Find the equation of two straight lines which are parallel to x + 7y +

1.	Find the equation of two straight lines which are parallel to x + 7y + 2 = 0 and at unit distance from the point (1, -1).
Answer» Given: The equation is parallel to x + 7y + 2 = 0 and at unit distance from the point (1, -1) The equation of given line is x + 7y + 2 = 0 … (1) The equation of a line parallel 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 = 1 – 7 + λ1 + 49 λ – 6 = ± 5√2 λ = 6 + 5√2, 6 – 5√2 Now, substitute the value 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

Find the equation of two straight lines which are parallel to x + 7y + 2 = 0 and at unit distance from the point (1, -1).

Answer»

Given:

The equation is parallel to x + 7y + 2 = 0 and at unit distance from the point (1, -1)

The equation of given line is

x + 7y + 2 = 0 … (1)

The equation of a line parallel 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 = 1 – 7 + λ1 + 49

λ – 6 = ± 5√2

λ = 6 + 5√2, 6 – 5√2
Now, substitute the value 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