Straight Lines and Pair of Straight Lines

Coordinate of $P$ is $\left(\frac{1+3}{2}, \frac{3+7}{2}\right)=(2,5)$

Coordinate of $Q$ is $\left(\frac{3+7}{2}, \frac{7+15}{2}\right)=(5,11)$

Let coordinates of $R$ be $(x, y)$,

$\begin{aligned} & (A C)^2+(Q R)^2=(P R)^2 \\ & \Rightarrow(7-1)^2+(15-3)^2+(5-x)^2+(11-y)^2 \\ & \quad=(2-x)^2+(5-y)^2 \\ & \Rightarrow \quad 36+144+25+x^2-10 x+121+y^2-22 y \\ & \quad=4+x^2-4 x+25+y^2-10 y \\ & \therefore \quad 6 x+12 y=297 \end{aligned}$

The lines formed by $|x+y|=2$ are $x+y=2$ and $x+y=-2$

$A B C D$ is the rhombus with both diagonals 4 units.

$\begin{aligned} \text { Area of rhombus } & =\frac{1}{2} \times d_1 \times d_2 \\ & =\frac{1}{2} \times 4 \times 4 \\ & =8 \text { sq units } \end{aligned}$

Let the perpendicular bisectors $x-y+5=0$ and $x+2 y=0$ of the sides $A B$ and $A C$ intersect at $D$ and $E$, respectively.

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

Thus, the coordinates of $D=\left(\frac{x_1+1}{2}, \frac{y_1-2}{2}\right)$ and the coordinates of $E=\left(\frac{x_2+1}{2}, \frac{y_2-2}{2}\right)$

The point $D$ lies on the line $x-y+5=0$

$x_1-y_1+13=0 \quad \text{.... (i)}$

The point $E$ lies on the line $x+2 y=0$

$x_2+2 y_2-3=0 \quad \text{... (ii)}$

Since, side $A B$ perpendicular to line $x-y+5=0$

$\begin{aligned} & \Rightarrow \quad 1 \times\left(\frac{y_1+2}{x_1-1}\right)=-1 \\ & \Rightarrow \quad x_1+y_1+1=0 \quad \text{... (iii)} \end{aligned}$

Similarly, side $A C$ is perpendicular to the line

$\begin{aligned} x+2 y & =0 \\ \Rightarrow \quad 2 x_2-y_2 & =4 \quad \text{... (iv)} \end{aligned}$

Solving Eqs. (i) and (iii), we get $x_1=-7$, and $y_1=6$

Solving Eqs. (ii) and (iv), we get $x_2=\frac{11}{5}, y_2=\frac{2}{5}$ Thus, coordinate of $C$ are $\left(\frac{11}{5}, \frac{2}{5}\right)$ and coordinate of $B$ are $(-7,6)$.

Thus, equation of the line $B C$ is

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

Since, given that $9 x^2+a x y+4 y^2+6 x+b y-3=0$ represents a pair of parallel lines, we have

$\begin{aligned} \left(\frac{a}{2}\right)^2 & =9 \times 4 \\ \Rightarrow \quad a^2 & =144 \Rightarrow a= \pm 12 \end{aligned}$

$\begin{aligned} & \text { and }\left|\begin{array}{ccc} 9 & \frac{a}{2} & 3 \\ \frac{a}{2} & 4 & \frac{b}{2} \\ 3 & \frac{b}{2} & -3 \end{array}\right|=0 \\ & \Rightarrow\left\{-12-\frac{b^2}{4}\right\}+\frac{a}{2}\left(\frac{3 b}{2}-\frac{3 a}{2}\right)+3\left(\frac{a b}{4}-12\right)=0 \end{aligned}$

If $a=12$, then

$\begin{aligned} & 9\left(\frac{b}{2}\right)^2-36\left(\frac{b}{2}\right)+36=0 \\ \Rightarrow \quad & b=4 \end{aligned}$

If $a=-12$, then $b=-4$

Thus, $a= \pm 12$ and $b= \pm 4$

Equation of the line through $P(2,3)$ is $\frac{x-2}{\cos \theta}=\frac{y-3}{\sin \theta}$ where $\theta=30^{\circ}$

If $P A=r_1, P B=r_2$ then $r_1, r_2$ are the roots of the equation,

$\begin{aligned} & (2+r \cos \theta)^2-2(2+r \cos \theta) \cdot(3+r \sin \theta) -(3+r \sin \theta)^2=0 \\ & \Rightarrow r^2(\cos 2 \theta-\sin 2 \theta)-2 r(\cos \theta+5 \sin \theta)-17=0 \\ \end{aligned}$

So, $P A \cdot P B=r_1 r_2=\frac{-17}{\cos 2 \theta-\sin 2 \theta}$

$\begin{aligned} & =\frac{-17}{\cos 60^{\circ}-\sin 60^{\circ}} \\ & =17(\sqrt{3}+1) \end{aligned}$

$\begin{aligned} & \quad A B=2 \text { units } \\ & \text { and } O B=3 \text { units } \\ & \text { In } \triangle B O C, \angle C+\angle B+\angle O=180^{\circ} \\ & \Rightarrow \quad 45^{\circ}+\angle B+90^{\circ}=180^{\circ} \quad\left[\because \text { slope }=\tan 45^{\circ}=1\right] \\ & \Rightarrow \quad \angle B=45^{\circ} \end{aligned}$

By cosine formula,

$\begin{array}{ll} & \cos 45^{\circ}=\frac{O B^2+A B^2-O A^2}{2 O B \times A B} \\ \Rightarrow & \frac{1}{\sqrt{2}}=\frac{9+4-O A^2}{2 \times 3 \times 2} \\ \Rightarrow & \frac{12}{\sqrt{2}}=13-O A^2 \\ \Rightarrow & 6 \sqrt{2}=13-O A^2 \\ \Rightarrow & O A^2=13-6 \sqrt{2} \\ \Rightarrow & O A=\sqrt{13-6 \sqrt{2}} \end{array}$

$A(-3,4), C(0,1)$

Let $R \equiv(x, y)$

$\begin{aligned} & A B+B C=\text { minimum } \\ & \sqrt{(x+3)^2+(y-4)^2}+\sqrt{x^2+(y-1)^2} \quad \text{.... (i)} \end{aligned}$

is minimum

$B$ is on $2 x+y=7 \Rightarrow y=7-2 x$

$\sqrt{(x+3)^2+(7-2 x-4)^2}+\sqrt{x^2+(7-2 x-1)^2} \quad \text{... (ii)}$

$\begin{aligned} \Rightarrow \sqrt{x^2+9+6 x+9+4 x^2-12 x} & \text { minimum } \\ +\sqrt{x^2+36+4 x^2-24 x} & \text { minimum } \\ \Rightarrow \sqrt{5 x^2-6 x+18}+\sqrt{5 x^2-24 x+36} & \text { minimum } \end{aligned}$

On differentiting w.r.t. $x$,

$\begin{aligned} & L^{\prime}=\left(\frac{1}{2}\right) \frac{10 x-6}{\sqrt{5 x^2-6 x+18}}+\frac{1}{2} \frac{10 x-24}{\sqrt{5 x^2-24 x+36}}=0 \\ &=(10 x-6) \sqrt{5 x^2-24 x+36} \\ & \quad+(10 x-24) \sqrt{5 x^2-6 x+18}=0 \end{aligned}$

$\begin{aligned} & \Rightarrow(10 x-6) \sqrt{5 x^2-24 x+36} \\ & \quad=-(10 x-24) \sqrt{5 x^2-6 x+18} \\ & \Rightarrow \quad(10 x-6)^2\left(5 x^2-24 x+36\right) \\ & \quad=(10 x-24)^2\left(5 x^2-6 x+18\right) \\ & \Rightarrow \quad\left(25 x^2+9-30 x\right)\left(5 x^2-24 x+36\right) \\ & \quad=\left(25 x^2+144-120 x\right)\left(5 x^2-6 x+18\right) \\ & \Rightarrow \quad 25 x^2-192 x+252=0 \\ & \Rightarrow \quad 25 x^2-186 x-6 x+252=0 \\ & \Rightarrow \quad x=6 \text { or } x=\frac{84}{50}=\frac{42}{25} \\ & \because \quad 2 x+y=7 \\ & \text { For } x=6 \Rightarrow 2(6)+y=7 \Rightarrow y=-5 \\ & \text { For } x=\frac{42}{25} \Rightarrow 2\left(\frac{42}{25}\right)+y=7 \\ & \Rightarrow \quad \frac{84}{25}+y=7 \Rightarrow y=\frac{91}{25} \end{aligned}$

When we substitute the value $(x, y)=(6,-9)$ and $\left(\frac{42}{25}, \frac{91}{25}\right)$ in Eq. (i) or when we substitute $x=6$ and $x=\frac{42}{25}$ in Eq. (ii), we get Minimum value occurs when $x=42 / 25$

$\begin{aligned} \therefore \quad B & \equiv\left(\frac{42}{25}, \frac{91}{25}\right) \\ \text { and } A B & =\sqrt{\left(\frac{42}{25}+3\right)^2+\left(\frac{91}{25}-4\right)^2} \\ & =\sqrt{\left(\frac{117}{25}\right)^2+\left(\frac{-9}{25}\right)^2}=\sqrt{\frac{13770}{25}} \end{aligned}$

For points inside the triangle,

$\begin{aligned} & x+y<10, x>0 \text { and } y>0 \\ & x=1 \Rightarrow(1, y) \\ & \text { i.e. } y=1,2,3,4,5,6,7,8 \Rightarrow 8 \text { points } \\ & \quad x=2 \Rightarrow(2, y) \\ & \text { i.e. } y=1,2,3,4,5,6,7 \Rightarrow 7 \text { points } \\ & x=3 \text { gives } 6 \text { points } \\ & x=4 \text { gives } 5 \text { points } \end{aligned}$

$\begin{aligned} x & =5 \text { gives } 4 \text { points } \\ x & =6 \text { gives } 3 \text { points } \\ x & =7 \text { gives } 2 \text { points } \\ x & =8 \text { gives } 1 \text { point } \\ x & =9 \text { gives no points } \\ \Rightarrow 8+7 & +6+5+4+3+2+1=36 \end{aligned}$

Given,

$a x^2+2 h x y+b y^2+2 g x+2 f y+c=0 \quad \text{.... (i)}$

This represents pair of straight lines when

$a b c+2 f g h-a f^2-b g^2-c h^2=0 \quad \text{... (ii)}$

When the pair of lines intersect $X$-axis, $y=0$.

Put $y=0$ in Eq. (i),

$a x^2+2 g x+c=0$

This is a quadratic equation in $x$. Since lines meet at $X$-axis, the roots must be equal.

So, $b^2-4 a c=0$

$\Rightarrow 4 g^2-4 a c=0 \Rightarrow g^2=a c$

Put $g^2=a c$ in Eq. (ii), we get

$\begin{aligned} & a b c+2 f g h-a f^2-b a c-c h^2=0 \\ \Rightarrow \quad & 2 f g h=a f^2+c h^2 \quad \text{.... (iii)} \end{aligned}$

Eq. (iii) is possible when $f=h$ and $a+c=2 g$

$\begin{aligned} & [x \cos \theta-y] \\ & {[(\cos \theta+\tan \alpha) x-(1-\cos \theta \tan \alpha) y]=0} \\ & \Rightarrow \cos \theta(\cos \theta+\tan \alpha) x^2-\left(\cos \theta-\cos ^2 \theta \tan \alpha\right) x y \\ & -(\cos \theta+\tan \alpha) y x+(1-\cos \theta \tan \alpha) y^2=0 \\ & \Rightarrow \cos \theta(\cos \theta+\tan \alpha) x^2+x y\left(-\cos \theta+\cos ^2 \theta\right. \\ & \tan \alpha-\cos \theta-\tan \alpha)+(1-\cos \theta \tan \alpha) y^2=0 \\ & \Rightarrow \cos \theta(\cos \theta+\tan \alpha) x^2+(-2 \cos \theta \\ & \left.-\sin ^2 \theta \tan \alpha\right) x y+(1-\cos \theta \tan \alpha) y^2=0 \\ & \Rightarrow \cos \theta(\cos \theta+\tan \alpha) x^2-\left(2 \cos \theta+\sin ^2 \theta \tan \alpha\right) x y \\ & \quad+(1-\cos \theta \tan \alpha) y^2=0 \end{aligned}$

We know, $\tan \theta=\left|\frac{2 \sqrt{h^2}-a b}{a+b}\right|$

$ = \left| {2{{\sqrt {{{\left( {{{2\cos \theta + {{\sin }^2}\theta \tan \alpha } \over 2}} \right)}^2} - \cos \theta (\cos \theta + \tan \alpha )(1 - \cos \theta \tan \alpha )} } \over {\cos \theta (\cos \theta + \tan \alpha ) + (1 - \cos \theta \tan \alpha )}}} \right|$

$\begin{aligned} & =\frac{\tan \alpha \sqrt{\cos ^4 \theta+2 \cos ^2 \theta+1}}{1+\cos ^2 \theta}=\tan \alpha \\ & \Rightarrow \tan \theta=\tan \alpha \\ & \Rightarrow \theta=\alpha \end{aligned}$

Given equation,

$9 x^2+4 y^2+10 x+12 y+1=0$ .... (i)

Let origin $(0,0)$ shifted to $(h, k)$, then

$\begin{gathered} x \rightarrow x-h \text { and } y=y-k \text {, putting in Eq. (i), we get } \\ \begin{aligned} & 9(x-h)^2+4(y-k)^2+10(x-h)+12(y-k)+1=0 \\ & \Rightarrow 9 x^2+4 y^2+9 h^2+4 k^2-18 x h-8 y k \\ &+10 x-10 h+12 y-12 k+1=0 \\ & \Rightarrow\left(9 x^2+4 y^2\right)-x(18 h-10)-y(8 k-12) \\ &+\left(9 h^2+4 k^2-10 h-12 k+1\right)=0 \end{aligned} \end{gathered}$

If equation is without $x$ and $y$ term, then

$\begin{aligned} 18 h-10 & =0 \\ \Rightarrow h & =\frac{5}{9} \\ 8 k-12 & =0 \\ \Rightarrow k & =\frac{3}{2} \\ \end{aligned}$

$\therefore (h, k) =\left(\frac{5}{9}, \frac{3}{2}\right) $.

Let ABCD be square, then AC be diagonal.

Equation of AC be

$\begin{array}{ll} & \frac{y-3}{5-3}=\frac{x-1}{7-1} \\ \Rightarrow & y-3=\frac{2(x-1)}{6} \\ \Rightarrow & y-3=\frac{x-1}{3} \\ \Rightarrow \quad & 3 y-9=x-1\end{array}$

or, $\quad 3y=x+8$ ..... (i)

Slope of line $AC=\frac{1}{3}$

Let the equation of side passes through A be

$y=mx+c$ ...... (ii)

Since, angle between Eqs. (i) and (ii) is $45^{\circ}$

$\begin{aligned} \tan 45^{\circ} & =\left|\frac{\frac{1}{3}-m}{1+m / 3}\right| \\ \Rightarrow \quad & =\left|\frac{1-3 m}{3+m}\right|=1 \end{aligned}$

i.e. $\frac{1-3 m}{3+m}=1$

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

Equation of line become,

$y=\frac{-1}{2} x+C$

and $y=2 x+C$

Since, it passes through $A(1,3)$, we obtain

$\begin{gathered} (3)=\frac{-1}{2}+C \Rightarrow C=\frac{-7}{2} \\ 3=2+C \Rightarrow C=1 \end{gathered}$

$\therefore$ Equation of side become

$\begin{aligned} & y=\frac{-1}{2} x+\frac{7}{2} \quad \Rightarrow \quad 2 y+x-7=0 \\ & y=2 x+1 \quad \Rightarrow \quad 2 x-y+1=0 \end{aligned}$

Centroid of $\Delta ABC$ having coordinates (x, y) is

$\begin{aligned} & x=\frac{2+3+1}{3}=2 \\ & y=\frac{4-1+3}{3}=2 \end{aligned}$

$\therefore$ Centroid $(x, y)=(2,2)$

Point of intersection of line $x+3 y-1=0$ and $3 x-y+1=0$ is obtained by solving these equation for $x$ and $y$, we obtain

$P(\alpha, \beta)=P\left(\frac{-1}{5}, \frac{2}{5}\right)$

Equation of line passes through $C(2,2)$ and $P\left(\frac{-1}{5}, \frac{2}{5}\right)$ is given as

$\begin{aligned} \frac{y-2}{\frac{2}{5}-2} & =\frac{x-2}{-\frac{2}{5}-2} \\ \Rightarrow \quad \frac{y-2}{-3} & =\frac{x-2}{-11} \\ \Rightarrow \quad 11(y-2) & =8(x-2) \end{aligned}$

or $\quad 11y-8x=6$ ...... (i)

Among the given option only $(-9,-6)$ satisfy Eq. (i).

Let point (1, 2) be denoted as P(1, 2)

Given line, $3x-y=7$ have slope 3.

Given, $P Q \| 3 x-y=7$

$\Rightarrow \quad$ Slope of $P Q=3$

Equation of line $P Q$ which passes through $(1,2)$ and have slope 3 is

$\begin{aligned} y-2 & =3(x-1) \\ \Rightarrow \quad y-3 x+1 & =0 \end{aligned}$

Then point of intersection of $P Q$ and $x+y+5=0$ is obtained by solving

$\begin{gathered} x+y=-5 \\ 3 x-y=1 \end{gathered}$

Solve, we get $x=-1, y=-4$

$\begin{aligned} & \therefore \text { Point } Q(-1,-4) \\ & \text { Distance } P Q=\sqrt{(-1-1)^2+(-4-2)^2} \\ & =\sqrt{4+36}=\sqrt{40} \end{aligned}$

Given equation,

$\begin{aligned} & x \cos \theta-y \sin \theta=2 \cos 2 \theta \\ & \Rightarrow \quad y=x \cot \theta-2 \cos 2 \theta / \sin \theta \end{aligned}$

$\therefore$ Slope of line $=\cot \theta$

Then, slope of line perpendicular to given line $ =\frac{-1}{\cot \theta}=-\tan \theta$

Let equation of line which is perpendicular to $x \cos \theta-y \sin \theta=2 \cos 2 \theta$, is

$\begin{aligned} y & =m x+C \quad ..... (\text{i})\\ \Rightarrow \quad y & =(-\tan \theta) x+C\quad ..... (\text{ii})\end{aligned}$

[ $\because$ its slope is $-\tan \theta$ ]

Since, Eq. (i) passes through $\left(2 \cos ^3 \theta, 2 \sin ^3 \theta\right)$.

From, Eq. (ii),

$\Rightarrow \quad \begin{aligned}2 \sin ^3 \theta & =-\tan \theta \cdot 2 \cos ^3 \theta+C \\ \Rightarrow \quad C & =2 \sin ^3 \theta+2 \sin \theta \cos ^2 \theta \end{aligned}$

$\therefore$ Eq. (ii), becomes

$\begin{aligned} & \qquad \begin{aligned} y & =(-\tan \theta) x+2 \sin ^3 \theta+2 \sin \theta \cos ^2 \theta \\ y & =-\frac{x \sin \theta}{\cos \theta}+2 \sin ^3 \theta+2 \sin \theta \cos ^2 \theta \end{aligned} \\ & \Rightarrow y \operatorname{cosec} \theta=-x \sec \theta+2\left(\sin ^2 \theta+\cos ^2 \theta\right) \\ & \Rightarrow y \operatorname{cosec} \theta+x \sec \theta=2 \\ & \text { i.e. } x \sec \theta+y \operatorname{cosec} \theta=2 \end{aligned}$

Given equation,

$x^2+p x y+y^2-5 x-7 y+6=0$ ..... (i)

General equation is

$a x^2+b y^2+2 h x y+2 g y+2 f x+c=0$ ...... (ii)

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

$a=1, b=1, h=\frac{p}{2}, g=\frac{-7}{2}, f=\frac{-5}{2}, c=6$

If Eq. (i) represent pair of equation, then

$\begin{aligned} & \left|\begin{array}{lll} a & h & g \\ h & b & f \\ g & f & c \end{array}\right|=0 \\ & \Rightarrow \quad\left|\begin{array}{ccc} 1 & \frac{p}{2} & \frac{-7}{2} \\ \frac{p}{2} & 1 & \frac{-5}{2} \\ \frac{-7}{2} & \frac{-5}{2} & 6 \end{array}\right|=0 \\ & \Rightarrow 6-\frac{25}{4}-\frac{p}{2}\left(\frac{6 p}{2}-\frac{35}{4}\right) \frac{-7}{2}\left(\frac{-5 p}{4}+\frac{7}{2}\right)=0 \\ & \Rightarrow \quad \frac{-1}{4}-\frac{6 p^2}{4}+\frac{35 p}{8}+\frac{35 p}{8}-\frac{49}{4}=0 \\ & \Rightarrow \quad \frac{35 p}{4}-\frac{6 p^2}{4}=\frac{50}{4} \\ & \Rightarrow \quad 6 p^2-35 p+50=0 \\ & \Rightarrow \quad(2 p-5)(3 p-10)=0 \\ & \Rightarrow \quad p=\frac{5}{2} \text { and } p=\frac{10}{3} \\ & \end{aligned}$

$a x^2+2 h x y+b y^2=0$ represent a homogeneous line.

Let the two lines by $y=m_1 x, y=m_2 x$

The line $y=m_1 x$ passes through $(2,3)$ is

$\begin{aligned} 3 & =m_1 \times 2 \\ \Rightarrow \quad m_1 & =3 / 2 \end{aligned}$

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

Similarly, equation of the second line which passes through $(4,5)$ is

$5 x-4 y=0$

Thus, the equation of pair of lines is

$\begin{aligned} & (5 x-4 y)(3 x-2 y)=0 \\ & 15 x^2-22 x y+8 y^2=0 \end{aligned}$

Compare it with the given equation

$a=15, h=-11, b=8$

Hence, $a+2 h+8=15-22+8=1$

Given pair of lines

$2 x^2-p x y+2 y^2=0$ ...... (i)

Divide Eq. (i) by $x^2$

$\begin{aligned} & \quad 2-p \frac{y}{x}+2\left(\frac{y}{x}\right)^2=0 \\ & \Rightarrow \quad 2-p m+2 m^2=0\left(\because y=m x \Rightarrow \frac{y}{x}=m\right) \\ & \Rightarrow \quad 2 m^2-p m+2=0 \\ & \Rightarrow \quad m=\frac{p \pm \sqrt{p^2-16}}{4} \\ & \text { If roots are real, then } p^2-16 \geq 0 \\ & \Rightarrow \quad p^2 \geq 16 \\ & \Rightarrow \quad p<-4 \text { and } p>4 \\ & \therefore(-\infty,-4) \cup(4, \infty) \text { is the desired interval. } \end{aligned}$

Let the original coordinates of point P be (x, y) and the new coordinates after rotation be (x', y').

We know that the transformation equations for rotating the axes by an angle θ are:

$x' = x \cos θ + y \sin θ$

$y' = -x \sin θ + y \cos θ$

In this case, θ = 45° and (x', y') = (1, -1). Substituting these values into the transformation equations, we get:

$1 = x \cos 45° + y \sin 45°$

$-1 = -x \sin 45° + y \cos 45°$

Simplifying these equations, we get:

$1 = \frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}}$

$-1 = -\frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}}$

Adding these equations, we get:

$0 = \frac{2y}{\sqrt{2}}$

$y = 0$

Substituting y = 0 into the first equation, we get:

$1 = \frac{x}{\sqrt{2}}$

$x = \sqrt{2}$

Therefore, the original coordinates of point P are ($\sqrt{2}$, 0).

So the answer is Option B.

Equation of line intercept form is

$\frac{x}{a}+\frac{y}{b}=1$ .... (i)

If it passes through $(5,6)$, then

${{ - 5} \over a} + {6 \over a} = 1 \quad {[\because a=b]}$

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

From Eq. (i), $x+y=1$

Equation of the line : $y=m x+4$

Distance from origin

$\frac{|m \cdot 0-0+4|}{\sqrt{1+m^2}}=\frac{4}{\sqrt{1+m^2}}$

$x+y=2$

Vertex : $(2,-1)$

Clearly, $2-1 \neq 2$. So, vertex does not lie on $x+y=2$

$\therefore$ Distance $=\frac{2}{\sqrt{3}}\left(\frac{|2-1-2|}{\sqrt{1^2+1^2}}\right)=\frac{2}{\sqrt{3}} \times \frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{\sqrt{3}}$

If slope of required line $=m$

$m\left(-\frac{\sec \theta}{\operatorname{cosec} \theta}\right)=-1 \Rightarrow m=\frac{\operatorname{cosec} \theta}{\sec \theta}=\cot \theta$

Equation required line

$\begin{aligned} &\left(y-a \sin ^3 \theta\right)=\cot \theta\left(x-a \cos ^3 \theta\right) \\ & \Rightarrow y=x \cot \theta-a \cot \theta \cos ^3 \theta+a \sin ^3 \theta \\ & \Rightarrow y \sin \theta=x \cos \theta-a\left(\cos ^4 \theta-\sin ^4 \theta\right) \\ & \Rightarrow y \sin \theta=x \cos \theta-a\left(\cos ^2 \theta-\sin ^2 \theta\right)\left(\cos ^2 \theta+\sin ^2 \theta\right) \\ & \Rightarrow \quad x \cos \theta-y \sin \theta=a \cos 2 \theta \\ \end{aligned}$

$6 x^2+11 x y-10 y^2 =0$

$\Rightarrow (3 x-2 y)(2 x+5 y)=0$

$\Rightarrow y =\left(\frac{3}{2}\right) x$ or $y=\left(\frac{-2}{5}\right) x$

$\begin{aligned} \tan \theta & =\frac{\frac{3}{2}+\frac{2}{5}}{1-\frac{3}{2} \cdot \frac{2}{5}}=\frac{\frac{19}{10}}{\frac{2}{5}}=\frac{19}{4} \\ \Rightarrow \quad \theta & =\tan ^{-1}\left(\frac{\sqrt{361}}{4}\right)\end{aligned}$

$\begin{aligned} & C: 2 x^2-2 x y+3 y^2+2 x-y-1=0 \\ & L: x+2 y=k \\ & \quad \frac{x+2 y}{k}=1 \\ & \begin{aligned} \Rightarrow \quad\left(2 x^2-2 x y+3 y^2\right)+(2 x-y) \cdot 1-1^2=0 \\ \Rightarrow \quad\left(2 x^2-2 x y+3 y^2\right)+(2 x-y) \frac{(x+2 y)}{k} \\ \quad-\left(\frac{x+2 y}{k}\right)^2=0 \\ k^2\left(2 x^2-2 x y+3 y^2\right)+k(2 x-y)(x+2 y)-(x+2 y)^2=0 \\ \Rightarrow x^2\left(2 k^2+2 k-1\right)+x y\left(-2 k^2+3 k-4\right) \\ \quad+y^2\left(3 k^2-2 k-4\right)=0 \end{aligned} \end{aligned}$

These lines are mutually perpendicular

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

$\begin{aligned} \Rightarrow\left(2 k^2+2 k-1\right)+\left(3 k^2-2 k-4\right) & =0 \\ \Rightarrow \quad 5 k^2-5 & =0 \\ k & = \pm 1 \Rightarrow k^2=1 \end{aligned}$

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

Comparing it with $a x^2+2 h x y+b y^2=0$

$a=3,2 h=-5 \text { and } b=4$

Equation of the bisector of the angle between the pair of lines is

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

Bisectors of $a x^2+2 h x y+b y^2=0$ is

$\begin{gathered} \frac{x^2-y^2}{a-b}=\frac{x y}{h} \\ \Rightarrow \quad h\left(x^2-y^2\right)=(a-b) x y \\ \Rightarrow \quad h x^2-(a-b) x y-h y^2=0 \\ \text { Here, } x^2-2 m x y-y^2=0 \\ \quad a=1, b=-1, h=-m \\ \text { Bisector } \Rightarrow-m x^2-2 x y+m y^2=0 \end{gathered}$

Given bisector is $x^2-2 n x y-y^2=0$

$\begin{array}{rlrl} & \frac{-m}{1}=\frac{-2}{-2 n}=\frac{m}{-1} \\ \Rightarrow &-m n =1 \\ \Rightarrow & 1+m n =0 \end{array}$

Equation of line passing through $A(4,7)$ and $B(-7,8)$ is $\frac{x-4}{-11}=\frac{y-7}{1}$

Or $x+11y-81=0$ ..... (i)

Mid-point of $A B=\left(\frac{-3}{2}, \frac{15}{2}\right)$

Let equation of line perpendicular to line (i) is

$11 x-y+C=0$

This line passes through $\left(\frac{-3}{2}, \frac{15}{2}\right)$

$\therefore \quad 11 \times \frac{-3}{2}-\frac{15}{2}+C=0 \Rightarrow C=24$

$\therefore$ Required equation of perpendicular bisector of $A B$ is

$11 x-y+24=0$

Let the straight line

$3 x+4 y=6$ .... (i)

divides the line joining the points $(2,-1)$ and $(1,1)$ in the ratio $\lambda: 1$. Then,

$x=\frac{2+\lambda}{\lambda+1}, y=\frac{-1+\lambda}{\lambda+1}$

This point lies on Eq. (i)

$\begin{aligned} \frac{3(2+\lambda)}{\lambda+1}+4 \frac{(-1+\lambda)}{\lambda+1} & =6 \\ 6+3 \lambda-4+4 \lambda & =6 \lambda+6 \\ \lambda & =4 \\ \Rightarrow \quad \lambda: 1 & =4: 1 \end{aligned}$

Let the equation of line be $\frac{x}{a}+\frac{y}{b}=1$

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

Area with the axes $=\frac{1}{2} a b=\frac{1}{2} \cdot \frac{3 a^2}{a-4}$

Area is minimum

$\begin{array}{rlrl} & \Rightarrow \frac{d}{d a}(\text { area }) =0 \\ & \Rightarrow \frac{d}{d x}\left(\frac{a^2}{a-4}\right) =0 \\ \Rightarrow & (a-4) \cdot 2 a-a^2 =0 \Rightarrow a^2-8 a=0 \\ \Rightarrow & a =8 \Rightarrow b=6 \\ \end{array}$

$\therefore$ Equation of line $\frac{x}{8}+\frac{y}{6}=1$

$\Rightarrow \quad \frac{3 x+4 y}{24}=1 \text { or } 3 x+4 y=24$

In $\Delta ABC$

$AB:2x + 3y - 1 = 0$

$AC:x + 2y + 1 = 0$

$BC:ax + by - 1 = 0$

$AD:(2x + 3y - 1) + \lambda (x + 2y + 1) = 0$

AD passes through origin

$ \Rightarrow - 1 + \lambda = 0$

or $\lambda = 1$

$\therefore$ $AD,3x + 5y = 0$

$\because$ $AD \bot BC \Rightarrow \left( {{{ - 3} \over 5}} \right)\left( {{{ - a} \over b}} \right) = - 1$ $[{m_1}\,.\,{m_1} = - 1]$

$ \Rightarrow 3a + 5b = 0$ .,...... (i)

Again, $BE,(2x+3y-1)+\propto(ax+by-1)=0$

BE passes through origin

$\Rightarrow -1-\propto=0\Rightarrow \propto = -1$

$BE,(2 - a)x + (3 - b)y = 0$

$ \Rightarrow \left( {{{ - 1} \over 2}} \right) - \left( {{{2 - a} \over {3 - b}}} \right) = - 1$

$ \Rightarrow a + 2b = 8$ ...... (ii)

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

$a = - 40,b = 24$

$\therefore$ ${1 \over a} + {1 \over b} = {{ - 1} \over {40}} + {1 \over {24}} = {1 \over {60}}$

Comparing given equation,

$8 x^2-24 x y+18 y^2-6 x+9 y-5=0$

On comparing with general form

$\begin{aligned} & \quad a x^2+2 h x y+b y^2+2 g x+2 f y+c=0 \\ & a=8, h=-12, b=18, g=-3, f=9 / 2, c=-5 \\ & \because \quad h^2=144 \text { and } a b=8 \times 18=144 \\ & \Rightarrow \quad h^2=a b \\ & \text { and } a f^2=8\left(\frac{9}{2}\right)^2=8 \times \frac{81}{4}=2 \times 81=162 \\ & \qquad b g^2=18(-3)=18 \times 9=162 \Rightarrow a f^2=b g^2 \\ & \because h^2=a b \text { and } a f^2=b g^2 \end{aligned}$

$\therefore$ Given pair of lines are parallel.

Given pair of line

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

where as $a=1, h=2, b=1$

Angle between lines

$\begin{aligned} \tan \theta & =\left|\frac{2 \sqrt{h^2-a b}}{a+b}\right| \Rightarrow \tan \theta=\left|\frac{2 \sqrt{4-1}}{2}\right| \\ \tan \theta & =\sqrt{3} \\ \Rightarrow \quad \theta & =60 \Upsilon \end{aligned}$

Angle between pair of lines

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

is given by

$\begin{array}{rlrl} \tan \theta & =\left|\frac{2 \sqrt{h^2-a b}}{a+b}\right| \\ \because & =\frac{\pi}{4} \\ \tan \frac{\pi}{4} & =\left|\frac{2 \sqrt{h^2-a b}}{a+b}\right| \\ \Rightarrow \quad\left(\frac{a+b}{2}\right)^2 & =h^2-a b \\ \Rightarrow \quad a^2+b^2+2 a b & =4 h^2-4 a b \\ \Rightarrow \quad 4 h^2 & =a^2+b^2+6 a b \end{array}$

Given pair of lines,

$\begin{aligned} & \cos \theta(\cos \theta+1) x^2-\left(2 \cos \theta+\sin ^2 \theta\right) x y \\ &+(1-\cos \theta) y^2=0 \end{aligned}$

Comparing with

$\begin{gathered} a x^2+2 h x y+b y^2=0 \\ a=\cos \theta(\cos \theta+1) \\ h=-\frac{\left(2 \cos \theta+\sin ^2 \theta\right)}{2}, \\ b=1-\cos \theta \\ a+b=\cos ^2 \theta+1 \\ h^2-a b=\frac{4 \cos ^2 \theta+\sin ^4 \theta+4 \cos \theta \sin ^2 \theta}{4} \\ =\frac{4 \cos ^2 \theta+\left(1-\cos ^2 \theta\right)^2+4 \cos \theta \sin ^2 \theta-4 \cos \theta \sin ^2 \theta}{4} \\ =\frac{\cos ^4 \theta+2 \cos ^2 \theta+1}{4}=\left(\frac{\left(\cos ^2 \theta+1\right)}{2}\right)^2 \end{gathered}$

If $\theta$ be the acute angle between the lines, then

$\begin{aligned} \tan \theta & =\frac{2 \sqrt{h^2-a b}}{|a+b|}=\frac{2\left(\frac{\cos ^2 \theta+1}{2}\right)}{\left(\cos ^2 \theta+1\right)}=1 \\ \Rightarrow \quad \theta & =\frac{\pi}{4} \end{aligned}$

Given, coordinate $(x, y)=(2 \sqrt{2},-3 \sqrt{2})$

Let new coordinates be $\left(x^{\prime}, y^{\prime}\right)$. Then,

$\begin{aligned} & x^{\prime}=x \cos \theta+y \sin \theta \\ & \text { and } y^{\prime}=-x \sin \theta+y \cos \theta \\ & \text { Here, } \theta=45 \Upsilon \\ & \therefore \quad x^{\prime}=2 \sqrt{2} \cos 45 Y+(-3 \sqrt{2}) \sin 45 Y \\ & =2 \sqrt{2} \cdot \frac{1}{\sqrt{2}}-3 \sqrt{2} \cdot \frac{1}{\sqrt{2}}=2-3=-1 \\ & \text { and } y^{\prime}=-2 \sqrt{2} \sin 45 Y+(-3 \sqrt{2}) \cos 45 Y \\ & =-2 \sqrt{2} \cdot \frac{1}{\sqrt{2}}-3 \sqrt{2} \cdot \frac{1}{\sqrt{2}}=-2-3=-5 \\ \end{aligned}$

New coordinate is $(-1,-5)$.

Given line,

$5x-2y=10$

${{5x} \over {10}} - {{2y} \over {10}} = 1 \Rightarrow {x \over 2} + {y \over { - 5}} = 1$

Intercept on X-axis = 2

Intercept on Y-axis = $-$5

Sum of the squares of the intercepts = 2$^2$ + ($-$5)$^2$ = 29

$a, b$ and $c$ are three consecutive odd integers.

$\begin{array}{lc} \Rightarrow & a, b \text { and } c \text { are in AP. } \\ \Rightarrow & 2 b=a+c \\ \Rightarrow & a-2 b+c=0 \end{array}$

which is similar to the line $a x+b y+c=0$ passing through $(1,-2)$.

$\begin{array}{lc} \therefore & (\alpha, \beta)=(1,-2) \\ \Rightarrow & \alpha^2+\beta^2=1^2+(-2)^2=5 \end{array}$

The distance of point A from the line will be equal to the distance of B from line.

$\therefore$ ${{|2 + 6 + 4|} \over {\sqrt {13} }} = {{|2\alpha + 3\beta + 4|} \over {\sqrt {13} }}$

$ \Rightarrow 2\alpha + 3\beta + 4 = \pm 12$

$ \Rightarrow 2\alpha + 3\beta = 8$ ..... (i)

$2\alpha + 3\beta = - 16$ ..... (ii)

Also, the line joining A and B is perpendicular to

$2x + 3y + 4 = 0$

$\Rightarrow$ Product of slopes = $-1$

$\therefore$ ${{\beta - 2} \over {\alpha - 1}} \times \left( {{{ - 2} \over 3}} \right) = - 1$

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

$ \Rightarrow 2\beta - 3\alpha = 1$ .... (iii)

Solving Eqs. (ii) and (iii), we have,

$\alpha = {{ - 35} \over {13}},\beta = {{ - 46} \over {13}}$

$\therefore$ $13\alpha + 13\beta = - 35 - 46 = - 81$

The equation of the pair of lines perpendicular to the pair $a x^2+2 h x y+b y^2+2 g x+2 f y+c=0$ and passes through origin is given by

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

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

$\therefore$ Required pair of lines $2 x^2-3 x y+2 y^2=0$

Given lines,

$x+y =k$ ..... (i)

and $2 y^2+5 x y-3 x^2 =0$

or $2 y^2+6 x y-x y-3 x^2 =0$

or $2 y(y+3 x)-x(y+3 x) =0$

or $(y+3 x)(2 y-x) =0$

$\Rightarrow y+3 x =0$ .... (ii)

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

Solving Eqs. (i), (ii) and (iii), we get vertices of triangle, which are $(0,0),\left(\frac{2 k}{3}, \frac{k}{3}\right)$ and $\left(\frac{-k}{2}, \frac{3 k}{2}\right)$

$\therefore \text { Centroid }=\left(\frac{k}{18}, \frac{11 k}{18}\right)$

Comparing with given centroid $\left(\frac{1}{18}, \frac{11}{18}\right)$, we have $k=1$

$\begin{aligned} & \text { } 5 x^2-8 x y+3 y^2=0 \\ & \Rightarrow(5 x-3 y)(x-y)=0 \\ & \Rightarrow \quad 5 x-3 y=0 \\ & x-y=0 \\ & \text { Slopes } m_1=\frac{5}{3}, m_2=1 \\ & m_1: m_2=\frac{5}{3}: 1 \\ & =5: 3 \\ & \end{aligned}$

(c) Let $m$ be the slope of one of the lines given by $a x^2+2 h x y+b y^2=0$

Then, the other line has slope $m^2$.

$\begin{array}{rlrl} & \therefore m+m^2=\frac{-2 h}{b} \\ & \text { and } \quad m \cdot m^2=\frac{a}{b} \\ \because \quad & \left(m+m^2\right)^3 =m^3+\left(m^2\right)^3+3 m \cdot m^2\left(m+m^2\right) \\ & \left(-\frac{2 h}{b}\right)^3=\frac{a}{b}+\frac{a^2}{b^2}+3.\frac{a}{b}. \left(-\frac{2 h}{b}\right)& =\frac{a}{b}+\frac{a^2}{b^2}+3 \cdot \frac{a}{b} \cdot\left(-\frac{2 h}{b}\right) \\ \Rightarrow \quad \frac{-8 h^3}{b^3} & =\frac{a}{b}+\frac{a^2}{b^2}-\frac{6 a h}{b^2} \end{array}$

$\begin{array}{ll}\Rightarrow & \frac{a}{b}+\frac{a^2}{b^2}-\frac{6 a h}{b^2}+\frac{8 h^3}{b^3}=0 \\ \text { or } & a^2 b+a b^2+8 h^3=6 a b h \\ \text { or } & a b(a+b)+8 h^3=6 a b h \\ \text { or } & \frac{a+b}{h}+\frac{8 h^2}{a b}=6 \\ \text { or } & \left| {{{a + b} \over h} + {{8{h^2}} \over {ab}}} \right| = 6\end{array}$