Straight Line and Pair of Straight Lines 1 Question 5
5. If the two lines $x+(a-1) y=1$ and $2 x+a^{2} y=1,(a \in R-{0,1})$ are perpendicular, then the distance of their point of intersection from the origin is
(a) $\frac{2}{5}$
(b) $\frac{\sqrt{2}}{5}$
(c) $\frac{2}{\sqrt{5}}$
(d) $\sqrt{\frac{2}{5}}$
(2019 Main, 9 April II)
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Solution:
Key Idea
(i) If lines are perpendicular to each other, then product of their slopes is -1 , i.e. $m _1 m _2=-1$
(ii) Distance between two points $\left(x _1, y _1\right)$ and $\left(x _2, y _2\right)$
$$ =\sqrt{\left(x _2-x _1\right)^{2}+\left(y _2-y _1\right)^{2}} $$
Given, lines $x+(a-1) y=1$
and $2 x+a^{2} y=1$, where $a \in R-{0,1}$
are perpendicular to each other
$\therefore \quad-\frac{1}{a-1} \times-\frac{2}{a^{2}}=-1$
$[\because$ If lines are perpendicular, then product of their slopes is -1 ]
$\Rightarrow \quad a^{2}(a-1)=-2 \Rightarrow a^{3}-a^{2}+2=0$
$\Rightarrow \quad(a+1)\left(a^{2}-2 a+2\right)=0 \Rightarrow a=-1$
$\therefore$ Equation of lines are
$$ x-2 y=1 $$
and $\quad 2 x+y=1$
On solving Eq. (i) and Eq. (ii), we get
$$ x=\frac{3}{5} \text { and } y=-\frac{1}{5} $$
$\therefore$ Point of intersection of the lines (i) and (ii) is $P \frac{3}{5},-\frac{1}{5}$.
Now, required distance of the point $P \frac{3}{5},-\frac{1}{5}$ from origin $=\sqrt{\frac{9}{25}+\frac{1}{25}}=\sqrt{\frac{10}{25}}=\sqrt{\frac{2}{5}}$
