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The equation of the line x+y+z-1=0, 4x+y...

The equation of the line `x+y+z-1=0`, `4x+y-2z+2=0` written in the symmetrical form is

A

`(x+1)/(1)=(y-2)/(-2)= (z-0)/(1)`

B

`(x)/(1)= (y)/(-2)= (z-1)/(1)`

C

`(x+1//2)/(1)= (y-1)/(-2)= (z-1//2)/(1)`

D

`(x-1)/(2)= (y+2)/(-1)= (z-2)/(2)`

Text Solution

Verified by Experts

`" "x+y+z-1=0`
`" "4x+y-2z+2=0`
Therefore, direction ratios of the line are `(-3, 6, -3)` i.e., `(1, -2, 1)`
Let `z= k`, then `x=k-1, y=2-2k`.
i.e., `(k-1, 2-2k, k)` is any point on the line
Therefore, `(-1, 2, 0), (0, 0, 1) and (-(1)/(2), 1, (1)/(2))` are points on the lines.
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