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If the two lines x+(a-1)y=1 and 2x+a^(2)...

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

A

`(2)/(sqrt(5))`

B

`(sqrt(2))/(5)`

C

`sqrt((2)/(5))`

D

`(2)/(5)`

Text Solution

Verified by Experts

The correct Answer is:
C
x + (a - 1) y = 1
`2x + a^(2) y = 1` `a in R - {0, 1}`, `m_(1) m_(2) = - 1`
`- (1)/(a - 1). (-2)/(a^(2)) = - 1` `-2 = a^(3) - a^(2)` , `a^(3) - a^(2) + 2 = 0 implies` a = - 1
`:.` lines are x - 2y = 1 and 2x + y = 1 `:.` point of intersection of lines is `((3)/(5), - (1)/(5))`
Distance from origin `= sqrt((9)/(25) + (1)/(25)) = sqrt((10)/(25)) = sqrt((2)/(5))`
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