Question

The perpendicular distance of a line from the origin is 5 units and its slope is - 1. Find the equation of the line.

Answer 1

$\text{Let c be the intercept on the y-axis.}$

$\text{Then, the equation of the line is}$

$y=-x+c[\because m=-1]$

$\Rightarrow x+y=c$

$\Rightarrow \frac{x}{\sqrt{1^2+1^2}}+\frac{y}{\sqrt{1^2+1^2}}=\frac{c}{\sqrt{1^2+1^2}}$

$\left[\text{Dividing both sides by}\sqrt{(\text{coefficient of x}{)}^2+(\text{coefficient of y}{)}^2}\right]$

$\Rightarrow \frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}=\frac{c}{\sqrt{2}}$

$\text{This is the normal form of the given line.}$

$\text{Therefore,}\frac{c}{\sqrt{2}}\text{denotes the length of the perpendicular from the origin.}$

$\text{But, the length of the perpendicular is 5 units.}$

$\therefore \left|\frac{c}{\sqrt{2}}\right|=5$

$\Rightarrow c=\pm 5\sqrt{2}$

$\text{Thus, substituting}c=\pm 5\sqrt{2}iny=-x+$

$c,\text{we get the equation of line to be}y=-x+5\sqrt{2}or,x+y-5\sqrt{2}=0$