If two lines $x+(a-1) y=1$ and $2 x+\mathrm{a}^2 y=1(\mathrm{a} \in \mathrm{R}-\{0,1\})$ are perpendicular,…

If two lines $x+(a-1) y=1$ and $2 x+\mathrm{a}^2 y=1(\mathrm{a} \in \mathrm{R}-\{0,1\})$ are perpendicular, then the distance of their point of intersection from the origin is
  1. $\frac{2}{5}$
  2. $\frac{\sqrt{2}}{5}$
  3. $\frac{2}{\sqrt{5}}$
  4. $\sqrt{\frac{2}{5}}$

Solution

Given equations of lines $\begin{aligned} & x+(a-1) y=1 \\ & 2 x+a^2 y=1 \end{aligned}$
Slope of $x+(a-1) y=1$ is $\frac{-1}{a-1}$ and slope of $2 x+a^2 y=1$ is $\frac{-2}{a^2}$ Given lines are perpendicular $\begin{aligned} & \therefore \quad \text { Product of slope }=-1 \\ & \therefore \quad \frac{-1}{(a-1)} \times \frac{-2}{a^2}=-1 \\ & \Rightarrow \frac{2}{\mathrm{a}^2(\mathrm{a}-1)}=-1 \\ & \Rightarrow \mathrm{a}^3-\mathrm{a}^2=-2 \\ & \Rightarrow \mathrm{a}^3-\mathrm{a}^2+2=0 \\ & \Rightarrow(\mathrm{a}+1)\left(\mathrm{a}^2-2 \mathrm{a}+2\right)=0 \\ & \Rightarrow \mathrm{a}=-1, \mathrm{a}^2-2 \mathrm{a}+2=0 \\ & \Rightarrow a^2-2 a+2 \neq 0 \ldots[\because a \in \dot{R}-\{0,1\}] \\ & \therefore \quad a=-1 \end{aligned}$ $\therefore \quad$ Equations will be $\begin{aligned} & x-2 y=1 \\ & 2 x+y=1 \end{aligned}$ $\therefore \quad$ Point of intersection is $\left(\frac{3}{5}, \frac{-1}{5}\right)$ Distance of point of intersection from origin $\begin{aligned} & =\sqrt{\left(0-\frac{3}{5}\right)^2+\left(0-\left(\frac{-1}{5}\right)\right)^2} \\ & =\sqrt{\frac{9}{25}+\frac{1}{25}} \\ & =\sqrt{\frac{10}{25}}=\sqrt{\frac{2}{5}} \end{aligned}$

Asked in: MHT CET 2024 (02 May Shift 1)

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