Question

The equation of the straight line in the normal form which is parallel to the lines $x+2 y+3=0$ and $x+2 y+8=0$ 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 1

Answer: (B)

Let the equation of required line is
$x+2 y+c=0 \,.....(i)$
According to the question,
$\frac{2|C-3|}{\sqrt{5}} =\frac{|8-C|}{\sqrt{5}} $
$\Rightarrow 2(C-3)=8-C \Rightarrow 3 C=14 $
$\Rightarrow C=\frac{14}{3} $
So, equation will be
$x+2 y+\frac{14}{3}=0\,....(ii)$
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}} $
$\Rightarrow \alpha=\pi+\tan ^{-1} 2 \text { and } p=\frac{14}{3 \sqrt{5}} \Rightarrow p=\frac{14}{\sqrt{45}}$
So, required equation is
$x \cos \alpha+y \sin \alpha=\frac{14}{\sqrt{45}}, \alpha=\pi+\tan ^{-1} 2$