Question

Find the equations of lines with slope $3$ and length of perpendicular distance from origin is equal to $\sqrt{6}$ units.

Answer 1

The equation of the family of lines with slope $3$ is $y=3x+c$

or $3x-y+c=0$ .......$\left(1\right)$

For the required members of the family, the perpendicular distance from the origin $(0,0)$ is $\sqrt{6}$ units.

$\Rightarrow \frac{|3\times 0-0+c|}{\sqrt{{\left(3\right)}^2+{(-1)}^2}}=\sqrt{6}$

$\Rightarrow \frac{\left|c\right|}{\sqrt{9+1}}=\sqrt{6}$

$\Rightarrow c=\pm \sqrt{60}=\pm \sqrt{4\times 15}=\pm 2\sqrt{15}$

$\therefore$ the equations of the lines are $y=3x+2\sqrt{15}$ or $y=3x-2\sqrt{15}$