A straight line which makes equal intercepts on positive $X$ and $Y$ axes and which is at a distance 1 unit…
- $3+\sqrt{2}$
- $\sqrt{2}-1$
- 1
- 0
Solution

$\frac{x}{a}+\frac{y}{a}=1$ ie, $\quad x+y=a$ ...(i) Distance of Eq. (i) from origin $=1$ $\left|\frac{0+0-a}{\sqrt{1+1}}\right|=1, \frac{-a}{\sqrt{2}} \mid=1$ $\Rightarrow \quad a=\sqrt{2}$ From Eq. (i), $\quad x+y=\sqrt{2}$ ...(ii) Also given line, $2 x-y=-3-\sqrt{2}$ ...(iii) The intersection point of line (ii) and line (iii) is $\left(x_0, y_0\right)=(-1, \sqrt{2}+1)$ So, $\quad 2 x_0+y_0=2(-1)+\sqrt{2}+1$ $=-2+\sqrt{2}+1=(\sqrt{2}-1)$ Hence, $\quad 2 x_0+y_0=\sqrt{2}-1$
Asked in: AP EAMCET 2010