QJEEM19Ma5b0c: JEE Main 2019 Mathematics - Class 11 JEEM19Ma5b0c Solution: If the two lines x + (a – 1) y = 1 and 2x + a 2 y = 1 (a R –…

Line 1 : x + (a – 1) y = 1

Slope of this line (m₁) = ${{ - 1} \over {a - 1}}$

Line 2 : 2x + a²y = 1

Slope of this line (m₂) = $- {2 \over {{a^2}}}$

Line 1 and Line 2 are perpendicular to each other.

$\therefore$ m₁ m₂ = -1

$\Rightarrow$ $\left( {{{ - 1} \over {a - 1}}} \right)\left( { - {2 \over {{a^2}}}} \right) = - 1$

$\Rightarrow$ ${a^3} - {a^2} + 2 = 0$

$\Rightarrow$ $\left( {a + 1} \right)$ $\left( {{a^2} - 2a + 2} \right)$ = 0

$\therefore$ ${a = - 1}$

So lines are

Line 1 : x - 2y + 1 = 0

Line 2 : 2x + y - 1 = 0

Solving these equation we get point of intersection

P $\left( {{3 \over 5}, - {1 \over 5}} \right)$

Now distance of P from origin

OP = $\sqrt {{{\left( {{3 \over 5}} \right)}^2} + {{\left( { - {1 \over 5}} \right)}^2}}$

= $\sqrt {{{10} \over {25}}}$

= $\sqrt {{2 \over 5}}$