Question

The equation of the line which is parallel to $3\cos \theta +4\sin \theta +\frac{5}{r}=0$, $3\cos \theta +4\sin \theta +\frac{10}{r}=0$ and equidistant from these lines is

• A. $3\cos \theta +4\sin \theta -\frac{5}{r}=0$
• B. $3\cos \theta +4\sin \theta +\frac{15}{r}=0$
• C. $6\cos \theta +8\sin \theta +\frac{15}{r}=0$
• D. $6\cos \theta +8\sin \theta -\frac{15}{r}=0$

Answer 1

Answer: (C)

$6\cos \theta +8\sin \theta +\frac{15}{r}=0$

Given equation of lines
$3\cos \theta +4\sin \theta +\frac{5}{r}=0$
$3\cos \theta +4\sin \theta +\frac{10}{r}=0$
Equation of line parallel to given line is
$3\cos \theta +4\sin \theta +\frac{k}{r}=0$
According to question,
$\frac{|k-5|}{5}=\frac{|k-10|}{5}$
$\Rightarrow k-5=\pm (k-10)$
$\Rightarrow 2k=15$
or $k=\frac{15}{2}$
So, the equation of required line is
$6r\cos \theta +8r\sin \theta +15=0$