Question

A line through $(0,-2)$ and is perpendicular to the line $5x-y+7=0$. Then the equation of the line parallel to it and at a distance of $3$ units from it is/are

• A. $x+5y+10-3\sqrt{26}=0$
• B. $x+5y+10-\sqrt{13}=0$
• C. $x+5y+10+3\sqrt{26}=0$
• D. $x+5y+10+\sqrt{13}=0$

Answer 1

Answer: (A), (C)

$x+5y+10+3\sqrt{26}=0$

The line through $(0,-2)$ and perpendicular to the line $5x-y+7=0$ is

$y+2=-\frac{1}{5}(x-0)$
$\Rightarrow x+5y+10=0$

Now, required line which makes a distance of $3$ units and parallel to it is $x+5y+k=0$

$\therefore \frac{|k-10|}{\sqrt{26}}=3$
$\Rightarrow k=10\pm 3\sqrt{26}$

$\therefore$ The required line is $x+5y+10\pm 3\sqrt{26}=0$