(b) 3x – y = 0 Given lines are 3x – y – 3 = 0 and 3x – y + 5 = 0. Line parallel to the given lines can be written as 3x – y + c = 0 ...(i) Let us taken a point, say, (0, c) on (i) (Putting x = 0 in (i), we get y = c)∴ \(rac{ ext{Distance of (0,c) from}\,3x-y-3}{ ext{Distance of (0,c) from}\,3x-y+5}\) = \(rac{3}{5}\)⇒ \(rac{rac{|3 imes0-c-3|}{\sqrt{3^2+1^2}}}{rac{|3 imes0-c+5|}{\sqrt{3^2+1^2}}}\) = \(rac{3}{5}\)⇒ \(rac{c+3}{-c+5}\) = \(rac{3}{5}\)⇒ 5c + 15 = – 3c + 15 ⇒ 8c = 0 ⇒ c = 0. Substituting c = 0 in (i), the required equation is 3x – y = 0.