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

The equation of line in intercept form is

$\frac{x}{a}+\frac{y}{b}=1$

Intercept on y-axis = 2a (given)

$\therefore$ equation is $\frac{x}{2a}+\frac{y}{a}=1$

$ax+2ay=2a^2...\left(1\right)$

Now, perpendicular distance of (1) from origin is given unity

$\Rightarrow \frac{|a{x}_1+b{y}_1+c|}{\sqrt{a^2+b^2}}=1$

$a=a,b=2a,c=-2a^2,x_1=0,y_1=0$

$=\frac{\left|a\right(0)+2a(0)-2a^2|}{\sqrt{(2a{)}^2+(a{)}^2}}=1$

$\Rightarrow -2a^2=\sqrt{5}a$

$\Rightarrow 4a^4=a^{2}5$

$a^2=\frac{5}{4}\Rightarrow a=\pm \frac{\sqrt{5}}{4}$

$\therefore$ The intercept form of straight line is

$\frac{x}{2a}+\frac{y}{a}=1$

$\frac{x}{\pm \frac{2\sqrt{5}}{4}}+\frac{y}{\pm \frac{\sqrt{5}}{4}}=1$

$x+2y=\pm \sqrt{5}$

$x+2y\pm \sqrt{5}=0$