Question

A straight line which makes equal intercepts on positive $X$ and $Y$ axes and which is at a distance 1 unit from the origin intersects the straight line $y=2 x+3+\sqrt{2}$ at $\left(x_{0}, y_{0}\right) .$ Then $2 x_{0}+y_{0}$ is equal to

• A. $3+\sqrt{2}$
• B. $\sqrt{2}-1$
• C. 1
• D. 0

Answer 1

Answer: (B)

The equation of line $A B$ which makes an equal intercepts on positive $x$ and $y$ axes are

$\frac{x}{a}+\frac{y}{a}=1$
ie, $x +y=a$ ...(i)
Distance of Eq. (i) from origin $=1$
$\left|\frac{0+0-a}{\sqrt{1+1}}\right|=1,\left|\frac{-a}{\sqrt{2}}\right|=1$
$\Rightarrow a=\sqrt{2}$
From Eq. (i), $x +y=\sqrt{2}$ ...(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,$2 x_{0}+y_{0} =2(-1)+\sqrt{2}+1$
$=-2+\sqrt{2}+1=(\sqrt{2}-1)$
Hence, $2 x_{0}+ y_{0}=\sqrt{2}-1$