Question

The perpendicular distance from $(5,{\tan }^{-1}\frac{4}{3})$ to the line $3\cos \theta +4\sin \theta +\frac{5}{r}=0$ is

• A. $1$
• B. $3$
• C. $5$
• D. $6$

Answer 1

The correct option is D $6$

Given point $(5,{\tan }^{-1}\frac{4}{3})$
$r=5,\theta ={\tan }^{-1}\frac{4}{3}$
$\Rightarrow \sin \theta =\frac{4}{5},\cos \theta =\frac{3}{5}$
Given equation is
$3\cos \theta +4\sin \theta +\frac{5}{r}=0$
$\Rightarrow 3r\cos \theta +4r\sin \theta +5=0$
So, $p=\frac{|9+16+5|}{5}$
$\Rightarrow p=6$