Straight Lines and Pair of Straight Lines

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

From Eqs. (i) and (ii), $x=-3, y=0$

From Eqs. (ii) and (iii), $x=3, y=-2$

From Eqs. (i) and (iii), $x=0, y=1$

∴ Vertices of given triangle are ( $-3,0$ ), ( $3,-2$ ) and $(0,1)$.

$ \begin{aligned} \therefore \text { Centroid } & =\left(\frac{-3+3+0}{3}, \frac{0-2+1}{3}\right) \\ & =\left(0, \frac{-1}{3}\right) \end{aligned} $

Given, equation of pair of lines

$ \begin{aligned} & 5 x^2+\frac{40}{3} x y+k y^2=0 \\ \Rightarrow & k\left(\frac{y}{x}\right)^2+\frac{40}{3}\left(\frac{y}{x}\right)+5=0 \end{aligned} $

Here, $y / x=3$ satisfies the equation

$ \therefore \quad 9 k+40+5=0 \Rightarrow k=-5 $

∴ Pair of straight line is

$ 5 x^2+\frac{40}{3} x y-5 y^2=0 $

As coefficients of $x^2$ and $y^2$ are negative of each other, so angle between the lines is right angle.

$\because 6 x^2-x y-12 y^2=0$

$ \begin{aligned} & \Rightarrow \quad(3 x+4 y)(2 x-3 y)=0 \\ & \text { and } \quad 15 x^2+14 x y-8 y^2=0 \\ & \Rightarrow \quad(3 x+4 y)(5 x-2 y)=0 \\ & \therefore \quad 3 x+4 y=0 \text { is common line } \end{aligned} $

∴ Combined equation of other two lines is

$ \begin{aligned} & (2 x-3 y)(5 x-2 y)=0 \\ \Rightarrow \quad & 10 x^2-19 x y+6 y^2=0 \end{aligned} $

Common line of two pairs is

$ 3 x+4 y=0 $

∴ Equation of $L$ is $(y-1)=\frac{-3}{4}(x+1)$

$ \Rightarrow L \text { is } 3 x+4 y-1=0 \Rightarrow 3 x+4 y=1 $

Homoginising given curve with

$ 3 x+4 y-1=0 $

we get

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

∴ Required equation of pair of straight lines is

$ x^2-y^2=0 $

Let $A(5,-3) B(3,-2) C(-1,5)$ be three points.

Let $P(h, k)$ satisfying

$ \begin{aligned} & (P A)^2+2(P B)^2=3(P C)^2 \\ & \begin{aligned} &(h-5)^2+(k+3)^2+2(h-3)^2+2(k+2)^2 \\ & \quad=3(h+1)^2+3(k-5)^2 \\ & \Rightarrow-10 h+25+6 k+9-12 h +18+8 k+8 \end{aligned} \end{aligned} $

$ \begin{aligned} & =6 h+3-30 k+75 \\ \Rightarrow \quad- & 28 h+44 k-18=0 \\ \Rightarrow \quad & 14 h-22 k+9=0 \end{aligned} $

Locus $14 x-22 y+9=0$

Only option (d), $\left(2, \frac{37}{22}\right)$ satisfies the locus.

Given, lines

$ \begin{aligned} & \frac{x}{a}+\frac{y}{b}=1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & \frac{x}{b}+\frac{y}{a}=1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Slope of line (i), $m_1=\frac{-1 / a}{1 / b}=\frac{-b}{a}$

Slope of line (ii), $m_2=\frac{-1 / b}{1 / a}=\frac{-a}{b}$

Then, $\tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right|$

$ =\left|\frac{-\frac{b}{a}+\frac{a}{b}}{1+1}\right|=\left|\frac{a^2-b^2}{2 a b}\right| $

Then, $\sin \theta=\left|\frac{a^2-b^2}{a^2+b^2}\right|$.

Let $A$ be the point of intersection of

$ \begin{gathered} x-y+1=0 \text { and } 2 x+2 y+3=0 \\ A\left(-\frac{5}{4},-\frac{1}{4}\right) \end{gathered} $

$B$ is the point of intersection of

$ \begin{gathered} x-y+1=0 \text { and } 3 x+3 y+2=0 \\ B\left(\frac{-5}{6}, \frac{1}{6}\right) \\ \therefore A B=\sqrt{\left(\frac{-5}{6}+\frac{5}{4}\right)^2+\left(\frac{1}{6}+\frac{1}{4}\right)^2} \end{gathered} $

$ \begin{aligned} &\text { Sides of triangle are } x=2\\ &\begin{aligned} & 4 x+3 y+7=0 \\ & y=3 \\ & B C=a=\sqrt{6^2+8^2}=10 \\ & A C=b=6 \\ & A B=c=8 \end{aligned} \end{aligned} $

$ \begin{aligned} & \left(\frac{a x_1+b x_2+c x_2}{a+b+c}, \frac{a y_1+b y_2+c y_3}{a+b+c}\right) \\ & I\left(\frac{(10 \times 2)+(6 \times 2)+(8 \times-4)}{10+6+8}\right. \\ & \left.\frac{(10 \times 3)+(6 \times-5)+(8 \times 3)}{10+6+8} I(0,1)\right) \end{aligned} $

As $\triangle A B C$ is right angled triangle.

So, mid-point of $B C$ is circumcentre

$ \begin{aligned} & S(-1,-1) \\ & I S=\sqrt{1^2+2^2}=\sqrt{5} \end{aligned} $

Equation $a x^2-4 x y-2 y^2=0$

represents pair of lines.

Also, $\cos \theta=\frac{1}{5} \Rightarrow \tan \theta=2 \sqrt{6}$

$ \begin{aligned} \because \quad \tan \theta & =\frac{2 \sqrt{h^2-a b}}{a+b} \\ 2 \sqrt{6} & =\frac{2 \sqrt{4+2 a}}{a-2} \end{aligned} $

Squaring both sides,

$ \begin{array}{rlrl} & & (a-2)^2 \cdot 6 & =4+2 a \\ \Rightarrow & & (a-2)^2 \cdot 3 & =2+a \\ \Rightarrow & & 3 a^2-12 a+12 & =2+a \\ \Rightarrow & & 3 a^2-13 a+10 & =0 \\ \Rightarrow & & (3 a-10)(a-1) & =0 \\ & & a=1, \frac{10}{3} \end{array} $

$ \begin{aligned} & \text { As } a_2>a_1 \quad \Rightarrow \quad a_1=1, a_2=\frac{10}{3} \\ & \therefore \quad a_1+3 a_2=11 \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text {Lines represented by } \\ & 4 x^2-5 x y+3 y^2=0 \text { are } L_1 \text { and } L_2 \\ & 3\left(\frac{y}{x}\right)^2-5\left(\frac{y}{x}\right)+4=0 \end{aligned} \end{aligned} $

$ \begin{aligned} m_1+m_2 & =\frac{5}{3} \\ m_1 \cdot m_2 & =\frac{4}{3} \\ \frac{1}{m_1}+\frac{1}{m_2} & =\frac{m_1+m_2}{m_1 m_2}=\frac{5}{4} \end{aligned} $

Let the slope of line $L_1$ is $m_1$ and the slope of line $L_2$ is $m_2$.

$L_3$ is perpendicular to $L_1 \Rightarrow$ So, slope of

$ L_3=-\frac{1}{m_1} $

$L_4$ is perpendicular to $L_2 \Rightarrow$ So, slope of

$ L_4=-\frac{1}{m_2} $

Let the equation of $L_3: y=\frac{-1}{m_1} x+c_1$

It passes through $(4,3): 3=\frac{-4}{m_1}+c_1$

$ \begin{aligned} & \Rightarrow c_1=3+\frac{4}{m_1} \quad L_3: y=\frac{-1}{m_1} x+3+\frac{4}{m_1} \\ & \Rightarrow y+\frac{x}{m_1}-\left(3+\frac{4}{m_1}\right)=0 \end{aligned} $

and Equation of $L_4: y=-\frac{x}{m_2}+c_2$

Passes through $(4,3): 3=-\frac{4}{m_2}+c_2$

$ \begin{aligned} & \Rightarrow \quad c_2=3+\frac{4}{m_2} \\ & L_4: y=-\frac{x}{m_2}+3+\frac{4}{m_2} \\ & \Rightarrow \quad y+\frac{x}{m_2}-\left(3+\frac{4}{m_2}\right)=0 \end{aligned} $

Combined equation of $L_3$ and $L_4$

$ \left[y+\frac{x}{m_1}-\left(3+\frac{4}{m_1}\right)\right] $

$ \begin{aligned} & {\left[y+\frac{x}{m_2}-\left(3+\frac{4}{m_2}\right)\right]=0 } \\ & \Rightarrow y^2+\frac{x y}{m_2}-y\left(3+\frac{4}{m_2}\right)+\frac{x y}{m_1}+\frac{x^2}{m_1 m_2} \\ &- \frac{x}{m_1}\left(3+\frac{4}{m_2}\right)-\frac{x}{m_2}\left(3+\frac{4}{m_1}\right) \\ &-y\left(3+\frac{4}{m_1}\right)+\left(3+\frac{4}{m_1}\right)\left(3+\frac{4}{m_2}\right)=0 \\ & \Rightarrow y^2+\frac{x^2}{m_1 m_2}+x y\left(\frac{1}{m_1}+\frac{1}{m_2}\right) \end{aligned} $

$ \begin{array}{r} -x\left(\frac{8}{m_1 m_2}+3\left(\frac{1}{m_1}+\frac{1}{m_2}\right)\right) \\ -y\left(6+4\left(\frac{1}{m_1}+\frac{1}{m_2}\right)\right) \\ +\left(9+12\left(\frac{1}{m_1}+\frac{1}{m_2}\right)+\frac{16}{m_1 m_2}\right)=0 \\ \Rightarrow y^2+\frac{5}{4} x y+\frac{3}{4} x^2-x\left(6+3 \times \frac{5}{4}\right) \\ -y\left(6+4 \times \frac{5}{4}\right) \end{array} $

$ \begin{array}{r} +\left(9+12 \times \frac{5}{4}+16 \times \frac{3}{4}\right)=0 \\ \Rightarrow y^2+\frac{5}{4} x y+\frac{3}{4} x^2-\frac{39}{4} x-11 y+36=0 \\ \Rightarrow 4 y^2+5 x y+3 x^2-39 x-44 y+144=0 \end{array} $

$ \begin{aligned} &\text { On comparing with }\\ &\begin{aligned} & a x^2+2 h x y+b y^2+2 g x+2 f y+c=0 \\ & a=4, h=\frac{5}{2}, b=3, g=\frac{-39}{2}, f=-22 \text { and } \\ & c=144 \\ & a f+b g+c h=-88-\frac{117}{2}+360=213.5 \end{aligned} \end{aligned} $

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

On comparing with

$ \begin{aligned} & a x^2+2 h x y+b y^2+2 g x+2 f y+c=0 \\ & a=1, h=0, b=-1, g=\frac{a}{2}, f=0 \text { and } \\ & c=b \end{aligned} $

If it represents a pair of straight line

Then, $D=0$

$ \begin{aligned} & \Rightarrow p b c+2 f g h-a f^2-b g^2-c h^2=0 \\ & -b+0-0+\frac{a^2}{4}-0=0 \Rightarrow a^2=4 b \end{aligned} $

From the options, $a=6$ and $b=9$ satisfy above relation.

So, pair of $(a, b) \equiv(6,9)$

Let the line $P Q$ be

$y-5=m(x-1) \quad \text{... (i)}$

Substituting $y=m x+5-m$ in the equation

$5 x-y-4=0$

We have, $5 x-m x-5+m-4=0$

$\Rightarrow \quad(5-m) x+m-9=0$

Therefore, $x=\frac{9-m}{5-m}$ and $y=m\left(\frac{9-m}{5-m}\right)+5-m$

$=\frac{25-m}{5-m}$

Here, $Q=\left(\frac{9-m}{5-m}, \frac{25-m}{5-m}\right)$

Substituting $y=m x+5-m$ in the equation

$3 x+4 y-4=0$

We have, $3 x+4(m x+5-m)-4=0$

$(3+4 m) x+16-4 m=0$

Therefore, $x=\frac{4 m-16}{4 m+3}$

$\begin{aligned} \text{and} \quad & y=\frac{m(4 m-16)}{4 m+3}+5-m \\ & y=\frac{m+15}{4 m+3} \end{aligned}$

Hence, $P=\left(\frac{4 m-16}{4 m+3}, \frac{m+15}{4 m+3}\right)$

Since, $m(1,5)$ is the mid-point of $P Q$, we have

$\begin{aligned} 1 & =\frac{1}{2}\left[\frac{9-m}{5-m}+\frac{4 m-16}{4 m+3}\right] \quad \text{... (ii)}\\ \text { and } 5 & =\frac{1}{2}\left[\frac{25-m}{5-m}+\frac{m+15}{4 m+3}\right] \quad \text{... (ii)} \end{aligned}$

From Eq. (ii), we get

$\begin{aligned} & 2(5-m)(4 m+3) \\ & =(9-m)(4 m+3)+(5-m)(4 m-16) \\ & \Rightarrow 2\left(-4 m^2+17 m+15\right)=\left(-4 m^2+33 m+27\right) \\ & +\left(-4 m^2+36 m-80\right) \\ & \Rightarrow-8 m^2+34 m+30=-8 m^2+69 m-53 \\ & \Rightarrow \quad 35 m=83 \\ & \Rightarrow \quad m=\frac{83}{35} \\ & \end{aligned}$

Length of intercept $=$

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

Given, lines

$\begin{aligned} & y-m_1 x=0 \quad \text{... (i)}\\ & y-m_2 x=0 \quad \text{... (ii)} \end{aligned}$

$\begin{aligned} & \Rightarrow\left(y-m_1 x\right)\left(y-m_2 x\right)=0 \\ & y^2-\left(m_1+m_2\right) x y+m_1 m_2 x^2=0 \\ & \text { Now, } a x^2+2 h x y+b y^2=0 \\ & \left(\frac{a}{b}\right) x^2+\frac{2 h}{b} x y+y^2=0 \Rightarrow m_1 \cdot m_2=\frac{a}{b} \\ & \therefore \quad-2=\frac{a}{b} \Rightarrow a=2 b \\ & \Rightarrow \quad \frac{a}{12}=\frac{b}{6} \\ & \end{aligned}$

$\begin{aligned} y^2-8 x y-9 x^2 & =0 \\ y^2-9 x y+x y-9 x^2 & =0 \\ (y-9 x)(y+x) & =0 \end{aligned}$

The two given lines are $y=9 x$ and $y=-x$

Slope of line $y=9 x$ is 9 and slope of line $y=-x$ is -1.

Let $A B$ and $B C$ be the line perpendicular to $y=9 x$ and $y=-x$ respectively.

Slope of $A B=\frac{-1}{9}$ and slope of line $B C$ is 1.

Equation of $A B$ is $y-\beta=\frac{-1}{9}(x-\alpha)$

$\begin{aligned} 9 y-9 \beta & =-x+\alpha \\ \Rightarrow \quad x+9 y-\alpha-9 \beta & =0 \end{aligned}$

$M$ is Mid-point of the $A B$ and $y=9 x$

So, $x+9(9 x)+\alpha+9 \beta$

$\begin{array}{r} 82 x=\alpha+9 \beta \\ x=\frac{\alpha+9 \beta}{82} \end{array}$

Coordinate of $M$ is $\left(\frac{\alpha+9 \beta}{82}, \frac{9 \alpha+81 \beta}{82}\right)$.

$M$ is Mid-point of $A B$,

Let coordinate of $A$ be $h, k$

$\begin{aligned} & \frac{\alpha+9 \beta}{82}=\frac{\alpha+h}{2} \text { and } \frac{9 \alpha+81 \beta}{82}=\frac{\beta+k}{2} \\ & \Rightarrow 2 \alpha+18 \beta=82 \alpha+82 h \\ & \Rightarrow \quad 82 h=-80 \alpha+18 \beta \\ & \Rightarrow \quad h=\frac{-80 \alpha+18 \beta}{82} \\ & \text { and } 18 \alpha+162 \beta=82 \beta+82 k \\ & \Rightarrow \quad 82 k=18 \alpha-80 \beta \\ & \Rightarrow \quad k=\frac{18 \alpha-80 \beta}{82} \\ \end{aligned}$

Equation of $B C$ is $y-\beta=1(x-\alpha)$

$x-y=\alpha-\beta$

$N$ is intersection point of $B C$ and $y=-x$

$\begin{array}{lr} \therefore & x+x=\alpha-\beta \\ \Rightarrow & 2 x=\alpha-\beta \\ \Rightarrow & x=\frac{\alpha-\beta}{2} \\ \Rightarrow & y=\frac{\beta-\alpha}{2} \end{array}$

Coordinate of $N$ is $\left(\frac{\alpha-\beta}{2}, \frac{\beta-\alpha}{2}\right)$.

$N$ is Mid-point of $B C$.

Let coordinate of $C$ be $(a, b)$

$\frac{\alpha-\beta}{2}=\frac{\alpha+a}{2} \text { and } \frac{\beta-\alpha}{2}=\frac{\beta+b}{2}$

$\Rightarrow a=-\beta$ and $b=-\alpha$

$\therefore$ Coordinate of $C$ is $(-\beta,-\alpha)$.

Centroid of $A B C$ is

$\left(\frac{h+a+\alpha}{3}, \frac{k+b+\beta}{3}\right)$

$=\left(\frac{1}{2}\left[\frac{-80 \alpha+18 \beta}{52}-\beta+\alpha\right], \frac{1}{2}\left[\frac{18 \alpha-80 \beta}{82}-\alpha+\beta\right]\right)$

$\begin{aligned} & =\left(\frac{1}{2}\left[\frac{-80 \alpha+18 \beta-82 \beta+82 \alpha}{82}\right],\right. \\ & \left.\frac{1}{3}\left[\frac{18 \alpha-80 \beta-82 \alpha+82 \beta}{82}\right]\right) \end{aligned}$

$\begin{aligned} & =\left(\frac{1}{3}\left[\frac{2 \alpha-64 \beta}{82}\right], \frac{1}{3}\left[\frac{-64 \alpha+2 \beta}{82}\right]\right) \\ & =\frac{1}{123}(\alpha-32 \beta,-32 \alpha+\beta) \end{aligned}$

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

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