Question

Find the equation of the straight line which cuts off intercepts on x-axis twice that on y-axis and is at a unit distance from the origin.

Answer 1

Let the intercepts on x-axis and y-axis be 2a and a, respectively.
So, the equation of the line with intercepts 2a on x-axis and a on y-axis be \[\frac{x}{2a} + \frac{y}{a} = 1\]

\[\Rightarrow x + 2y = 2a\]           ... (1)
Let us change equation (1) into normal form.

\[\frac{x}{\sqrt{1 + 2^2}} + \frac{2y}{\sqrt{1 + 2^2}} = \frac{2a}{\sqrt{1 + 2^2}}\]

\[\frac{x}{\sqrt{5}} + \frac{2y}{\sqrt{5}} = \frac{2a}{\sqrt{5}}\]

Thus, the length of the perpendicular from the origin to the line (1) is \[p = \left| \frac{2a}{\sqrt{5}} \right|\] 

Given:
p = 1 

\[\therefore \left| \frac{2a}{\sqrt{5}} \right| = 1\]

\[ \Rightarrow a = \pm \frac{\sqrt{5}}{2}\]

Required equation of the line: \[x + 2y = \pm \frac{2\sqrt{5}}{2}\]

\[ \Rightarrow x + 2y \pm \sqrt{5} = 0\]