Straight Lines and Pair of Straight Lines

Let point $P:(h,\,k)$

${(h - 1)^2} + {(k - 2)^2} + {(h + 2)^2} + {(k - 1)^2} = 14$

$2{h^2} + 2{k^2} + 2h - 6k - 4 = 0$

Locus of $P:{x^2} + {y^2} + x - 3y - 2 = 0$

Intersection with x-axis,

${x^2} + x - 2 = 0$

$ \Rightarrow x = - 2,\,1$

Intersection with y-axis,

${y^2} - 3y - 2 = 0$

$ \Rightarrow y = {{3\, \pm \,\sqrt {17} } \over 2}$

Area of the quadrilateral ACBD is

$ = {1 \over 2}(|{x_1}| + |{x_2}|)(|{y_1}| + |{y_2}|)$

$ = {1 \over 2} \times 3 \times \sqrt {17} = {{3\sqrt {17} } \over 2}$

${L_1}:3x - 4y + 12 = 0$

${L_2}:8x + 6y + 11 = 0$

Equation of angle bisector of L₁ and L₂ of angle containing origin

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

$2x + 14y - 13 = 0$ ...... (i)

${{3\alpha - 4\beta + 12} \over 5} = 1$

$ \Rightarrow 3\alpha - 4\beta + 7 = 0$ ...... (ii)

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

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

Points A($\alpha$, $-$3), B(2, 0) and C(1, $\alpha$) are collinear.

$\therefore$ Slope of AB = Slope of BC

$ \Rightarrow {{0 + 3} \over {2 - \alpha }} = {{\alpha - 0} \over {1 - 2}}$

$ \Rightarrow - 3 = \alpha (2 - \alpha )$

$ \Rightarrow - 3 = 2\alpha - {\alpha ^2}$

$ \Rightarrow {\alpha ^2} - 2\alpha - 3 = 0$

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

$ \Rightarrow \alpha (\alpha - 3) + 1(\alpha - 3) = 0$

$ \Rightarrow (\alpha + 1)(\alpha - 3) = 0$

$ \Rightarrow \alpha = - 1,\,3$

Given, ${\alpha _1} < {\alpha _2}$

$\therefore$ ${\alpha _1} = -1$ and ${\alpha _2} = 3$

$\therefore$ $\left( {{\alpha _1},\,{\alpha _2}} \right) = ( - 1,\,3)$

Now, equation of the line passing through ($-$1, 3) and making angle ${\pi \over 3}$ with positive x-axis is

$(y - {y_1}) = m(x - {x_1})$

$ \Rightarrow y - 3 = \left( {\tan {\pi \over 3}} \right)(x + 1)$

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

$ \Rightarrow \sqrt 3 x - y + \sqrt 3 + 3 = 0$

For point A,

2x $-$ y $-$ 1 = 0 ...... (1)

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

Solving (1) and (2), we get

x = 1, y = 1.

$\therefore$ Point A = (1, 1)

Altitude from B to line AC is perpendicular to line AC.

$\therefore$ Equator of altitude BH is

2x + y + $\lambda$ = 0 ...... (3)

It passes through point $H\left( {{7 \over 3},{7 \over 3}} \right)$ so it satisfy the equation (3).

${{14} \over 3} + {7 \over 3} + \lambda = 0$

$\Rightarrow$ $\alpha$ = $-$7

$\therefore$ Altitude BH = 2x + y $-$ 7 = 0 ...... (4)

Solving equation (1) and (4), we get

x = 2, y = 3.

$\therefore$ Point B = (2, 3)

Altitude from C to line AB is perpendicular to line AB.

$\therefore$ Equation of altitude CH is

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

It passes through point $H\left( {{7 \over 3},{7 \over 3}} \right)$ so it satisfy equation (5).

${7 \over 3} + {{14} \over 3} + \lambda = 0$

$\Rightarrow$ $\lambda$ = $-$7

$\therefore$ Altitude CH = x + 2y $-$ 7 = 0 ...... (6)

Solving equation (2) and (6), we get

x = 3, y = 2

$\therefore$ Point C = (3, 2)

Centroid G (x, y) of triangle A (1, 1), B (2, 3) and C (3, 2) is

$x = {{1 + 2 + 3} \over 3} = 2$, $y = {{1 + 2 + 3} \over 3} = 2$

Now, distance of point G (2, 2) from center O (0, 0) is

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

$\left. \matrix{ 2x + y = 4 \hfill \cr 2x + 6y = 14 \hfill \cr} \right\}y = 2,\,x = 3$

B(1, 2)

Let C(k, 4 $-$ 2k)

Now $A{B^2} = A{C^2}$

${5^2} + {( - 1)^2} = {(6 - k)^2} + {( - 3 + 2k)^2}$

$ \Rightarrow 5{k^2} - 24k + 19 = 0$

$(5k - 19)(k - 1) = 0 \Rightarrow k = {{19} \over 5}$

$C\left( {{{19} \over 5}, - {{18} \over 5}} \right)$

Centroid ($\alpha$, $\beta$)

$\alpha = {{6 + 1 + {{19} \over 5}} \over 3} = {{18} \over 5}$

$\beta = {{1 + 2 - {{18} \over 5}} \over 3} = - {1 \over 5}$

Now $15(\alpha + \beta )$

$15\left( {{{17} \over 5}} \right) = 51$

Let, side of triangle = a.

$h = {{|3 + 7 - 5|} \over {\sqrt {{1^2} + {1^2}} }}$

$ = {5 \over {\sqrt 2 }}$

From figure, $h = a\sin 60^\circ $

$ \Rightarrow a = {{2h} \over {\sqrt 3 }}$

$ = {2 \over {\sqrt 3 }} \times {5 \over {\sqrt 2 }}$

$ = {{10} \over {\sqrt 6 }}$

$\therefore$ Area $ = {3 \over 4}{\left( {{{10} \over {\sqrt 6 }}} \right)^2}$

$ = {{25} \over {2\sqrt 3 }}$

$\because$ A(1, $\alpha$), B($\alpha$, 0) and C(0, $\alpha$) are the vertices of $\Delta$ABC and area of $\Delta$ABC = 4

$\therefore$ $\left| {{1 \over 2}\left| {\matrix{ 1 & \alpha & 1 \cr \alpha & 0 & 1 \cr 0 & \alpha & 1 \cr } } \right|} \right| = 4$

$ \Rightarrow \left| {1(1 - \alpha ) - \alpha (\alpha ) + {\alpha ^2}} \right| = 8$

$ \Rightarrow \alpha = \, \pm \,8$

Now, $(\alpha ,\, - \alpha ),\,( - \alpha ,\alpha )$ and $({\alpha ^2},\beta )$ are collinear

$\therefore$ $\left| {\matrix{ 8 & { - 8} & 1 \cr { - 8} & 8 & 1 \cr {64} & \beta & 1 \cr } } \right| = 0 = \left| {\matrix{ { - 8} & 8 & 1 \cr 8 & { - 8} & 1 \cr {64} & \beta & 1 \cr } } \right|$

$ \Rightarrow 8(8 - \beta ) + 8( - 8 - 64) + 1( - 8\beta - 8 \times 64) = 0$

$ \Rightarrow 8 - \beta - 72 - \beta - 64 = 0$

$ \Rightarrow \beta = - 64$

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

Equation of $A C$ in slope intercept form,

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

Similarly, equation of $B D$,

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

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

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

Given,

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

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

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

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

Now, Required Equation

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

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

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

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

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

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

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

∴ All three lines are concurrent.

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

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

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

$(a, b)$ is equidistance from

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

For ( + )ve sign.

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

For ( - ) ve sign,

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

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

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

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

∵ Lines are at right angled.

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

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

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

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

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

∴ By homogenisation,

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

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

According to the question,

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

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

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

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

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

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

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

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

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

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

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

also $x$-coordinate will be equal

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

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

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

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

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

$x$-intercept of line $=5$

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

$y$-intercept of line $=7$

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

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

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

Equation of line w.r.t. old axis

$ :7 x+5 y=35 $

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

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

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

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

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

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

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

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

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

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

Let $B$ and $D$ be other two coordinates of the square lying on the line

$ \begin{array}{r} 7 x-y+8=0 \\ B\left(x_1, 7 x_1+8\right) \\ D\left(x_2, 7 x_2+8\right) \end{array} $

Mid-point of $B$ and $D$ is $\left(-\frac{1}{2}, \frac{9}{2}\right)$.

$ \begin{array}{ll} \therefore & \frac{x_1+x_2}{2}=\frac{-1}{2} \\ \Rightarrow & x_1+x_2=-1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{array} $

Also,

$ \begin{aligned} & (A M)^2=(B M)^2 \\ & \Rightarrow\left(-4+\frac{1}{2}\right)^2+\left(5-\frac{9}{2}\right)^2=\left(x_1+\frac{1}{2}\right)^2 +\left(7 x_1+8-\frac{9}{2}\right)^2 \\ & \Rightarrow \frac{49}{4}+\frac{1}{4}=\left(x_1+\frac{1}{2}\right)^2+49\left(x_1+\frac{1}{2}\right)^2 \\ & \Rightarrow 50\left(x_1+\frac{1}{2}\right)^2=\frac{50}{4} \\ & \Rightarrow \quad x_1+\frac{1}{2}= \pm \frac{1}{2} \\ & x_1=0,-1 \\ & x_1=0, y_1=8 \\ & x_1=-1, y_1=1 \end{aligned} $

∴ Required coordinates of the two vertices are $(0,8),(-1,1)$.

Let $P(\alpha, 2 \alpha-1)$ and its image with respect to the line $3 x-2 y+4=0$ is $Q(h, k)$

$ \begin{aligned} & \begin{aligned} \Rightarrow \frac{h-\alpha}{3} & =\frac{k-(2 \alpha-1)}{-2} \\ & =-2\left(\frac{3 \alpha-2(2 \alpha-1)+4}{3^2+(-2)^2}\right) \\ \Rightarrow \frac{h-\alpha}{3} & =\frac{k-(2 \alpha-1)}{-2}=\frac{2 \alpha-12}{13} \\ \Rightarrow \quad h & =\alpha+\frac{6 \alpha-36}{13}=\frac{19 \alpha-36}{13} \\ \text { and } \quad k & =2 \alpha-1+\frac{24-4 \alpha}{13}=\frac{22 \alpha+11}{13} \\ \Rightarrow \quad \alpha & =\frac{13 h+36}{19}, \alpha=\frac{13 k-11}{22} \end{aligned} \end{aligned} $

So, $\quad \frac{13 h+36}{19}=\frac{13 k-11}{22}$

$ \begin{aligned} & \Rightarrow \quad 22(13 h+36)=19(13 k-11) \\ & \text { Locus: } 22(13 x+36)=19(13 y-11) \end{aligned} $

Let the foot of perpendicular be $M(x, y)$.

$ \begin{aligned} & \frac{x-5}{3}=\frac{y+7}{-5}=\frac{-(3 \times 5-5 \times(-7)+1)}{3^2+(-5)^2} \\ & \frac{x-5}{3}=\frac{y+7}{-5}=\frac{51}{34}=-\frac{3}{2} \\ & \Rightarrow x=5-\frac{9}{2}=\frac{1}{2}, y=\frac{15}{2}-7=\frac{1}{2} \end{aligned} $

So, $M\left(\frac{1}{2}, \frac{1}{2}\right)$

Now perpendicular distance of $M$ to the line $2 x+5 y-3=0$ is given by

$ \begin{aligned} &\text { ∴ Perpendicular distance }\\ &=\left|\frac{2 \times \frac{1}{2}+5 \times \frac{1}{2}-3}{\sqrt{4+25}}\right|=\frac{1}{2 \sqrt{29}} \end{aligned} $

Given, $A(-1,3), B(3,5), C(5,7)$.

Let $P(x, y)$ is equidistant from $A, B, C$.

So,  $ P A^2=P B^2 $

$ \begin{aligned} & (x+1)^2+(y-3)^2=(x-3)^2+(y-5)^2 \\ & \Rightarrow x^2+2 x+1+y^2-6 y+9 \\ & \quad=x^2-6 x+9+y^2-10 y+25 \end{aligned} $

$ \begin{array}{lrl} \Rightarrow & & 8 x+4 y=24 \\ \Rightarrow & & 2 x+y=6 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \text { Again, } & P A^2=P C^2 \end{array} $

$ \begin{aligned} &\begin{array}{lrl} \Rightarrow & (x+1)^2+(y-3)^2 & =(x-5)^2+(y-7)^2 \\ \Rightarrow & 12 x+8 y & =64 \\ \Rightarrow & 3 x+2 y & =16\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{array}\\ &\text { From Eqs. (i) and (ii), } x=-4, y=14\\ &\begin{aligned} \therefore P(-4,14) & \\ \text { Hence, } P A & =\sqrt{(-4+1)^2+(14-3)^2} \\ & =\sqrt{9+121}=\sqrt{130} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { (A) } 5 x^2+2 \sqrt{7} x y-y^2=0\\ &\begin{aligned} & \tan \theta=\frac{2 \sqrt{h^2-a b}}{a+b} \\ & =\frac{2 \sqrt{(\sqrt{7})^2-5 \times(-1)}}{5-1}=\frac{2 \sqrt{12}}{4} \\ & =\sqrt{3} \\ & \Rightarrow \cos \theta=\frac{1}{2} \\ & \text { } \begin{aligned}(B)\, & x^2+\sqrt{11} x y+2 y^2=0 \\ & \tan \theta=\frac{2 \sqrt{\left(\sqrt{11} / 2^2-1 \times 2\right.}}{1+2} \\ &= \frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}} \Rightarrow \cos \theta=\frac{\sqrt{3}}{2} \end{aligned} \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { (C) } x^2+2 \sqrt{2} x y+y^2=0 \\ & \Rightarrow \tan \theta=\frac{2 \sqrt{(\sqrt{2})^2-1 \times 1}}{1+1}=1 \\ & \Rightarrow \cos \theta=\frac{1}{\sqrt{2}} \\ & \text { (D) } 3 x^2+4 \sqrt{2} x y+y^2=0 \\ & \tan \theta=\frac{2 \sqrt{(2 \sqrt{2})^2-3 \times 1}}{3+1}=\frac{\sqrt{5}}{2} \\ & \Rightarrow \cos \theta=\frac{2}{3} \end{aligned} $

Given, $a x^2+6 x y-2 y^2=0$ represent a pair of perpendicular lines.

$ \begin{array}{ll} \text { So, } & m_1 m_2=-1 \\ \Rightarrow & \frac{a}{-2}=-1 \\ \Rightarrow & a=2 \end{array} $

Given, $9 x^2+2 h x y+4 y^2=0$ represent a pair of coincident lines.

$ \begin{array}{ll} \therefore & 4\left(\frac{y}{x}\right)^2+2 h\left(\frac{y}{x}\right)+9=0 \\ \Rightarrow & 4 m^2+2 h m+9=0 \end{array} $

For coincident lines, $D=0$

$ \begin{aligned} \Rightarrow & & 4 h^2-4 \times 4 \times 9 & =0 \\ \Rightarrow & & h & =3 \times 2 \\ \therefore& & h & =3 a \end{aligned} $

Line $x+2 y=k$ meets the curve $2 x^2-2 x y+3 y^2+2 x-y-1=0$ at $A$ and $B$.

So, combined equation of $O A$ and $O B$ is

$ \begin{gathered} 2 x^2-2 x y+3 y^2+2 x\left(\frac{x+2 y}{k}\right) \\ -y\left(\frac{x+2 y}{k}\right)-\left(\frac{x+2 y}{k}\right)^2=0 \\ \Rightarrow 2 x^2-2 x y+3 y^2 \\ +\frac{2 x^2+4 x y-y x-2 y^2}{k} \\ -\left(\frac{x^2+4 y^2+4 x y}{k^2}\right)=0 \end{gathered} $

$ \begin{aligned} &\begin{aligned} \left(2 k^2+2 k-1\right) x+ & \left(-2 k^2+3 k-4\right) x y \\ & +\left(3 k^2-2 k-4\right) y^2=0 \end{aligned}\\ &\text { For } O A \text { and } O B \text { perpendicular }\\ &\begin{array}{lcl} & & a+b=0 \\ \Rightarrow & & 2 k^2+2 k-1+3 k^2-2 k-4=0 \\ \Rightarrow & & 5 k^2-5=0 \\ \Rightarrow & & k= \pm 1 \end{array} \end{aligned} $

Let the line $L$ intersect $X$-axis at a point $(h, 0)$.

Slope of line $L\left(m_1\right)=\frac{0+3}{h-5}=\frac{3}{h-5}$

Slope of line $\left(m_2\right), \sqrt{3} x+y-9=0$ is $-\sqrt{3}$

$ \begin{aligned} & \tan 60^{\circ}=\left|\frac{\frac{3}{h-5}+\sqrt{3}}{1-\frac{3 \sqrt{3}}{h-5}}\right| \\ & \Rightarrow \quad \sqrt{3}=\left|\frac{3+\sqrt{3}(h-5)}{h-5-3 \sqrt{3}}\right| \\ & \Rightarrow \quad \sqrt{3}= \pm \frac{3+\sqrt{3}(h-5)}{h-5-3 \sqrt{3}} \\ & \text { When } \sqrt{3}=\frac{3+\sqrt{3}(h-5)}{h-5-3 \sqrt{3}} \\ & \Rightarrow \quad \sqrt{3} h-5 \sqrt{3}-9=3+\sqrt{3} h-5 \sqrt{3} \end{aligned} $

(Not possible)

When $\sqrt{3}=\frac{-(3+\sqrt{3}(h-9)}{h-5-3 \sqrt{3}}$

$ \begin{array}{llrl} \Rightarrow & & \sqrt{3} h-5 \sqrt{3}-9 & =-3-\sqrt{3} h+5 \sqrt{3} \\ \Rightarrow & & 2 \sqrt{3} h & =10 \sqrt{3}+6 \\ \Rightarrow & & h & =5+\sqrt{3} \end{array} $

The line $L$ intersect $X$-axis at point $(5+\sqrt{3}, 0)$.

Equation of line $L$ is

$ \begin{array}{ll} & y+3=\frac{3}{\sqrt{3}}(x-5) \\ \Rightarrow & y+3=\sqrt{3}(x-9) \\ \Rightarrow & \sqrt{3} x-y-5 \sqrt{3}-3=0 \end{array} $

Given, $a \alpha+\beta b+\gamma c=0$

$ \Rightarrow \quad b=-\frac{a \alpha-\gamma c}{\beta} $

The equation which represent the family of concurrent straight line is

$ \begin{aligned} & a x+b y+c=0 \\ & \quad a x+\left(\frac{-a \alpha-\gamma c}{\beta}\right) y+c=0 \\ & \Rightarrow a\left(x-\frac{\alpha y}{\beta}\right)+c\left(1-\frac{\gamma y}{\beta}\right)=0 \end{aligned} $

It represent a family of lines passing through the intersect of both the line

$ \begin{aligned} &\begin{array}{rlrl} \Rightarrow & \gamma y / \beta =1 \\ \Rightarrow & y =\beta / \gamma \\ \text { Also, } & x-\frac{\alpha y}{\beta} =0 \\ \Rightarrow & x-\frac{\alpha}{\beta}\left(\frac{\beta}{\gamma}\right) =0 \\ & x =\alpha / \gamma \end{array}\\ &\text { ∴ The required point is }\left(\frac{\alpha}{\gamma}, \frac{\beta}{\gamma}\right) \text {. } \end{aligned} $

Let the equation of line be

$ a x+b y+c=0 $

Sum of distance from $(2,0),(0,2)$ and $(1,1)$ is 0

$ \begin{aligned} & \frac{2 a+c}{\sqrt{a^2+b^2}}+\frac{2 b+c}{\sqrt{a^2+b^2}}+\frac{a+b+c}{\sqrt{a^2+b^2}}=0 \\ \Rightarrow \quad & 3 a+3 b+3 c=0 \\ \Rightarrow \quad & a+b+c=0 \end{aligned} $

For the line $a x+b y+c=0$, if $a+b+c=0$, then the line passes through the point $(1,1)$.

Given,

$ \begin{gathered} 6 a^2-3 b^2-c^2+7 a b-a c+4 b c=0 \\ \Rightarrow c^2+c(a-4 b)+3 b^2-6 a^2-7 a b=0 \end{gathered} $

which is the quadratic equation, from the quadratic formula,

$ \begin{aligned} c & =\frac{4 b-a \pm \sqrt{\begin{array}{c} a^2+16 b^2-8 a b-12 b^2 \\ +24 a^2+28 a b \end{array}}}{2} \\ & =\frac{4 b-a \pm \sqrt{25 a^2+4 b^2+20 a b}}{2} \\ c & =\frac{(4 b-a) \pm(5 a+2 b)}{2} \end{aligned} $

$ \begin{array}{lrl} \Rightarrow & & 2 c=4 b-a+5 a+2 b \\ \text { and } & & 2 c=4 b-a-5 a-2 b \\ \Rightarrow & & 4 a+6 b-2 c=0 \\ \text { and } & & 6 a-2 b+2 c=0 \\ \Rightarrow & & 2 a+3 b-c=0 \\ \text { and } & & 3 a-b+c=0 \end{array} $

For determining the point of concurrency with $a x+b y+c=0$, compare both equations with

$ \begin{aligned} & a x+b y+c=0 \\ & \frac{x}{2}=\frac{y}{3}=\frac{1}{-1} \text { and } \frac{x}{3}+\frac{y}{(-1)}=\frac{1}{1} \\ & \Rightarrow x=-2, y=-3 \text { and } x=3, y=-1 \end{aligned} $

∴ The lines are concurrent at $(-2,-3)$ and $(3,-1)$.

Given, $H \equiv a x^2-x y+b y^2=0$

$ \begin{aligned} \tan \theta & =\frac{2 \sqrt{h^2-a b}}{|a+b|} \\ \Rightarrow \quad 5 & =\frac{\sqrt[2]{\left(-\frac{1}{2}\right)^2-a b}}{|a+b|} \\ \Rightarrow \quad 5|a+b| & =2 \sqrt{\frac{1}{4}-a b} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Also, $(1,-1)$ is a point on $H=0$

$ a+1+b=0 \Rightarrow a=-(b+1) $

Substitute value of $a$ in Eq. (i),

$ \begin{aligned} & & 5|-1| & =2 \sqrt{\frac{1}{4}+(b+1) b} \\ \Rightarrow & & 25 & =4\left(\frac{1}{4}+b^2+b\right) \\ & \Rightarrow & 25 & =1+4 b^2+4 b \end{aligned} $

$ \begin{aligned} \Rightarrow & & 4 b^2+4 b-24 & =0 \\ \Rightarrow & & b^2+b-6 & =0 \\ \Rightarrow & & b^2+3 b-2 b-6 & =0 \\ \Rightarrow & & b & =2,-3 \end{aligned} $

If $b=2$, then $a=-3$

If $b=-3$, then $a=2$

$ a^2+a b+b^2=4-6+9=7 $

Given, line are

$ \begin{aligned} & 3 x^2-4 x y+5 y^2=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & a=3, b=5, h=-2 \end{aligned} $

Line is passing $(2,3)$ and perpendicular to Eq. (i),

$ \begin{aligned} & \therefore 5(x-2)^2+4(x-2)(y-3)+3(y-3)^2=0 \\ & \Rightarrow 5 x^2+20-20 x+4(x y-3 x-2 y+6) \\ & \quad+3 y^2+27-18 y=0 \\ & \begin{aligned} \Rightarrow 5 x^2+4 x y+3 y^2-32 x-26 y+71=0 \\ \therefore a=5, b=3, c=71, f=-13, \\ \quad g=-16, h=2 \\ \therefore a+b+c+f+g+h \\ \quad=5+3+71-13-16+2 \\ \quad=8+71-27=8+44=52 \end{aligned} \end{aligned} $

Given, the equation of line $4 x-6 y-2=0$ and the equation of curve

$ 3 x^2-4 x y+5 y^2-2 x+y-6=0 $

The equation of line $4 x-6 y-2=0$ can be written as $2 x-3 y=1$

The equation of curve can be written as

$ \left(3 x^2-4 x y+5 y^2\right)-(2 x-y)(1)-6(1)^2=0 $

Now, substitute $1=2 x-3 y$ in equation of curve

$ \begin{aligned} & \left(3 x^2-4 x y+5 y^2\right)-(2 x-y)(2 x-3 y) -6(2 x-3 y)^2=0 \\ & \Rightarrow 3 x^2-4 x y+5 y^2-4 x^2+8 x y-3 y^2 \end{aligned} $

$ \begin{array}{rlrl} & -24 x^2-54 y^2+72 x y=0 \\ & - 25 x^2+76 x y-52 y^2=0 \\ & f(x, y) =25 x^2-76 x y+52 y^2 \\ f(1,-1) & =25+76+52=153 \\ & f(-1,-1) =25-76+52=1 \\ & \therefore \quad \frac{f(1,-1)}{f(-1,-1)} =153 \end{array} $

Given, $2 x-y=4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

(a, b) lies on line $2 x-y-4=0$

$ \begin{aligned} b & =2 a-4 \\ \left(\frac{-4 m-n}{m+n}\right) & =2\left(\frac{m+2 n}{m+n}\right)-4 \\ -4 m-n & =2 m+4 n-4 m-4 n \\ 2 m+n & =0 \\or\,\,\,\,\,\,\,\, n & =-2 m \end{aligned} $

Equation of line joining the points $(2,-1)$ and $(1,-4)$ is

$ \begin{aligned} y+4 & =\left(\frac{-1+4}{2-1}\right)(x-1) \\ y+4 & =3(x-1) \\ y+4 & =3 x-3 \\ 3 x-y & =7 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Subtracting Eq. (i) from Eq. (ii), we get

$ \begin{aligned} & x=3 \text { and } y=2 \\ & \therefore a=3, b=2 \\ & \text { So, } 4\left(a-b\left(\frac{m}{n}\right)^2\right)=4\left[3-2\left(\frac{-1}{2}\right)^2\right] \\ & \quad=4\left[3-\frac{1}{2}\right]=4 \times \frac{5}{2}=10 \end{aligned} $

Given, $x+(5 \lambda+1) y+1-3 \lambda=0$

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

The given family of lines contains two lines i.e. $x+y+1=0$

and    $ 5 y-3=0 $

$ \begin{gathered} 5 y-3=0 \Rightarrow y=\frac{3}{5} \\ x+\frac{3}{5}+1=0 \Rightarrow x=-\frac{8}{5} \end{gathered} $

The point of concurrent of family of lines

$ \begin{aligned} & x+(5 \lambda+1) y+1-3 \lambda \text { is }\left(\frac{-8}{5}, \frac{3}{5}\right) \\ & (5 \mu+2) x-3 y+3+6 \mu=0 \\ & 2 x-3 y+3+\mu(5 x+6)=0 \end{aligned} $

The family of lines contain

$ \begin{aligned} & 2 x-3 y+3=0 \\and\,\,\,\, & 5 x+6=0 \\ & x=\frac{-6}{5} \text { and } y=\frac{1}{5} \end{aligned} $

The point of concurrent of family of lines

$ (5 \mu+2) x-3 y+3+6 \mu=0 \text { is }\left(\frac{-6}{5}, \frac{1}{5}\right) $

∴ Required distance

$ \begin{aligned} & =\sqrt{\left(\frac{-8}{5}+\frac{6}{5}\right)^2+\left(\frac{3}{5}-\frac{1}{5}\right)^2} \\ & =\sqrt{\frac{4}{25}+\frac{4}{25}} \\ & =\sqrt{\frac{8}{25}}=\frac{2 \sqrt{2}}{5} \end{aligned} $

The two given lines $2 x-3 y+4=0$ and $6 x-9 y+7=0$ are parallel.

Slope of these lines $=\frac{2}{3}$ and slope of line $L=\frac{-3}{2}$

Equation of line $L$ is

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

Solving equation $3 x+2 y=5$ and $2 x-3 y+4=0$ to determine the point $A$.

$ x=\frac{7}{13}, y=\frac{22}{13} $

Solving equation $3 x+2 y=5$ and $6 x-9 y+7=0$ to get the point $B$.

$ x=\frac{31}{39} \text { and } y=\frac{17}{13} $

Coordinates of point $A$ is $\left(\frac{7}{13}, \frac{22}{13}\right)$ and point $B$ is $\left(\frac{31}{39}, \frac{17}{13}\right)$.

Let $P$ divides in the ratio $K: 1$

$ \begin{array}{cc} & 1=\frac{K \times \frac{31}{39}+1 \times \frac{7}{13}}{K+1} \\ \Rightarrow & 39(K+1)=31 K+21 \\ \Rightarrow & 39 K+39=31 K+21 \\ \Rightarrow & 8 K=-18 \\ & K=\frac{-9}{4} \end{array} $

$\therefore P$ divides $A B$ in the ratio $9: 4$ externally.

$ \text { Slope of line } B C=\frac{5+1}{-3-5} $

$ =\frac{6}{-8}=\frac{-3}{4} $

Slope of line $A D=\frac{4}{3}\,\,\,\,\, [\because A D \perp B C] $

Equation of $A D$ is

$ \begin{aligned} y-3 & =\frac{4}{3}(x-1) \\ 3 y-9 & =4 x-4 \\ 4 x-3 y & =-5 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Slope of $A C=\frac{3+1}{1-5}=-1$

Slope of $B E=1$         $[B E \perp A C]$

Equation of line $B E$ is

$ \begin{aligned} y-5 & =1(x+3) \\ y & =x+3+5 \\ y & =x+8 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Substitute value of $y$ from Eq. (ii) to Eq. (i),

$ \begin{gathered} 4 x-3(x+8)=-5 \\ 4 x-3 x-24=-5 \\ x=19 \Rightarrow y=27 \end{gathered} $

∴ The coordinate of orthocentre is $(19,27)$.

$ \begin{aligned} & \text { Given, } \alpha x^2+2 \gamma x y+\beta y^2=0 \\ & \left(y-m_1 x\right)\left(y-m_2 x\right)=0 \\ & y^2-\left(m_1+m_2\right) x y+m_1 m_2 x^2=0 \\ & \Rightarrow \beta=1,2 r=-\left(m_1+m_2\right) \text { and } \alpha=m_1 m_2 \\ & \text { Again, } b h x^2+a b x y+a h y^2=0 \\ & \left(y+\frac{1}{m_1} x\right)\left(y+\frac{1}{m_2} x\right)=0 \end{aligned} $

$ \begin{array}{r} \left(m_1 y+x\right)\left(m_2 y+x\right)=0 \\ m_1 m_2 y^2+\left(m_1+m_2\right) x y+x^2=0 \end{array} $

Here, $a h=m_1 m_2, a b=m_1+m_2 a h=1$

Here, $a h=\beta, a b=-2 r$ and $\alpha=a h$

$ \frac{\alpha \beta}{r^2}=\frac{a h \times b h}{\left(\frac{a b}{-2}\right)^2}=\frac{4 h^2}{a b} $

$ \begin{aligned} \text {Given, } \frac{x^2}{a}+\frac{x y}{h}+\frac{y^2}{b} & =0 \\ \Rightarrow \quad h b x^2+a b x y+a h y^2 & =0 \\ (y-m x)^2 & =0 \\ y^2+m^2 x^2+2 m x y & =0 \end{aligned} $

Here, $h b=m^2, a b=2 m$ and $a h=1$

$ \begin{aligned} h b & =\left(\frac{a b}{2}\right)^2 \Rightarrow 4 h b=a^2 b^2 \\ 4 h & =a^2 b \Rightarrow 4 h \times h=a b \times a h \\ 4 h^2 & =a b \quad[\because a h=1] \end{aligned} $

Given, curve : $x^2+y^2-2(x+2 y)$

(1) $+2(1)^2=0$

Given, straight line $\frac{x+y}{k}=1$

Now, the line joining the origin to point of intersection of the given line and given curve is

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

If the two homogenous lines are perpendicular, then coefficient of $x^2+$ Coefficient of $y^2=0$

$ \begin{aligned} k^2-2 k+2+k^2-4 k+2 & =0 \\ 2 k^2-6 k+4 & =0 \\ k^2-3 k+2 & =0 \\ k & =1,2 \end{aligned} $

∴ Sum of value of $k=1+2=3$

Let origin is shifted to $(p, q)$

$ \begin{aligned} & 3(X+p)^2+4(X+p)(Y+q)+(Y+q)^2 \\ & -8(X+p)-4(Y+q)-4=0 \\ & \Rightarrow 3 X^2+4 X Y+Y^2+(6 p+4 q-8) X \\ & +(4 p+2 q-4) Y+\left(3 p^2+4 p q+q^2\right. \\ & \quad-8 p-4 q-4)=0 \end{aligned} $

Now, $6 p+4 q-8=0,4 p+2 q-4=0$

$ \begin{array}{lrl} \Rightarrow & p & =0, q=2 \\ \therefore & f(X, Y) & =3 X^2+4 X Y+Y^2-8 \\ \Rightarrow & f(1,1) & =3+4+1-8=0 \end{array} $

$ \begin{aligned} &\text { According to the questions, }\\ &\begin{aligned} & \frac{m}{n}=-\frac{(2 \times(-2)-3 \times 3+4)}{(2 \times 3-3(-2)+4}=\frac{9}{16} \\ \therefore & \frac{-4 m}{3 n}=\frac{-4}{3} \times \frac{9}{16}=\frac{-3}{4} \end{aligned} \end{aligned} $

$ \begin{aligned} P & \equiv\left(\frac{3 \times 3-4 \times(-2)}{3-4}, \frac{3 \times(-2)-4(3)}{3-4}\right) \\ & =(-17,18) \end{aligned} $

$ \begin{aligned} &\text { According to question, }\\ &\left|\begin{array}{ccc} 2 & 1 & 3 \\ k & 2 & -3 \\ 3 & -2 & 1 \end{array}\right|=0 \end{aligned} $

$ \Rightarrow \quad 2(-4)-1(k+9)+3(-2 k-6)=0 $

$ \Rightarrow \quad-8-k-9-6 k-18=0 $

$ \begin{array}{rlrl} \Rightarrow &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, -7 k =35 \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,k =-5 \end{array} $

$ \therefore \quad L_2:-5 x+2 y-3=0 $

$ \begin{aligned} &\text { Let } \theta \text { be the angle between } L_2 \text { and }\\ &\begin{aligned} & 2 x-5 y+7=0 \\ & \therefore \tan \theta=\left|\frac{\frac{5}{2}-\frac{2}{5}}{1+\frac{5}{2} \times \frac{2}{5}}\right|=\frac{21}{20} \Rightarrow \cos \theta=\frac{20}{29} \end{aligned} \end{aligned} $

$ \text { According to the question, } $

$ \begin{aligned} &\begin{array}{rlrl} \frac{h-1}{1} & =\frac{k-1}{-1}=-2 \frac{(1-1+1)}{\left(1^2+1^2\right)}=-1 \\ & h =0, k=2 \\ ∴& Q =(0,2) \end{array}\\ &\text { ∴ Required length of perpendicular }\\ &=\frac{|3 \times 0-4 \times 2+3| \mid}{\sqrt{3^2+4^2}}=\frac{5}{5}=1 \end{aligned} $