Question

The equation of the line passing through $(5,{\tan }^{-1}\frac{3}{4})$and perpendicular to $\cos \theta +2\sin \theta =\frac{4}{r}$ is

• A. $2\cos \theta -\sin \theta =\frac{5}{r}$
• B. $2\cos \theta -\sin \theta =\frac{1}{r}$
• C. $2\cos \theta -\sin \theta =\frac{2}{r}$
• D. $2\cos \theta -\sin \theta =\frac{3}{r}$

Answer 1

The correct option is A $2\cos \theta -\sin \theta =\frac{5}{r}$

Given point is $(5,{\tan }^{-1}(\frac{3}{4}\left)\right)$
Let $\theta ={\tan }^{-1}\left(\frac{3}{4}\right)$
$\Rightarrow \tan \theta =\frac{3}{4}$
$\sin \theta =\frac{3}{5},\cos \theta =\frac{4}{5}$

Given equation is
$cos\theta +2sin\theta =\frac{4}{r}$
Equation of required line perpendicular to given line is
$cos(\frac{\pi }{2}+\theta )+2sin(\frac{\pi }{2}+\theta )=\frac{k}{r}$
$\Rightarrow -\sin \theta +2\cos \theta =\frac{k}{r}$
Since, it passes through $(5,{\tan }^{-1}\frac{3}{4})$
$\Rightarrow k=5$
So, the required equation is
$\Rightarrow 2\cos \theta -\sin \theta =\frac{5}{r}$
$\Rightarrow 2r\cos \theta -r\sin \theta =5$
Alternative Method:
Given point is $(5,{\tan }^{-1}(\frac{3}{4}\left)\right)$
Let $\theta ={\tan }^{-1}\left(\frac{3}{4}\right)$
$\Rightarrow \tan \theta =\frac{3}{4}$
$\sin \theta =\frac{3}{5},\cos \theta =\frac{4}{5}$
So, the point is $(4,3)$
Given equation is
$cos\theta +2sin\theta =\frac{4}{r}$
$\Rightarrow x+2y=4$
Slope of this line is $-\frac{1}{2}$
Slope of required line (perpendicular to given line) is $2.$
Equation of line passing through $(4,3)$ and slope $2$ is
$y-3=2(x-4)$
$\Rightarrow 2x-y=5$
$\Rightarrow 2rcos\theta -r\sin \theta =5$
$\Rightarrow 2cos\theta -\sin \theta =\frac{5}{r}$