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प्रश्न
The perpendicular distance of a line from the origin is 5 units and its slope is − 1. Find the equation of the line.
उत्तर
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, \[x + y - 5\sqrt{2} = 0\]
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