Question

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

Answer 1

$\begin{array}{c}Let\\ Y-interceptbe\left(0,n\right)andX-interceptbe\left(2n,0\right)\\ \therefore slope=\frac{0-n}{2n-0}=\frac{-1}{2}\\ Equationofline=y-y_1=m\left(x-x_1\right)\\ y-0=\frac{-1}{2}\left(x-2n\right)\\ 2y=-x+2n\\ x+2y-2n=0\end{array}$

Distance is $1$ away away from the origin

Distance from a line of equation

$=\frac{\left|A{x}_1+B{y}_{1}C\right|}{\sqrt{A^2+B^2}}=\frac{\left|-2n\right|}{\sqrt{5}}$

$Butdistance=1\frac{\left|-2n\right|}{\sqrt{3}}=n=\pm \frac{\sqrt{3}}{2}$

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