Question

The equation of the straight line in 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 1

Answer: (B)

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),
$\frac{\cos \alpha }{-1}=\frac{\sin \alpha }{-2}=\frac{p}{\frac{14}{3}}$
$\Rightarrow \alpha =\pi +{\tan }^{-1}2\text{ and }p=\frac{14}{3\sqrt{5}}\Rightarrow p=\frac{14}{\sqrt{45}}$
So, required equation is
$x\cos \alpha +y\sin \alpha =\frac{14}{\sqrt{45}},\alpha =\pi +{\tan }^{-1}2$