Discussion Forum



1)

The equation of the straight lin ein the normal form which is parallel to the lines x+2y+3=0 and x+2y+8=0 and dividing  the distance between these two lines in the ratio 1:2 internally is 


A) $x\cos \alpha+y\sin \alpha=\frac{10}{\sqrt{45}}, \alpha=\tan^{1}\sqrt{2}$

B) $x\cos \alpha+y\sin \alpha=\frac{14}{\sqrt{45}}, \alpha=\pi+\tan^{-1}{2}$

C) $x\cos \alpha+y\sin \alpha=\frac{14}{\sqrt{45}}, \alpha=\tan^{1}{2}$

D) $x\cos \alpha+y\sin \alpha=\frac{10}{\sqrt{45}}, \alpha=\pi+\tan^{1}{\sqrt{2}}$

Answer:

Option B

Explanation:

 Let  the equation  of required line is

    $x+2y+c=0  $   .................(i)

 According to the question,

    $\frac{2|C-3|}{\sqrt{5}}=\frac{|8-C|}{\sqrt{5}}$

  $\Rightarrow$      $2( C-3)=8-C  \Rightarrow   3C=14$

$\Rightarrow$       $C=\frac{14}{3}$

 So, equation will be

       $x+2y+\frac{14}{3}=0$.............(ii)

 Let the normal form is

   $x \cos \alpha+y \sin \alpha=p$  .....(iii)

 From Eqs.(ii) and (iii) ,

        $\frac{\cos \alpha}{-1}=\frac{\sin \alpha}{-2}=\frac{p}{\frac{14}{3}}$

 $\Rightarrow$    $\alpha=\pi+\tan^{-1}2 $  and   $p=\frac{14}{3\sqrt{5}}\Rightarrow p=\frac{14}{\sqrt{45}}$

 So, required equation is

 $ x \cos \alpha+y \sin \alpha$= $\frac{14}{\sqrt{45}}, \alpha= \pi+ \tan^{-1}2$