} \\ & \Rightarrow \frac{\alpha-1}{1}=\frac{\beta-2}{1}=-4 \Rightarrow \alpha=-3, \beta=-2 \end{aligned} \end{aligned} $

$ \begin{aligned} & \therefore Q=(-3,-2) \\ & \text { And } \frac{\gamma+3}{1}=\frac{\delta+2}{-1}=\frac{-2(-3+2-1)}{1+1} \\ & \Rightarrow \frac{\gamma+3}{1}=\frac{\delta+2}{-1}=2 \Rightarrow \gamma=-1, \delta=-4 \\ & R(-1,-4) \\ & \qquad M=\left(\frac{1-3}{2}, \frac{2-2}{2}\right)=(-1,0) \\ & \quad N=\left(\frac{-3-1}{2}, \frac{-2-4}{2}\right)=(-2,-3) \\ & \quad M N=\sqrt{1+9}=\sqrt{10} \end{aligned} $

$6 x^2+2 h x y+4 y^2=0$

Let slopes be $m_1$ of $m_2$

$ \begin{array}{ll} \therefore & \frac{m_1}{m_2}=\frac{2}{3} \\ \text { And } & m_1 m_2=\frac{a}{b}=\frac{6}{4} \\ & m_1 \cdot \frac{3 m_1}{2}=\frac{6}{4} \\ & m_1^2=1 \Rightarrow m= \pm 1 \quad \text { (+ ve angle) } \\ & m_1=1 \\ & m_2=\frac{3}{2} \\ \therefore & m_1+m_2=\frac{-2 h}{b} \\ \Rightarrow & 1+\frac{3}{2}=\frac{-2 h}{4} \\ \Rightarrow & h=-5 \end{array} $

Now, equation

$ 2 x^2+x y-6 y^2+k=0 $

Old equation

$ 2 x^2+x y-6 y^2-13 x+9 y+15=0 $

Origin is shifted to $(a, b)$

$ \begin{aligned} 2(x-a)^2+ & (x-a)(y-b)-6(y-b)^2 \ -13(x-a)+9(y-b)+15=0 \end{aligned} $

∴ Coefficient of $x=0$

$ 4 a+b-13=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Coefficient of $y=0$

$ a-12 b+9=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

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

$ a=3 \text { and } b=1 $

∴ Constant

$ \Rightarrow k=2 a^2+a b-6 b^2-13 a+9 b+15 $

Put $a=3$ and $b=1$

$ k=0 $

$ \text { } \begin{aligned} Point\,\,\,P & =\left(\frac{-20+12}{-5+4}, \frac{-30+20}{-5+4}\right) \\ & =\left(\frac{-8}{-1}, \frac{-10}{-1}\right)=(8,10) \end{aligned} $

$\because P$ lies on line $6 x+3 y+k=0$

$ \begin{array}{ll} \therefore \quad & 6(8)+3(10)+k=0 \\ & k=-78 \end{array} $

∴ Equation of line $6 x+3 y-78=0$

$ \Rightarrow 2 x+y-26=0 $

∴ Point of intersection of $2 x+y-26=0$ and $x-y+1=0$ is $(g, h)=(8,9)$

$ \begin{array}{ll} \therefore \quad & g-h+1=0 \\ & h=g+1 \end{array} $

$ \begin{gathered} x-1= \pm \frac{1}{2} \cos \theta \\ y-2= \pm \frac{1}{2} \sin \theta \end{gathered} $

$ \begin{aligned} &\begin{aligned} & x= \pm \frac{\cos \theta}{2}+1 \\ & y= \pm \frac{\sin \theta}{2}+2 \end{aligned}\\ &\begin{aligned} & \therefore \theta \text { lies on line } x+\sqrt{3} y-2 \sqrt{3}=0 \\ & \pm \frac{\cos \theta}{2}+1 \pm \frac{\sqrt{3}}{2} \sin \theta+2 \sqrt{3}-2 \sqrt{3}=0 \\ & \Rightarrow \quad \pm\left(\frac{\sqrt{3}}{2} \sin \theta+\frac{1}{2} \cos \theta\right)=-1 \\ & \Rightarrow \quad \sin \left(\theta+\frac{\pi}{6}\right)= \pm 1 \\ & \Rightarrow \quad \theta+\frac{\pi}{6}=\frac{\pi}{2} \text { or }-\frac{\pi}{2} \\ & \Rightarrow \quad \theta=\frac{\pi}{3} \text { or } \frac{-2 \pi}{3} \end{aligned} \end{aligned} $

$ \begin{aligned} & \left|\begin{array}{ccc} 1 & -2 & 1 \\ 2 & -3 & -1 \\ 3 & -1 & k \end{array}\right|=0 \\ & \Rightarrow 1(-3 k-1)+2(2 k+3)+1(-2+9)=0 \\ & \Rightarrow-3 k-1+4 k+6+7=0 \\ & \Rightarrow k=-12 \\ & L_1: 3 x-y+k=0 \\ & \Rightarrow 3 x-y-12=0 \Rightarrow m_1=3 \end{aligned} $

$ \begin{aligned} L_2: m x & -3 y+6=0 \\ \Rightarrow \quad m_2 & =\frac{m}{3} \\ \tan \theta & =\left|\frac{m_1-m_2}{1+m_1 m_2}\right| \\ \tan 45^{\circ} & =\left|\frac{3-\frac{m}{3}}{1+m}\right| \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } \quad a x^2+4 x y-2 y^2=0, h=2 a \\ & \tan \theta=\left|\frac{2 \sqrt{h^2-a b}}{a+b}\right| \\ & \Rightarrow \quad 2 \sqrt{10}=\left|\frac{2 \sqrt{4+2 a}}{a-2}\right| \\ & \Rightarrow \quad 10=\frac{4+2 a}{(a-2)^2} \\ & \Rightarrow \quad 5\left(a^2-4 a+4\right)=a+2 \\ & \Rightarrow \quad 5 a^2-21 a+18=0 \\ & \Rightarrow \quad 5 a^2-15 a-6 a+18=0 \\ & \Rightarrow \quad 5 a(a-3)-6(a-3)=0 \\ & \quad a=3, a=\frac{6}{5} \\ & \because \quad a \in Z \\ & \therefore \quad a=3 \\ & \therefore \quad m_1 m_2=\frac{a}{b}=\frac{3}{-2}=\frac{-3}{2} . \end{aligned} \end{aligned} $

Rotation by angle $\theta$ in the positive direction

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

Let $\left(x_2, y_2\right)=(-4 \sqrt{2},-2 \sqrt{2})$

Point before rotation $\left(x_1, y_1\right)$,

$ \begin{aligned} x_1 & =x_2 \cos \left(-\frac{\pi}{4}\right)+y_2 \sin \left(-\frac{\pi}{4}\right) \\ & =x_2\left(\frac{\sqrt{2}}{2}\right)-y_2\left(\frac{\sqrt{2}}{2}\right) \end{aligned} $

And $y_1=x_2\left(\frac{\sqrt{2}}{2}\right)+y_2\left(\frac{\sqrt{2}}{2}\right)$

$ \begin{aligned} \Rightarrow \quad x_1 & =\frac{\sqrt{2}}{2}(-4 \sqrt{2})-\frac{\sqrt{2}}{2}(-2 \sqrt{2}) \\ & =-4+2=-2 \\ \text { And } y_1 & =\frac{\sqrt{2}}{2}(-4 \sqrt{2})+\frac{\sqrt{2}}{2}(-2 \sqrt{2}) \\ & =-4-2=-6 \end{aligned} $

So, point before rotation is $(-2,-6)$ reflection about $y=-x$

$\Rightarrow$ Reflection of point $(x, y)$ is $(-y,-x)$ i.e., $(6,2)$

Translation by +3 in $X$-axis

$ \begin{array}{rlrl} \therefore & (\alpha, \beta) & =(6,-3,2) \\ & & =(3,2) \\ \therefore & \alpha+\beta & =3+2=5 \end{array} $

Let slope of line passing through the point $(4,-3)$ be $m$

And slope of the line joining point $(1,1)$ and $(2,3)$ is $=\frac{3-1}{2-1}=2$

$ \begin{array}{llrl} \therefore & & \tan 45^{\circ} & =\left|\frac{m-2}{1+2 m}\right| \\ \Rightarrow & & 1 & =\left|\frac{m-2}{1+2 m}\right| \end{array} $

taking positive sign, we get

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

Taking negative sign, we get

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

$ \Rightarrow \quad m=\frac{1}{3} $

Since, slope is negative.

$ \therefore \quad m=-3 $

Equation of line is

$ \begin{aligned} & & y+3 & =(-3)(x-4) \\ \Rightarrow & & y+3 & =-3 x+12 \\ \Rightarrow & & 3 x+y & =9 \end{aligned} $

$\therefore x$-intercept, when $y=0$, is $3 x=9$

$ \Rightarrow \quad x=3 $

and $y$-intercept, when $x=0$ is $y=9$

$\therefore$ Sum of intercepts $=3+9=12$

We have $O(0,0), B(-3,-1), C(-1,-3)$ are vertices of a $\triangle O B C$ and equation of $D E$ is $2 x+2 y+\sqrt{2}=0$

Slope of $B C$ is $\frac{-3-(-1)}{-1-(-3)}=\frac{-2}{2}=-1$

$\therefore$ Slope of attitude from 0 to $B C$ is 1 $(\perp$ to $B C)$

$\therefore$ Equation of altitude with slope passes through $(0,0)$

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

Equation of $B C$

$ y+1=(-1)(x+3) \Rightarrow y=-x-4 $

Point of intersection of line $B C$ and altitude

$ \begin{aligned} & & x & =-x-4 \\ \Rightarrow & & 2 x & =-4 \\ \Rightarrow & & x & =-2 \\ \therefore & & y & =-2 \end{aligned} $

$\therefore$ Foot of perpendicular $A=(-2,-2)$

Now, $2 x+2 y+\sqrt{2}=0$

$ \begin{array}{ll} \Rightarrow & x+y=-\frac{\sqrt{2}}{2} \\ \Rightarrow & x+x=\frac{-\sqrt{2}}{2} \quad(\text { altitude } y=x) \\ \Rightarrow & 2 x=\frac{-\sqrt{2}}{2} \\ \Rightarrow & x=\frac{-\sqrt{2}}{4} \end{array} $

Also, $\quad y=\frac{-\sqrt{2}}{4}$

$\therefore$ Intersection of altitude and $D E$ is P. $\left(\frac{-\sqrt{2}}{4}, \frac{-\sqrt{2}}{4}\right)$

$ \begin{aligned} & \therefore A P=\sqrt{\left(-2+\frac{\sqrt{2}}{4}\right)^2+\left(-2+\frac{\sqrt{2}}{4}\right)^2} \\ & \quad=\sqrt{2\left(-2+\frac{\sqrt{2}}{4}\right)^2}=\sqrt{2}\left(-2+\frac{\sqrt{2}}{4}\right) \\ & \text { And } P O=\sqrt{2\left(\frac{\sqrt{2}}{4}\right)^2} \\ & \quad=\sqrt{2 \times \frac{2}{16}}=\frac{1}{2} \\ & \therefore \frac{P O}{A P}=\frac{\frac{1}{2}}{\sqrt{2}\left(-2+\frac{\sqrt{2}}{4}\right)}=1: 4 \sqrt{2}-1 \end{aligned} $

Given equation of curve $3 x+2 y-3 x y=0$ every point on the curve is the centroid of a triangle formed by the coordinate axes and a line $(L)$ Since, $L$ intersect the coordinate axes

$ \begin{aligned} \therefore \text { let } A & =(a, 0) \\ B & =(0, b) \end{aligned} $

$\therefore$ Centroid of $\triangle O A B$,

$ \begin{aligned} & (x, y)=\left(\frac{0+0+a}{3}, \frac{0+b+0}{3}\right)=\left(\frac{a}{3}, \frac{b}{3}\right) \\ & \Rightarrow a=3 x, b=3 y \end{aligned} $

Equation of line

$ \begin{aligned} & \frac{X}{3 x}+\frac{Y}{3 y}=1 \\ \Rightarrow & \frac{X}{x}+\frac{Y}{y}=3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Since, centroid $(x, y)$ lies on given curve

$ \begin{aligned} & 3 x+2 y-3 x y=0 \\ & \Rightarrow \quad 3 x+2 y=3 x y \\ & \Rightarrow \quad 3 x y=3 x+2 y \Rightarrow 3=\frac{3 x}{x y}+\frac{2 y}{x y} \\ & \Rightarrow \quad \frac{X}{x}+\frac{Y}{y}=\frac{3}{y}+\frac{2}{x} \quad \text { (Using Eq. (i)) } \\ & \Rightarrow \quad \frac{X-2}{x}+\frac{Y-3}{y}=0 \end{aligned} $

$\Rightarrow$ Does not represents to a single line

Now, $3 x+2 y=3 x y$

$ \begin{aligned} & \Rightarrow \quad \frac{3 x}{3 x y}+\frac{2 y}{3 x y}=1 \\ & \Rightarrow \quad \frac{1}{y}+\frac{2}{3 x}=1 \\ & \Rightarrow \quad \frac{1}{\frac{b}{3}}+\frac{2}{3\left(\frac{a}{3}\right)}=1 \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad \frac{3}{b}+\frac{2}{a}=1 \Rightarrow 3 a+2 a=a b \\ & \Rightarrow \quad a b-3 a-2 b=0 \\ & \Rightarrow \quad a=\frac{2 b}{b-3} \end{aligned} $

Consider the equation of line $L$

$ \begin{aligned} & b X+a Y=a b \\ & \Rightarrow \quad b X+\frac{2 b}{b-3} Y=b\left(\frac{2 b}{b-3}\right) \\ & \Rightarrow \quad X+\frac{2}{b-3} Y=\frac{2 b}{b-3} \\ & \Rightarrow \quad X(b-3)+2 Y=2 b \\ & \Rightarrow \quad b(X-2)+(-3 X+2 Y)=0 \end{aligned} $

For all such lines $L$

$ X-2=0 \Rightarrow X=2 $

and $\quad-3 X+2 Y=0$

$\Rightarrow \quad Y=3$

$\Rightarrow \quad L$ is passes through $(2,3)$

$\Rightarrow \quad L$ are concurrent.

Given, equation

$ \left(a^2-3\right) x^2+16 x y-2 a y^2+4 x-8 y-2=0 $

represents a pair of perpendicular line

$ \begin{aligned} & \therefore\left(a^2-3\right)+(-2 a)=0 \\ & \Rightarrow \quad a^2-2 a-3=0 \\ & \Rightarrow \quad a^2-3 a+a-3=0 \\ & \Rightarrow \quad a(a-3)+1(a-3)=0 \\ & \Rightarrow \quad(a-3)(a+1)=0 \\ & \Rightarrow \quad a=3,-1 \end{aligned} $

$\Rightarrow a=3$, since for $a=-1$ lines does not exists.

Let the centroid be $G(x, y)$. The coordinates of the centroid of triangles with vertices $\left(x_1, y_1\right),\left(x_2, y_2\right)$ and $\left(x_3, y_3\right)$ are

$ x=\frac{x_1+x_2+x_3}{3} \text { and } y=\frac{y_1+y_2+y_3}{3} $

Substituting the coordinates of $A, B$ and $P$.

$ \begin{aligned} & x=\frac{2+3+t}{3}=\frac{5+t}{3}, \\ & y=\frac{3+2+t^2}{3}=\frac{5+t^2}{3} \end{aligned} $

$ \begin{array}{ll} \Rightarrow & 3 x-5=t, y=\frac{5+(3 x-5)^2}{3} \\ & (\because t=2 x-5) \\ \Rightarrow & 3 y=5+9 x^2+25-30 x \\ \Rightarrow & 9 x^2-30 x-3 y+30=0 \\ \Rightarrow & 3 x^2-10 x-y+10=0 \end{array} $

The equation of the locus of the centroid of $\triangle A B C$ is $3 x^2-10 x-y+10=0$

Let $S(x, y)=2 x^2-x y-y^2-3 x+3 y$

$ \frac{\partial S}{\partial x}=4 x-y-3, \frac{\partial S}{\partial y}=-x-2 y+3 $

Solve, $\frac{\partial S}{\partial x}=0$ and $\frac{\partial S}{\partial y}=0$, we get

$ 4 x-y-3=0 \text { and }-x-2 y+3=0 $

Solving these two equations for $x=h$ and $y=k$

$ \begin{aligned} & 4 h-k-3=0 \text { and }-h-2 k+3=0 \\ & \therefore h=1, k=1 \end{aligned} $

So, the new origin is $(h, k)=(1,1)$

We know that a general second-degree equation

$A x^2+B x y+C y^2+D x+E y+F=0$, the angle of rotation $\theta$ satisfies

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

Here, $S \equiv 2 x^2-x y-y^2-3 x+3 y=0$, we have

$ \begin{aligned} & A=2 B=-1 \text { and } C=-1 \\ & \therefore \tan 2 \theta=\frac{-1}{2-(-1)}=-\frac{1}{3} \end{aligned} $

Now, $h+k=1+1=2 h-k=1-1=0$,

$ h k=1 \times 1=1, \frac{h}{3 k}=\frac{1}{3} $

So, $\tan 2 \theta=\frac{-1}{3}=\frac{-h}{3 k}$

Give line is $5 x-12 y+6=0$

Slope of this line is $m_1=\frac{-5}{-12}=\frac{5}{12}$

Since, line $L$ is perpendicular to this line, the product of their slope is -1 . Let the slope of line $L$ be $m_L$.

Then, $m_L \times \frac{5}{12}=-1$

$ \Rightarrow \quad m_L=-\frac{12}{5} $

The equation of a line in normal form is $x \cos \theta+y \sin \theta=P$, where $P$ is the distance from the origin to the line and $\theta$ is the angle made by the perpendicular from the origin.

We have $P=2$ unit, So, the equation of line is $x \cos \theta+y \sin \theta=2$

Slope, $m_L=\frac{-\cos \theta}{\sin \theta}=-\cot \theta$

$ \therefore-\cot \theta=\frac{-12}{5} \Rightarrow \cot \theta=\frac{12}{5} $

For $y$-intercept, $x=0$ then $y \sin \theta=2$

$ \Rightarrow \quad y=\frac{2}{\sin \theta} $

And since $y$-intercept is positive, so

$ \begin{aligned} & \frac{2}{\sin \theta}>0 \\ & \Rightarrow \sin \theta>0 \end{aligned} $

$\Rightarrow \cot \theta=\frac{12}{5}>0, \theta$ must be in first quadrant.

Now, $\tan \theta=\frac{1}{\cot \theta}=\frac{5}{12}$

So, $\tan \theta+\cot \theta=\frac{5}{12}+\frac{12}{5}$

$ =\frac{25+144}{60}=\frac{169}{60} $

Let the angle made by Line $L$ with positive $X$-axis be $\theta$.

So, $B=(A x+A B \cos \theta, A y+A B \sin \theta)$

Given, $A(2,3)$ and $A B=4$, we have

$ B=(2+4 \cos \theta, 3+4 \sin \theta) $

Since, point $B$ lies on the line $4 x-3 y-19=0$, substitute the coordinate of $B$ into this equation.

$ \begin{aligned} & 4(2+4 \cos \theta)-3(3+4 \sin \theta)-19=0 \\ & \Rightarrow \quad 8+16 \cos \theta-9-12 \sin \theta-19=0 \\ & \Rightarrow \quad 16 \cos \theta-12 \sin \theta-20=0 \\ & \Rightarrow \quad 4 \cos \theta-3 \sin \theta-5=0 \\ & \Rightarrow \quad \frac{4}{5} \cos \theta-\frac{3}{5} \sin \theta=1 \end{aligned} $

Let $\cos \alpha=\frac{4}{5}$ and $\sin \alpha=\frac{3}{5}$ for some angle $\alpha$ in the first quadrant.

$\therefore \cos \alpha \cdot \cos \theta-\sin \alpha \cdot \sin \theta=1$

$ \begin{array}{ll} \Rightarrow & \cos (\alpha+\theta)=1 \\ \Rightarrow & \alpha+\theta=2 n \pi, \text { for some integer } n . \\ \Rightarrow & \alpha+\theta=0 \\ \Rightarrow & \alpha=-\theta \text { or } \alpha+\theta=2 \pi \\ \Rightarrow & \theta=2 \pi-\alpha \end{array} $

The angle made by the line $L$ with positive $X$-axis in the anti-clockwise direction is taken as a positive angle, where

$ \cos \theta=\frac{4}{5} \text { and } \sin \theta=-\frac{3}{5} $

This angle lies in the fourth quadrant.

So, $\tan \theta=\frac{\sin \theta}{\cos \theta}=\frac{-3}{4}$

$ \Rightarrow \quad \theta=\tan ^{-1}\left(\frac{-3}{4}\right)=-\tan ^{-1}\left(\frac{3}{4}\right) $

Let the equation of the line $L$ be $\frac{x}{a}+\frac{y}{b}=1$, where $a$ and $b$ are the $x$ and $y$ intercepts, respectively. So, $O A=a$ and $O B=b, a>0, b>0$.

Here, the line passes through the point $(4,9)$, So,

$ \frac{4}{a}+\frac{9}{b}=1 $

We want to minimize $a+b$. So consider

$ \begin{aligned} (a+b)\left(\frac{4}{a}+\frac{9}{b}\right) & =4+\frac{9 a}{b}+\frac{4 b}{a}+9 \\ & =13+\frac{9 a}{b}+\frac{4 b}{a} \end{aligned} $

By AM-GM inequality, for positive numbers,

$ \begin{aligned} & \frac{9 a}{b}+\frac{4 b}{a} \geq \sqrt[2]{\frac{9 a}{b} \cdot \frac{4 b}{a}} \\ \Rightarrow \quad & 2 \sqrt{36}=2 \cdot 6=12 \end{aligned} $

So, $13+\frac{9 a}{b}+\frac{4 b}{a}=13+12=25$

But, since $(a+b)\left(\frac{4}{a}+\frac{9}{b}\right)=(a+b) \cdot 1$

$ =a+b $     (Using Eq. (i))

$\therefore$ The minimum value of $a+b$ is 25 and this occurs when $\frac{9 a}{b}=\frac{4 b}{a}$

$ \begin{aligned} \Rightarrow & & 9 a^2 & =4 b^2 \text { or } 3 a=2 b \quad(\because a, b>0) \\ \Rightarrow & & b & =\frac{3}{2} a \end{aligned} $

So, $\frac{4}{a}+\frac{9}{b}=1$

$ \begin{aligned} & \Rightarrow \frac{4}{a}+\frac{9}{\frac{3}{2} a}=1 \Rightarrow \frac{4}{a}+\frac{18}{3 a}=1 \\ & \Rightarrow \frac{4}{a}+\frac{6}{a}=1 \Rightarrow \frac{10}{a}=1 \\ & \Rightarrow \quad a=10 \end{aligned} $

Then, $b=\frac{3}{2} a=\frac{3}{2} \times 10=15$

So, the minimum value of $O A+O B$ is $a+b=10+15=25$

Given equation,

$ 4 x^2+12 x y+9 y^2+2 g x+2 f y-1=0 $

Since, general second-degree equation $a x^2+2 h x y+b y^2+2 g x+2 f y+c=0$ represents a pair of straight lines, if

$ \Delta=a b c+2 f g h-b g^2-c h^2-a f^2=0 $

Here, $a=4, b=9, h=6$

So, $\Delta=4 \times 9 \times c+2 f g \times 6-9 g^2$

$ \begin{aligned} & -c(6)^2-4 f^2=0 \\ & =36 c+12 f g-9 g^2-36 c-4 f^2=0 \\ & =12 f g-9 g^2-4 f^2=0 \\ \Rightarrow \quad & -(3 g-2 f)^2=0 \Rightarrow 3 g-2 f=0 \\ \Rightarrow \frac{g}{f} & =\frac{2}{3} \text { and } \frac{f}{g}=\frac{3}{2} \end{aligned} $

So, $\frac{f}{g}+\frac{g}{f}=\frac{3}{2}+\frac{2}{3}=\frac{9+4}{6}=\frac{13}{6}$

To solve the problem of shifting the origin to the point $(a, b)$ in order to eliminate the first-degree terms from the equation $2x^2 - 3xy + 4y^2 + 5y - 6 = 0$, we start by considering new coordinates $(x+a, y+b)$.

Substituting these into the equation gives:

$ 2(x+a)^2 - 3(x+a)(y+b) + 4(y+b)^2 + 5(y+b) - 6 = 0 $

Expanding this, we get:

$ \begin{aligned} & 2(x^2 + 2ax + a^2) - 3(xy + ay + bx + ab) + 4(y^2 + 2by + b^2) \\ & \quad + 5y + 5b - 6 = 0 \end{aligned} $

Simplifying further, the remaining first-degree terms become:

$ (4a - 3b)x - (3a - 8b - 5)y = 0 $

For these terms to be eliminated, we require:

$ 4a - 3b = 0 \quad \text{and} \quad 3a - 8b - 5 = 0 $

Solving these equations simultaneously, we find:

$ a = -\frac{15}{23}, \quad b = \frac{-20}{23} $

Now, consider the transformed equation $ax^2 + 23ab xy + by^2 = 0$. We need to find the angle $\theta$ to rotate the axes such that the $xy$-term disappears. This involves calculating $\tan 2\theta$, given by:

$ \tan 2\theta = \frac{23ab}{a-b} $

Substituting the values of $a$ and $b$:

$ \tan 2\theta = \frac{23 \times \frac{-15}{23} \times \frac{-20}{23}}{-\frac{15}{23} + \frac{20}{23}} = 60 $

Therefore, $\tan 2\theta = 60$.

Slope of $ B C=\frac{-6}{1}=-6 $

Equation of $ L_{1} $

$ \begin{aligned} y+2 & =\frac{1}{6}(x-1) \\\\ \Rightarrow \quad 6 y+12 & =x-1 \\\\ 6 y & =x-13 \end{aligned} $

Slope of $ A B=\frac{5}{-3} $

Equation of $ L_{2} $

$ \begin{array}{l} y-\frac{1}{2}=\frac{3}{5}\left(x+\frac{1}{2}\right)=\frac{3 x}{5}+\frac{3}{10} \\\\ y=\frac{3 x}{5}+\frac{3}{10}+\frac{1}{2}=\frac{3 x}{5}+\frac{4}{5} \cdots \end{array} $

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

$ \begin{array}{ll} & 6\left(\frac{3 x}{5}+\frac{4}{5}\right)=x-13 \\\\ \Rightarrow & 18 x+24=5 x-65 \\\\ \Rightarrow & 13 x=-89 \\\\ \Rightarrow & l=-\frac{89}{13} \end{array} $

and $ 6 m=\frac{-89}{13}-13 $

$ =-\left(\frac{89+169}{13}\right)=-\frac{258}{13} $

$ \therefore \quad m=-\frac{43}{13} $

$ (l, m)=\left(\frac{-89}{13}, \frac{-43}{13}\right) $

Hence, $ 26 m-3=26\left(\frac{-43}{13}\right)-3 $

$ =-86-3=-89 $

Hence, $ 26 m-3=13 \mathrm{l} $

To find the area of the parallelogram formed by the given lines, we start by understanding the relationships between the lines. We have the lines:

$ L_1: \lambda x + 4y + 2 = 0 $

$ L_2: 3x + 4y - 3 = 0 $

Since $ L_1 $ is parallel to $ L_2 $, their slopes must be equal. Adjusting for slope, this gives us $\lambda = 3$.

From the equation of $ L_2: 3x + 4y - 3 = 0 $, we express $ y $ in terms of $ x $:

$ 4y = -3x + 3 \quad \Rightarrow \quad y = -\frac{3}{4}x + \frac{3}{4} $

So, the slope of $ L_2 $ (and therefore $ L_1 $) is $ m_1 = -\frac{3}{4} $. The y-intercepts of the parallel lines, $ L_1 $ and $ L_2 $, are $ c_1 = \frac{3}{4} $ and $ c_2 = -\frac{2}{4} $ respectively.

Next, consider the lines:

$ L_3: 2x + \mu y + 6 = 0 $

$ L_4: 2x + y + 3 = 0 $

Given these lines are parallel, solve for the slopes, setting $\mu = 1$ to obtain:

From $ L_4: 2x + y + 3 = 0 $:

$ y = -2x - 3 $

This gives $ m_2 = -2 $. The y-intercepts are $ d_1 = -3 $ for $ L_4 $ and $ d_2 = -6 $ for $ L_3 $.

Now, calculate the expressions needed for the area of the parallelogram.

The difference between y-intercepts of $ L_1 $ and $ L_2 $:

$ C_1 - C_2 = \frac{3}{4} + \frac{2}{4} = \frac{5}{4} $

The difference between y-intercepts of $ L_3 $ and $ L_4 $:

$ d_1 - d_2 = -3 + 6 = 3 $

Finally, calculate the area of the parallelogram using the formula:

$ \text{Area} = \frac{(C_1 - C_2) \times (d_1 - d_2)}{|m_1 - m_2|} $

Substituting the values:

$ \text{Area} = \frac{\frac{5}{4} \times 3}{|-\frac{3}{4} + 2|} = \frac{15}{4} \times \frac{4}{5} = 3 $

Therefore, the area of the parallelogram is $ 3 $.

To find the angle between the pair of lines given by the equation $ ax^2 + 4xy + 2y^2 = 0 $ and knowing that this angle is $45^\circ$, we start by using the formula for the angle $ \theta $ between two lines given by:

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

For this equation, we identify the coefficients:

$ h = 2 $

$ b = 2 $

Plugging these into the formula for $ \tan \theta $ and setting $\tan \theta = 1$ (since the angle is $45^\circ$):

$ \frac{2 \sqrt{4 - 2a}}{a + 2} = 1 $

Squaring both sides and rearranging gives:

$ 4(4 - 2a) = (a + 2)^2 $

Simplifying further:

$ 16 - 8a = a^2 + 4 + 4a $

Rearranging terms forms the quadratic equation:

$ a^2 + 12a - 12 = 0 $

To solve for $ a $, use the quadratic formula $ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $:

$ a = \frac{-12 \pm \sqrt{144 + 48}}{2} = \frac{-12 \pm \sqrt{192}}{2} $

Since $ \sqrt{192} = 8\sqrt{3} $, the solutions for $ a $ are:

$ a = \frac{-12 \pm 8\sqrt{3}}{2} = -6 \pm 4\sqrt{3} $

Therefore, the possible values of $ a $ are $ -6 \pm 4\sqrt{3} $.

To transform the equation $ y = x^3 - 9x^2 + cx - d $ to $ Y = X^3 $ by shifting the origin to the point $ (h, 5) $, we use the translation:

$ Y = y - 5 $

$ X = x - h $

Substituting, the equation becomes:

$ y - 5 = (x - h)^3 $

Expanding the right side, we have:

$ y = x^3 - h^3 - 3hx^2 + 3h^2x + 5 $

By comparing coefficients with the original equation $ y = x^3 - 9x^2 + cx - d $, we deduce:

$-3h = -9$

$c = 3h^2$

$d = h^3 - 5$

Solving these equations:

From $-3h = -9$, we find $h = 3$.

Substituting $h = 3$ in $c = 3h^2$, we get $c = 3 \times 3^2 = 27$.

Substituting $h = 3$ in $d = h^3 - 5$, we find $d = 3^3 - 5 = 22$.

Finally, we compute:

$ d - \frac{c}{h} = 22 - \frac{27}{3} = 22 - 9 = 13 $

Let the point $ P $ divide

then, $ P \equiv\left(\frac{-12-3}{4-3}, \frac{20-6}{4-3}\right) $

$ \Rightarrow \quad $ Slope $ =-\frac{2}{3} $

$ \begin{array}{c} \Rightarrow \quad \frac{y-14}{x+15}=-\frac{2}{3} \text { \{Point slope form \} } \\\\ 3 y-42=-2 x-30 \\\\ 2 x+3 y-12=0 \end{array} $

Let

$ A B: 7 x+y-24=0 $

Slope $ =m_{1}=-7 $

$ A C: x+7 y-24=0 $,

Slope $ =m_{2}=\frac{-1}{7} $

$ B C $ : passes through $ (-1,1) $

Let slope $ =m $

then equation of $ B C: y-1=m(x+1) $

$ \Rightarrow m x-y+m+1=0\quad \ldots \text { (i) } $

$ \therefore A B=A C \Rightarrow \angle A B C=\angle A C B $

$ \Rightarrow\left|\frac{m+7}{1-7 m}\right|=\left|\frac{m+\frac{1}{7}}{1-\frac{m}{7}}\right| $

$ \left(\because \tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|\right) $

$ \Rightarrow\left|\frac{m+7}{1-7 m}\right|=\left|\frac{7 m+1}{7-m}\right| $

$ \Rightarrow\left|49-m^{2}\right|=\left|49 m^{2}-1\right| $

$ \Rightarrow 49-m^{2}= \pm\left(49 m^{2}-1\right) $

$ \Rightarrow \quad m= \pm 1 $

So, equation of third line is $ x-y+2=0 $ or $ x+y=0 $

We are given two lines:

$ x + 2y - 3 = 0 $

$ 2x - y - 1 = 0 $

The intersection point of these lines is $(1, 1)$.

The area of the square is given as 16 square units, which implies the side length $a$ is 4, since:

$ a^2 = 16 \quad \Rightarrow \quad a = 4 $

The perpendicular distance from the center $(1, 1)$ to each side of the square is half the side length:

$ \text{Perpendicular distance} = \frac{a}{2} = 2 $

Next, we check the perpendicular distance of point $(1, 1)$ from the line $2x - y - 1 = 2\sqrt{5}$:

$ \text{Distance} = \left|\frac{2(1) - 1 - 2\sqrt{5}}{\sqrt{5}}\right| = \left|\frac{2 - 1 - 2\sqrt{5}}{\sqrt{5}}\right| = \left|\frac{1 - 2\sqrt{5}}{\sqrt{5}}\right| = |2| = 2 $

Then, we check the perpendicular distance of point $(1, 1)$ from the line $x + 2y - 3 + 2\sqrt{5} = 0$:

$ \text{Distance} = \left|\frac{1 + 2(1) - 3 + 2\sqrt{5}}{\sqrt{5}}\right| = \left|\frac{3 - 3 + 2\sqrt{5}}{\sqrt{5}}\right| = 2 $

Both distances are indeed equal to 2, confirming the side length of the square. This verifies the equations of the sides for this configuration:

$ (2x - y - 1 - 2\sqrt{5})(x + 2y - 3 + 2\sqrt{5}) = 0 $

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 $