Straight Lines and Pair of Straight Lines

Lines $x+2 a y+a=0$

$ \begin{aligned} & x+3 b y+b=0 \\ & x+4 c y+c=0 \end{aligned} $

These lines are concurrent

$ \begin{aligned} & \left|\begin{array}{lll} 1 & 2 a & a \\ 1 & 3 b & b \\ 1 & 4 c & c \end{array}\right|=0 \\ & \Rightarrow(3 b c-4 b c)-2 a(c-b)+a(4 c-3 b)=0 \\ & \Rightarrow-b c-2 a c+2 a b+4 a c-3 a b=0 \\ & \Rightarrow-b c+2 a c-a b=0 \\ & \Rightarrow 2 a c=a b+b c \end{aligned} $

$a, b, c$ are in HP

$ \begin{aligned} &\text { Foot of perpendicular from origin to }\\ &x-2 y+3=0 \end{aligned} $

$ \begin{gathered} \frac{x-0}{1}=\frac{y-0}{-2}=\frac{-3}{5} \\ x=\frac{-3}{5}, y=\frac{6}{5} \\ A \text { on }=\sqrt{\left(\frac{12}{5}\right)^2+\left(\frac{6}{5}\right)^2}=\frac{6 \sqrt{4+1}}{5}=\frac{6 \sqrt{5}}{5} \end{gathered} $

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

$ 3 y^2-5 x y-2 x^2=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

From Eq. (i), $x^2+2 x y-x y-2 y^2=0$

$ \begin{aligned} & (x-y)(x+2 y)=0 \\ & l_1=x-y=0 \\ & l_2=x+2 y=0 \end{aligned} $

From Eq. (ii), $3 y^2-5 x y-2 x^2=0$

$ \begin{aligned} \Rightarrow \quad & 3 y^2-6 x y+x y-2 x^2=0 \\ \Rightarrow \quad & (3 y+x)(y-2 x)=0 \\ & l_3 \equiv x+3 y=0 \\ & l_4 \equiv 2 x-y=0 \\ & l_2 \perp l_4 \end{aligned} $

$\therefore$ Non perpendicular lines

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

$ \begin{aligned} &\text { Given lines, }\\ &x+2 y+\lambda=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

$ 2 x^2-2 x y+3 y^2+2 x-y-1=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

From Eqs. (i) $\frac{x+2 y}{-\lambda}=1$

Now, homogenizing the Eq. (ii)

$ \begin{aligned} & \Rightarrow \quad 2 x^2-2 x y+3 y^2+(2 x-y)\left(\frac{x+2 y}{-\lambda}\right) \\ & \quad-\left(\frac{x+2 y}{-\lambda}\right)^2=6 \\ & \Rightarrow \quad\left(2 \lambda^2-2 \lambda-1\right) x^2+\left(-2 \lambda^2-3 \lambda-4\right) x y +\left(3 \lambda^2+2 \lambda-4\right) y^2=0 \end{aligned} $

For lines to be $\perp^{\prime} r$

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

Let $P(h, k)$

Given, $\left|\frac{2 h-3 k+1}{\sqrt{4+9}}\right|=2$

$ 2 h-3 k+1=2 \sqrt{13} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

and $\sqrt{(h-5)^2+(k-6)^2}=\sqrt{13}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

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

$ \left.2 h-3 k+1=2 \sqrt{\left[(h-15)^2+(k-6)^2\right.}\right] $

On squaring both sides, we get

$ \begin{aligned} & 4 h^2+9 k^2+1-12 h k-6 k \\ + & 4 h=4\left[h^2+25-10 h+k^2+36-12 k\right] \\ \Rightarrow & 5 k^2-12 h k+44 h+42 k-243=0 \\ \Rightarrow & 12 h k-5 k^2-44 h-42 k+243=0 \end{aligned} $

Taking locus of $P(h, k)$, we get

$ 12 x y-5 y^2-44 x-42 y+243=0 $

We have, $3 x^2+4 y^2-x y+k=0$ is transformed equation after shifting

$ \begin{aligned} &\text { origin to the point }(\alpha, \beta) \text { replacing }\\ &\begin{aligned} & \begin{aligned} & x \rightarrow x-\alpha \text { and } y \rightarrow y-\beta \\ & \therefore 3(x-\alpha)^2+4(y-\beta)^2-(x-\alpha)(y-\beta)+k=0 \\ & \Rightarrow 3\left(x^2+\alpha^2-2 \alpha x\right)+4\left(y^2+\beta^2-2 \beta y\right) \\ & \quad-x y+\alpha y+\beta x-\alpha \beta+k=0 \\ & \Rightarrow 3 x^2+4 y^2-x y+x(-6 \alpha+\beta) +y(-8 \beta+\alpha) \\ & \quad+3 \alpha^2+4 \beta^2-\alpha \beta+k=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} \end{aligned} \end{aligned} $

Comparing Eq. (i) with given equation,

$ \begin{aligned} & -6 \alpha+\beta=-5 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & \alpha-8 \beta=-7 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\\ & 3 \alpha^2+4 \beta^2-\alpha \beta+k=2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\end{aligned} $

On solving Eqs. (ii) and (iii), we get $\alpha=\beta=1$

By Eqs. (ii)

$ \begin{array}{ll} & 3+4-1+k=2 \\ \Rightarrow & k=-4 \\ \therefore & \alpha+\beta-k=1+1+4=6 . \end{array} $

Let the $L$ makes and angle $\theta$ from positive direction of $X$-axis.

$\therefore$ The equation of $L$ is

$ \frac{x-1}{\cos \theta}=\frac{y-5}{\sin \theta}=r \text { and }-r $

Let $A(r \cos \theta+1, r \sin \theta+5)$ and $B(-r \cos \theta+1,-r \sin \theta+5)$

be two points which lies in the given two lines $5 x-y-4=0$ and $3 x+4 y-4=0$

because $L$ bisects these lines.

$ \begin{aligned} & 5(r \cos \theta+1)-r \sin \theta-5-4=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & 3(-r \cos \theta+1)+4(-r \sin \theta+5)-4=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & r(5 \cos \theta-\sin \theta)=4 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\\ & r(-3 \cos \theta-4 \sin \theta)=-19 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\end{aligned} $

From Eqs. (iii) and (iv)

$ \begin{aligned} & -95 \cos \theta+19 \sin \theta=-12 \cos \theta-16 \sin \theta \\ & \Rightarrow \quad 35 \sin \theta=83 \cos \theta \\ & \therefore \quad \tan \theta=\frac{83}{35}=m \end{aligned} $

$\therefore$ Equation of $L$ is:

$ \begin{aligned} & y-5=\frac{83}{35}(x-1) \\ \Rightarrow \quad & 83 x-35 y+92=0 \end{aligned} $

We have, $\theta=60^{\circ}$

Equation of $L$ is

$ \frac{x-1}{\cos 60^{\circ}}=\frac{y-2}{\sin 60^{\circ}}=r $

Given, $A$ and $B$ are two points on $L$ which is at a distances of 4 units from $P$

$\because$ Take values of $r=4$ and -4

$ \begin{aligned} & A=\left(r \cos 60^{\circ}+1, r \sin 60^{\circ}+2\right) \text { and } \\ & B=\left(-r \cos 60^{\circ}+1,-r \sin 60^{\circ}+2\right) \end{aligned} $

$ \begin{aligned} &\begin{aligned} A & =\left(4 \cdot \frac{1}{2}+1,4 \frac{\sqrt{3}}{2}+2\right)=(3,2 \sqrt{3}+2) \\ B & =\left(-4 \cdot \frac{1}{2}+1,-4 \frac{\sqrt{3}}{2}+2\right) \\ & =(-1,-2 \sqrt{3}+2) \end{aligned}\\ &\text { Now area of } \triangle O A B \text { is } 4-2 \sqrt{3} \end{aligned} $

$(2 p-3) x^2+2 p x y-y^2=0$ represent a pair of distinct straight line. If

$ \begin{aligned} & h^2-a b>0 \\ \Rightarrow & p^2+2 p-3>0 \\ \Rightarrow & (p+3)(p-1)>0 \\ \Rightarrow & p<-3 \text { or } p>1 \\ \therefore & p \in R-[-3,1] \end{aligned} $

Let point be $P(h, k)$ and $A(2,-2$ given $P A=2 h$

$ \begin{aligned} & \left(h-2^2+\left(k+2)^2=4 h^2\right.\right. \\ \Rightarrow \quad & h^2+4-4 h+k^2+4+4 k=4 h^2 \\ \Rightarrow \quad & 3 h^2-k^2+4 h-4 k-8=0 \end{aligned} $

Taking locus of $P(h, k)$, we get

$ 3 x^2-y^2+4 x-4 y-8=0 $

Given transformed equation is

$ 2 x^2+3 x y-2 y^2-17 x+6 y+8=0 $

Original equation $x=X-\alpha, y=Y-\beta$

$ \begin{aligned} & 2(X-\alpha)^2+3(X-\alpha)(Y-\beta) \\ & -2(Y-\beta)^2-17(X-\alpha)+6(Y-\beta)+8=0 \\ & \Rightarrow 2\left(X^2+\alpha^2-2 \alpha X\right)+3(X Y-\beta X-\alpha Y+\alpha \beta) \\ & -2\left(Y^2+\beta^2-2 \beta Y\right)-17 X+17 \alpha+6 Y \\ & \quad-6 \beta+8=0 \\ & \Rightarrow 2 X^2+3 X Y-2 Y^2+X(-4 \alpha-3 \beta-17) \\ & \quad+Y(-3 \alpha+4 \beta+6) \\ & \quad+2 \alpha^2+3 \alpha \beta-2 \beta^2+17 \alpha-6 \beta+8=0 \end{aligned} $

Comparing it with

$ a X^2+2 h X Y+b Y^2+C=0 $

We get, $-4 \alpha-3 \beta-17=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

$ -3 \alpha+4 \beta+6=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Solving Eqs. (i) and (ii), we get $\beta=-3$ and $\alpha=-2$

$ \begin{aligned} & C=2 \alpha^2+3 \alpha \beta-2 \beta^2+17 \alpha-6 \beta+8 \\ & C=0 \end{aligned} $

$ \begin{aligned} \because 3 \alpha+C & =3(-2)+0=-6=2(-3) \\ & =2 \beta \end{aligned} $

We have a line $x-y-2=0$

Let $x=t$ is a point on the line $(t, t-2)$ be a parametric point on the line.

We have, $A$ and $B$ are two points on either side of $P$.

$ \because A=(6+a, 4+a) \text { and } B(6-a, 4-a) $

Given, $P A=4$

$ \begin{aligned} &(6+a-6)^2+(4+a-4)^2=16 \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, a^2+a^2=16 \\ &\,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a^2=8 \\ &\,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, a= \pm 2 \sqrt{2} \\ & \because \quad A=(6+2 \sqrt{2}, 4+2 \sqrt{2}) \text { and } \\ & B=(6-2 \sqrt{2}, 4-2 \sqrt{2}) \end{aligned} $

$ \begin{aligned} & A(\alpha, \beta) \text { and } B=(\gamma, \delta) \\ & \because \alpha=6+2 \sqrt{2}, \beta=4+2 \sqrt{2}, \gamma=6-2 \sqrt{2} \\ & \quad \begin{array}{l} \delta=4-2 \sqrt{2} \\ \because \alpha^2+\beta^2+\gamma^2+\delta^2 \\ \quad=(6+2 \sqrt{2})^2+(4+2 \sqrt{2})^2+(6-2 \sqrt{2})^2+(4-2 \sqrt{2})^2 \end{array} \\ & \quad=2(36+8)+2(16+8) \\ & \quad=88+48=136 \end{aligned} $

Let the slope of other line be $m$.

Now, point of intersection of the line $2 x+3 y+1=0$ and $3 x+2 y+4=0$ is $x=-2$ and $y=1$

Now, the $2 x+3 y+1=0$ makes equal angle to the line $3 x+2 y+4=0$ and other line.

$ \begin{aligned} \frac{m+\frac{2}{3}}{1-\frac{2 m}{3}} & =\frac{\frac{3}{2}-\frac{2}{3}}{1+\frac{3}{2} \cdot \frac{2}{3}} \\ \Rightarrow \quad \frac{3 m+2}{3-2 m} & =\frac{5}{12} \end{aligned} $

Solving this, we get $m=-\frac{9}{46}$

$\because$ Equation of other line

$ \begin{aligned} & Y-1=-\frac{9}{46}(x+2 \\ \Rightarrow & 46 y-46=-9 x-18 \\ \Rightarrow & 9 x+46 y-28=0 \end{aligned} $

Given, $x^2+2 h x y+6 y^2=0$

$ \begin{aligned} & \alpha=\tan ^{-1} \frac{2 \sqrt{h^2-a b}}{a+b}=\tan ^{-1}\left(\frac{2 \sqrt{h^2-6}}{7}\right) \\ & =\tan ^{-1} \frac{1}{7} \\ & \Rightarrow \quad 2 \sqrt{h^2-6}=1 \\ & \Rightarrow \quad h^2=6+\frac{1}{4}=\frac{25}{4} \\ & \quad h=\frac{5}{2} \text { and }-\frac{5}{2} \end{aligned} $

Now, the given equation $x^2 \pm 5 x y+6 y^2=0$ given slopes are positive its two equation be

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

Now, product of the perpendicular distances from the point $(1,0)$

$ \frac{1}{\sqrt{5}} \times \frac{1}{\sqrt{10}}=\frac{1}{\sqrt{50}}=\frac{1}{5 \sqrt{2}} $

Clearly, slope of one line be $m_1=1$

Let $m_2=m$

We know that $m_1+m_2=-\frac{2 h}{b}$

$ \begin{aligned} & m_1 m_2=\frac{a}{b} \\ & 1+m=\frac{-2 h}{b} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

$ \begin{aligned} &m=\frac{a}{b}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ &\text { From Eqs. (i) and (ii), we get }\\ &\begin{aligned} a+b & =-2 h \\ \because a+b+2 h & =0 \end{aligned} \end{aligned} $

The $x$-intercept is $-\frac{r}{p}=-\frac{8}{2}=-4$. and the $y$-intercept is $\frac{-r}{q}=\frac{-8}{4}=-2$

$ \begin{aligned} \text { Thus, the area of the triangle } & =\frac{1}{2} \times x_{\text {intercept }} \times y_{\text {intercept }} \\ & =\frac{1}{2} \times 4 \times 2=4 \end{aligned} $

$ \begin{aligned} &\text { The equations of the four lines are }\\ &\begin{aligned} & a x+b y+c=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & a x-b y+c=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & a x+b y-c=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\\ & a x-b y-c=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv) \end{aligned} \end{aligned} $

Clearly, lines (i) and (iii) are parallel. Also, lines (ii) and (iv) are also parallel.

So, the given lines form a parallelogram whose vertices are $A\left(\frac{c}{a}, 0\right), B\left(0, \frac{c}{b}\right), C\left(\frac{-c}{a}, 0\right)$ and $D\left(0, \frac{-c}{b}\right)$.

Clearly, diagonal $A C$ and $B D$ of this 11 gm are at right angle to each other. So, the given lines form a rhombus $A B C D$.

$ \begin{aligned} \therefore \text { Area of rhombus } A B C D & =\frac{1}{2}(A C \times B D) \\ & =\frac{1}{2}\left(\frac{2 c}{a} \times \frac{2 c}{b}\right)=\frac{2 c^2}{a b} \text { sq. units } \end{aligned} $

The equation $2 h x y+2 g x+2 f y+c=0$ will represents a pair of straight lines if

$ \begin{aligned} & 0+2 f g h-0 \times f^2-0 \times g^2-c h^2=0 \\ & \Rightarrow \quad 2 f g=c h \\ & \Rightarrow \quad c=\frac{2 f g}{h} \\ & \quad 2 h x y+2 g x+2 f y+\frac{2 f g}{h}=0 \\ & \Rightarrow h^2 x y+g h x+f h y+f g=0 \\ & \Rightarrow h y(h x+f)+g(h x+f)=0 \end{aligned} $

$ \Rightarrow(h x+f)(h y+g)=0 $

Thus, the equations of lines represented by the given equation are $h x+f=0$ and $h y+g=0$.

These two lines and the coordinate axes form a rectangle $O A B C$.

∴ Area of rectangle $O A B C=O A \times O C$

$ =\left|\frac{-f}{h}\right| \times\left|\frac{-g}{h}\right|=\frac{|f g|}{h^2} $

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

$\begin{aligned} & \text { Area of triangle }=\left|\frac{1}{2} \times a \times b\right| \\ & =\frac{1}{2} \times \frac{3 b}{b-5} \times b \\ & \Rightarrow f(b)=\frac{3 b^2}{2 b-10} \\ & \Rightarrow f^{\prime}(b)=0,(2 b-10) 6 b-2\left(3 b^2\right)=0 \\ & 12 b^2-60 b-6 b^2=0 \\ & 6 b^2-60 b=0 \\ & b^2-10 b=0 \\ & b(b-10)=0 \\ & b=0 \text { or } b=10 \end{aligned}$

So for minimum area, $b=10$

then $\frac{1}{2} \times a \times b=30$

$\begin{aligned} & P^{\prime} R: y+2=\frac{5}{3}(x-1) \\ & \text { For Point } Q \Rightarrow y=0 \\ & \frac{6}{5}=a-1 \Rightarrow a=\frac{11}{5} \end{aligned}$

Now, $P Q R S$ is parallelogram

$\begin{aligned} & \therefore \frac{h+a}{2}=\frac{4+1}{2} \Rightarrow h=5-\frac{11}{5}=\frac{14}{5} \\ & \text { and } \frac{2+3}{2}=\frac{k}{2} \Rightarrow K=5 \end{aligned}$

Now $h k^2=25 \times \frac{14}{5}=14 \times 5=70$

To solve this problem, we will use the concept of the slope and tangent of the angle subtended by the line segments at the origin.

Given points: $(5,2)$ and $(2,a)$

The slope of the line joining the origin and $(5,2)$ is:

The slope of the line joining the origin and $(2,a)$ is:

The line segment joining these points subtends an angle $\theta = \frac{\pi}{4}$ at the origin.

According to the tangent formula for the angle between two lines:

Here, $\theta = \frac{\pi}{4}$ so, $\tan \left( \frac{\pi}{4} \right) = 1$. Therefore,

This simplifies to:

Multiplying both the numerator and denominator by 10 to simplify:

This leads to two equations due to the absolute value:

1. $ 5a - 4 = 10 + 2a $

2. $ 5a - 4 = -(10 + 2a) $

Thus, the possible values of $a$ are $\frac{14}{3}$ and $-\frac{6}{7}$.

The product of these values is:

Therefore, the absolute value of the product of all possible values of $a$ is 4.

Answer: Option A

$\begin{aligned} & 2=\frac{2 x_2+x_1}{3}, \frac{-4}{3}=\frac{2 y_2+y_1}{3} \\ & 2 x_2+x_1=6,2 y_2+y_1=-4 \end{aligned}$

$\begin{aligned} & x_1=6-2 x_2 \quad \text{.... (1)}\\ & y_1=-4-2 y_2 \quad \text{.... (2)}\\ & 4 x_1+y_1=14 \quad \text{.... (3)}\\ & 3 x_2-2 y_2=5 \quad \text{.... (4)} \end{aligned}$

From here, $x_2=1, y_2=-1, x_1=4, y_1=-2$

$\begin{aligned} & B(4,-2) C(1,-1) \\ & y+2=\frac{-1+2}{1-4}(x-4) \\ & -3 y-6=x-4 \\ & x+3 y+2=0 \end{aligned}$

$\begin{aligned} & \therefore \frac{(h-2)^2+(k-1)^2}{(h-1)^2+(k-3)^2}=\frac{25}{16} \\ & \frac{h^2+k^2-4 h-2 k+5}{h^2+k^2-2 h-6 k+10}=\frac{25}{16} \\ \end{aligned}$

Replacing $h \rightarrow x$ and $k \rightarrow y$.

$\begin{aligned} & 16 x^2+16 y^2-64 x-32 y+80 \\ & =25 x^2+25 y^2-50 x-150 y+250 \\ & 9 x^2+9 y^2+14 x-118 y+170=0 \\ & \Rightarrow a=9, b=9, c=0, d=14, e=-118 \\ & a^2+2 b+3 c+4 d+e \\ & 81+18+56-118 \\ & \Rightarrow 37 \end{aligned}$

$\begin{aligned} & L:(y+9)=m(x-4) \\ & A:\left(4+\frac{9}{m}, 0\right) \\ & B:(0,-9-4 m) \\ & O A+O B]_{\min } \\ & \Rightarrow E_{\min }=4+\frac{9}{m}+9+4 m \\ & E=13+\frac{9}{m}+4 m \\ & \frac{d E}{d M}=0 \Rightarrow-\frac{9}{m^2}+4=0 \Rightarrow m= \pm \frac{3}{2} \end{aligned}$

$\begin{aligned} & \frac{d^2 E}{d M^2}=\frac{18}{m^3}>0 \text { for } \quad m=\frac{3}{2} \\ & \therefore E_{\min }=4+6+9+6=25 \end{aligned}$

$O P=O Q$ ($\triangle P Q R$ is isosceles triangle)

Let slope of line $O P \rightarrow m_1$

So, equation $\rightarrow y=m_1 x$

$\begin{aligned} &\begin{aligned} & \tan 45^{\circ}=\left|\frac{m_1-m_2}{m_1 m_2}\right| \\ & 1=\left|\frac{m_1+\frac{3}{4}}{1-\frac{3}{4} m_1}\right| \\ & \Rightarrow 1-\frac{3}{4} m_1=m_1+\frac{3}{4} \\ & \frac{1}{4}=\frac{7}{4} m_1 \Rightarrow m_1=\frac{1}{7} \end{aligned}\\ &\text { Equation } O P \rightarrow y=\frac{1}{7} x \end{aligned}$

Point of intersection of $O P$ & line $3 x+4 y=12$ is $P\left(\frac{84}{25}, \frac{12}{25}\right)$

$\begin{aligned} & \Rightarrow \quad O P^2=a^2=\left(\frac{84}{25}\right)^2+\left(\frac{12}{25}\right)^2=\frac{288}{25} \\ & \therefore \quad I=O P^2+P Q^2+Q O^2 \\ &=a^2+a^2+2 a^2 \\ &=4 a^2 \\ &=4 \times \frac{288}{25} \\ & I=46.08 \\ & {[I] }=46 \end{aligned}$

Equation of $A C: x+y=2$

Equation of $A B: x-y+4=0$

Equation of $B C: 3 x+5 y=4$

The line nearest to origin is parallel to $A C$ and inward. Let its equation is $x+y=C$.

$\therefore\left|\frac{C-2}{\sqrt{2}}\right|=1$

$\therefore \quad C=2-\sqrt{2}$

$\therefore$ required equation line is :

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

$\mathrm{A}(\mathrm{a}, \mathrm{b}), \quad \mathrm{B}(3,4), \quad \mathrm{C}(-6,-8)$

$\Rightarrow \mathrm{a}=0, \mathrm{~b}=0 \quad \Rightarrow \mathrm{P}(3,5)$

Distance from $\mathrm{P}$ measured along $\mathrm{x}-2 \mathrm{y}-1=0$

$\Rightarrow x=3+r \cos \theta, \quad y=5+r \sin \theta$

$\begin{aligned} & \text { Where } \tan \theta=\frac{1}{2} \\ & \mathrm{r}(2 \cos \theta+3 \sin \theta)=-17 \\ & \Rightarrow \mathrm{r}=\left|\frac{-17 \sqrt{5}}{7}\right|=\frac{17 \sqrt{5}}{7} \end{aligned}$

Let E is mid point of diagonals

$\frac{\alpha+\gamma}{2}=\frac{1+1}{2}$

$\alpha+\gamma=2$

& $\frac{\beta+\delta}{2}=\frac{2+0}{2}$

$\beta+\delta=2$

$2(\alpha+\beta+\gamma+\delta)=2(2+2)=8$

Cocus of point $\mathrm{P}(\mathrm{x}, \mathrm{y})$ whose distance from Gives $X+2 y+7=0$ & $2 x-y+8=0$ are equal is $\frac{x+2 y+7}{\sqrt{5}}= \pm \frac{2 x-y+8}{\sqrt{5}}$

$(x+2 y+7)^2-(2 x-y+8)^2=0$

Combined equation of lines

$\begin{aligned} & (x-3 y+1)(3 x+y+15)=0 \\ & 3 x^2-3 y^2-8 x y+18 x-44 y+15=0 \\ & x^2-y^2-\frac{8}{3} x y+6 x-\frac{44}{3} y+5=0 \\ & x^2-y^2+2 h x y+2 g x 2+2 f y+c=0 \\ & h=\frac{4}{3}, g=3, f=-\frac{22}{3}, c=5 \\ & g+c+h-f=3+5-\frac{4}{3}+\frac{22}{3}=8+6=14 \end{aligned}$

$\mathrm{Eq}^{\mathrm{n}}: y-0=\tan 15^{\circ}(x-9) \Rightarrow y=(2-\sqrt{3})(x-9)$

Solving lines $\mathrm{L}_1(3 \mathrm{x}+2 \mathrm{y}=14)$ and $\mathrm{L}_2(5 \mathrm{x}-\mathrm{y}=6)$ to get $\mathrm{A}(2,4)$ and solving lines $\mathrm{L}_3(4 \mathrm{x}+3 \mathrm{y}=8)$ and $\mathrm{L}_4(6 \mathrm{x}+\mathrm{y}=5)$ to get $\mathrm{B}\left(\frac{1}{2}, 2\right)$.

Finding Eqn. of $\mathrm{AB}: 4 \mathrm{x}-3 \mathrm{y}+4=0$

Calculate distance PM

$\Rightarrow\left|\frac{4(5)-3(-2)+4}{5}\right|=6$

Writing $P$ in terms of parametric co-ordinates $2+r$

$\begin{aligned} & \cos \theta, 3+\mathrm{r} \sin \theta \text { as } \tan \theta=\sqrt{3} \\ & \mathrm{P}\left(2+\frac{\mathrm{r}}{2}, 3+\frac{\sqrt{3} \mathrm{r}}{2}\right) \end{aligned}$

$\mathrm{P}$ must satisfy $2 \mathrm{x}-3 \mathrm{y}+28=0$

So, $2\left(2+\frac{r}{2}\right)-3\left(3+\frac{\sqrt{3} \mathrm{r}}{2}\right)+28=0$

We find $r=4+6 \sqrt{3}$

$\begin{aligned} & A D: D C=1: 2 \\ & \frac{4-\alpha}{6-\alpha}=\frac{10}{8} \\ & \alpha=\beta \\ & \alpha=14 \text { and } \beta=14 \end{aligned}$

$\begin{aligned} & \mathrm{P}\left(\mathrm{a}^2, \mathrm{a}+1\right) \\ & \mathrm{L}_1=3 \mathrm{x}-\mathrm{y}+1=0 \end{aligned}$

Origin and $\mathrm{P}$ lies same side w.r.t. $\mathrm{L}_1$

$\begin{aligned} & \Rightarrow \mathrm{L}_1(0) \cdot \mathrm{L}_1(\mathrm{P})>0 \\ & \therefore 3\left(\mathrm{a}^2\right)-(\mathrm{a}+1)+1>0 \end{aligned}$

$\begin{aligned} & \Rightarrow 3 \mathrm{a}^2-\mathrm{a}>0 \\ & \mathrm{a} \in(-\infty, 0) \cup\left(\frac{1}{3}, \infty\right) \quad \text{..... (1)} \end{aligned}$

Let $L_2: x+2 y-5=0$

Origin and $\mathrm{P}$ lies same side w.r.t. $\mathrm{L}_2$

$\begin{aligned} & \Rightarrow \mathrm{L}_2(0) \cdot \mathrm{L}_2(\mathrm{P})>0 \\ & \Rightarrow \mathrm{a}^2+2(\mathrm{a}+1)-5<0 \\ & \Rightarrow \mathrm{a}^2+2 \mathrm{a}-3<0 \\ & \Rightarrow(\mathrm{a}+3)(\mathrm{a}-1)<0 \end{aligned}$

$\therefore \mathrm{a} \in(-3,1)\quad \text{..... (2)}$

Intersection of (1) and (2)

$\mathrm{a} \in(-3,0) \cup\left(\frac{1}{3}, 1\right)$

Co-ordinates of $\mathrm{A}=\left(\frac{5}{3}, \frac{8}{3}\right)$

Co-ordinates of $\mathrm{B}=\left(\frac{10}{3}, \frac{4}{3}\right)$

Slope of $\mathrm{OA}=\mathrm{m}_1=\frac{8}{5}$

Slope of $\mathrm{OB}=\mathrm{m}_2=\frac{2}{5}$

$\begin{aligned} & \tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right| \\ & \tan \theta=\frac{\frac{6}{5}}{1+\frac{16}{25}}=\frac{30}{41} \\ & \tan \theta=\frac{30}{41} \end{aligned}$

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 $