If the two lines x + (a - 1) y = 1 and 2x + a2y = 1(a ∈ R - 0, 1 ) are perpendicular, then the distance of their point of intersection from the origin is :

Question

If the two lines x + (a - 1) y = 1 and 2x + a2y = 1(a ∈ R - 0, 1 ) are perpendicular, then the distance of their point of intersection from the origin is : (A) (2/5) (B) (2/√5) (C) (√2/5) (D) √(2/5)

Tardigrade Admin · Accepted Answer

((-1/a-1))((-2/a2) ) = - 1 2= -(a2)(a-1) a3 -a2 + 2 =0 (a+1)(a2 - 2a +2 ) = 0 ∴ a =- 1 . beginaligned L1: x -2 y + 1 = 0 L2: 2x +y - 1 = 0 endaligned rbrace 0(0,0) P((1/5), (3/5)) OP = √(1/25) + (9/25) = √(10/25) = √(2/5)

Q. 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 :

$\frac{2}{5}$

$\frac{2}{5}$

$\frac{2}{5}$

$\frac{2}{5}$

$(\frac{- 1}{a - 1}) (\frac{- 2}{a ^{2}}) = - 1$
$2 = - (a^{2}) (a - 1)$
$a^{3} - a^{2} + 2 = 0$
$(a + 1) (a^{2} - 2 a + 2) = 0$
$∴ a = - 1$
$L_{1} : x - 2 y + 1 = 0 L_{2} : 2 x + y - 1 = 0}$
$0 (0, 0) P (\frac{1}{5}, \frac{3}{5})$
$OP = \frac{1}{25} + \frac{9}{25} = \frac{10}{25} = \frac{2}{5}$

Q. If the two lines x+(a−1)y=1 and 2x+a2y=1(a∈R−{0,1}) are perpendicular, then the distance of their point of intersection from the origin is :

52​

5​2​

52​​

52​​

(a−1−1​)(a2−2​)=−1 2=−(a2)(a−1) a3−a2+2=0 (a+1)(a2−2a+2)=0 ∴a=−1 L1​:x−2y+1=0L2​:2x+y−1=0​} 0(0,0)P(51​,53​) OP=251​+259​​=2510​​=52​​

Q. 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 :

$\frac{2}{5}$

$\frac{2}{5}$

$\frac{2}{5}$

$\frac{2}{5}$

$(\frac{- 1}{a - 1}) (\frac{- 2}{a ^{2}}) = - 1$
$2 = - (a^{2}) (a - 1)$
$a^{3} - a^{2} + 2 = 0$
$(a + 1) (a^{2} - 2 a + 2) = 0$
$∴ a = - 1$
$L_{1} : x - 2 y + 1 = 0 L_{2} : 2 x + y - 1 = 0}$
$0 (0, 0) P (\frac{1}{5}, \frac{3}{5})$
$OP = \frac{1}{25} + \frac{9}{25} = \frac{10}{25} = \frac{2}{5}$