Straight Lines

Eqⁿ : y – 0 = tan45° (x – 9) Þ y = (x – 9)

Option (B) is correct

|r₁ – r₂| < c₁c₂ < r₁ + r₂

-> $\left|2-\sqrt[]{4{\lambda }^2-9}\right|<\left|2\lambda \right|<2+\sqrt[]{4{\lambda }^2-9}$

$\left|2\lambda \right|-2<\sqrt[]{4{\lambda }^2-9}$    

$4{\lambda }^2+4-8\left|\lambda \right|<4{\lambda }^2-9$

  $\lambda >\frac{13}{8},\lambda <-\frac{13}{8}$           

$\sqrt[]{4{\lambda }^2-9}>0$

$\lambda >\frac{3}{2},\lambda <-\frac{3}{2}$

$\lambda \in \left(-\infty ,-\frac{13}{8}\right)\cup \left(\frac{13}{8},\infty \right)$           

Now,

$\left|2-\sqrt[]{4{\lambda }^2-9}\right|<\left|2\lambda \right|$            

$4+4{\lambda }^2-9-4\sqrt[]{4{\lambda }^2-9}<4{\lambda }^2$  

$4\sqrt[]{4{\lambda }^2-9}>-5\Rightarrow \lambda \in R$  

$\lambda \in \left(-\infty ,-\frac{13}{8}\right)\cup \left(\frac{13}{8},\infty \right)$  

(y – 2) = m (x – 8)

⇒   x-intercept

⇒     $\left(\frac{-2}{m}+8\right)$

⇒   y-intercept

⇒   (–8m + 2)

⇒   OA + OB = $\frac{-2}{m^2}$  + 8 – 8m + 2

$f\text{'}\left(m\right)=\frac{2}{m^2}-8=0$  

-> $m^2=\frac{1}{4}$

-> $m=\frac{-1}{2}$

-> $f\left(\frac{-1}{2}\right)=18$

->Minimum = 18

Kindly consider the following figure

According to question,

$\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{1}{4}$

$\frac{1}{a}+\frac{1}{b}=\frac{1}{2}........(i)$

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

Obviously B (2, 2) satisfying condition (i)

  $h=\frac{cos\theta +3}{2}$          

$k=\frac{sin\theta +2}{2}$            

-> $cos\theta =2h-3\&sin\theta =2k-2$

-> ${\left(h-3/2\right)}^2+{\left(k-1\right)}^2=\frac{1}{4}$

$\therefore$ circle of radius r = $\frac{1}{2}$  

Let y = mx + c is the common tangent

              $soc=\frac{1}{m}=\pm \frac{3}{2}\sqrt[]{1+m^2}\Rightarrow m^2=\frac{1}{3}$  

              so equation of common tangents will be

              $y=\pm \frac{1}{\sqrt[]{3}}x\pm \sqrt[]{3}$  

     &nb

$?{\mathcal{l}}_{1}and{\mathcal{l}}_2$ are perpendicular, so

$3\times 1+\left(-2\right)\left(\frac{\alpha }{2}\right)+0\times 2=0$

⇒a = 3

Now angle between  ${\mathcal{l}}_{2}and{\mathcal{l}}_3$ ,

$cos\theta =\frac{1\left(-3\right)+\frac{\alpha }{2}\left(-2\right)+2\left(4\right)}{\sqrt[]{1+\frac{{\alpha }^2}{4}+4}.\sqrt[]{9+4+16}}$

$\Rightarrow cos\theta =\frac{\frac{2}{29}}{2}\Rightarrow \theta =co{s}^{-1}\left(\frac{4}{29}\right)=se{c}^{-1}\left(\frac{29}{4}\right)$

RM = $\left|\frac{3+7-5}{\sqrt[]{2}}\right|=\frac{5}{\sqrt[]{2}}$

$\mathcal{l}sin60°=\frac{5}{\sqrt[]{2}}\Rightarrow \mathcal{l}=5\sqrt[]{\frac{2}{3}}$

$\therefore Areaof\Delta PQR=\frac{\sqrt[]{3}}{4}{\mathcal{l}}^2=\frac{25}{2\sqrt[]{3}}$

$y+2x=\sqrt[]{11}+7\sqrt[]{7}$  ………(i)

              $2y+x=2\sqrt[]{11}+6\sqrt[]{7}$        ………(ii)

              $\Rightarrow x+y=\sqrt[]{11}+\frac{13}{3}\sqrt[]{7}$    ………(iii)

              Centre of the circle given by solving (i) & (ii)

&n