Find the equation of me line which cuts of intersection on $X$ axis twice as that of $Y$ axis and is at a unit distance from origin

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Solution

$\begin{matrix} L e t Y - i n t e r c e p t b e (0, n) a n d X - i n t e r c e p t b e (2 n, 0) ∴ s l o p e = \frac{0 - n}{2 n - 0} = \frac{- 1}{2} E q u a t i o n o f l i n e = y - y_{1} = m (x - x_{1}) y - 0 = \frac{- 1}{2} (x - 2 n) 2 y = - x + 2 n x + 2 y - 2 n = 0 \end{matrix}$

Distance is $1$ away away from the origin
Distance from a line of equation
$= \frac{| A x_{1} + B y_{1} C |}{\sqrt{A^{2} + B^{2}}} = \frac{| - 2 n |}{\sqrt{5}}$

$B u t d i s t a n c e = 1 \frac{| - 2 n |}{\sqrt{3}} = n = \pm \frac{\sqrt{3}}{2}$

Thus, equation become $x + 2 y \pm \sqrt{5} = 0$

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