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

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

\[ \Rightarrow x + y = c\]

\[ \Rightarrow \frac{x}{\sqrt{1^2 + 1^2}} + \frac{y}{\sqrt{1^2 + 1^2}} = \frac{c}{\sqrt{1^2 + 1^2}} \left[ \text { Dividing both sides by } \sqrt{\left( \text { coefficient of x } \right)^2 + \left( \text { coefficient of y }\right)^2} \right]\]

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

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.

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

\[ \Rightarrow c = \pm 5\sqrt{2}\]

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, \[x + y - 5\sqrt{2} = 0\]