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

Let c be the intercept on the y-axis.
Then, the equation of the line is

$$\begin{gathered} y=-x+c\left[\because m=-1\right] \\ \Rightarrow x+y=c \\ \Rightarrow \frac{x}{\sqrt{1^2+1^2}}+\frac{y}{\sqrt{1^2+1^2}}=\frac{c}{\sqrt{1^2+1^2}}\left[\mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}\sqrt{{\left(\mathrm{coefficient}\mathrm{of}x\right)}^2+{\left(\mathrm{coefficient}\mathrm{of}y\right)}^2}\right] \\ \Rightarrow \frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}=\frac{c}{\sqrt{2}} \end{gathered}$$

This is the normal form of the given line.
Therefore, $\frac{c}{\sqrt{2}}$ denotes the length of the perpendicular from the origin.
But, the length of the perpendicular is 5 units.

$$\begin{gathered} \therefore \left|\frac{c}{\sqrt{2}}\right|=5 \\ \Rightarrow c=\pm 5\sqrt{2} \end{gathered}$$

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