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

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Straight Lines

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:

Answer: (D)

The two lines are perpendicular when the product of their slopes is

- 1

(- \frac{1}{a - 1}) (- \frac{2}{a ^{2}}) = - 1

, that is

\frac{2}{a ^{2} ( a - 1 )} = - 1

, so

a^{3} - a^{2} + 2 = 0

. This factors as

(a + 1) (a^{2} - 2 a + 2) = 0

, and since the quadratic has no real root,

a = - 1

. The lines then become

x - 2 y = 1

and

2 x + y = 1

; solving them,

x = \frac{3}{5}

and

y = - \frac{1}{5}

, so they meet at

(\frac{3}{5}, - \frac{1}{5})

. Its distance from the origin is

\frac{9}{25} + \frac{1}{25} = \frac{10}{25} = \frac{2}{5}

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Single correct option