Question -
Answer -
Given:
The equation is parallelto x + 7y + 2 = 0 and at unit distance from the point (1, -1)
The equation of givenline is
x + 7y + 2 = 0 … (1)
The equation of a lineparallel to line x + 7y + 2 = 0 is given below:
x + 7y + λ =0 … (2)
The line x + 7y+ λ = 0 is at a unit distance from the point (1, − 1).
So,
1 =
λ – 6 = ± 5√2
λ = 6 + 5√2, 6 – 5√2
now, substitute thevalue of λ back in equation x + 7y + λ = 0, we get
x + 7y + 6 + 5√2 = 0 and x + 7y + 6 – 5√2
∴ The required lines:
x + 7y + 6 + 5√2 = 0 and x + 7y + 6 – 5√2