Question

If the two lines $x+(a-1)y=1$ and $2x+a^{2}y=1$ $(a\in R-\{0,1\left\}\right)$ are perpendicular, then the distance of their point of intersection from the origin is :

• A. $\frac{\sqrt{2}}{5}$
• B. $\frac{2}{\sqrt{5}}$
• C. $\sqrt{\frac{2}{5}}$
• D. $\frac{2}{5}$

Answer 1

The correct option is C $\sqrt{\frac{2}{5}}$

$x+(a-1)y=12x+a^{2}y=1$
The equations can be written as
$y=-\frac{x}{(a-1)}+\frac{1}{(a-1)}y=-\frac{2x}{a^2}+\frac{1}{a^2}$
Since the given lines are perpendicular,
$\therefore \frac{-1}{(a-1)}\times \frac{-2}{a^2}=-1$
$\Rightarrow a^2(a-1)=-2$
$\Rightarrow a^3-a^2+2=0$
$\Rightarrow (a+1)(a^2-2a+2)=0$
or, $a=-1$
Therefore, given equations can be represented as :
$x-2y+1=02x+y-1=0$
Point of intersection is $(\frac{1}{5},\frac{3}{5})$
Distance from origin $=\sqrt{\frac{1}{25}+\frac{9}{25}}=\sqrt{\frac{2}{5}}$