Find the equation of the line which is equidistant from the parallel l

1.	Find the equation of the line which is equidistant from the parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 = 0.
Answer» The given parallel lines are 9x + 6y – 7 = 0 …(1) 3x + 2y + 6 = 0. ⇒ 9x + 6y + 18 = 0 …(2) Let the line which is equidistant from (1) and (2) be 9x + 6y + \(\lambda\)= 0 …(3) Distance between (1) and (2) = \(^\frac{\|7 - \lambda\|}{\sqrt{9^2+6^2}}\) = \(^\frac{\|7+ \lambda\|}{\sqrt{117}}\) and distance between (2) and (3) = \(^\frac{\|18 - \lambda\|}{\sqrt{9^2+6^2}}\)=\(^\frac{\|18 - \lambda\|}{\sqrt{117}}\) Given, \(^\frac{\|7+ \lambda\|}{\sqrt{117}}\)=\(^\frac{\|18 - \lambda\|}{\sqrt{117}}\) ⇒ 7 + \(\lambda\) = 18 – \(\lambda\) ⇒ \(\lambda\) = \(\frac{11}{2}\) The equation of the required line is 9x + 6y +\(\frac{11}{2}\) = 0 ⇒ 18x + 12y + 11 = 0

Find the equation of the line which is equidistant from the parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 = 0.

Answer»

The given parallel lines are

9x + 6y – 7 = 0 …(1)

3x + 2y + 6 = 0.

⇒ 9x + 6y + 18 = 0 …(2)

Let the line which is equidistant from (1) and (2) be 9x + 6y + \(\lambda\)= 0 …(3)

Distance between (1) and (2) = \(^\frac{|7 - \lambda|}{\sqrt{9^2+6^2}}\) = \(^\frac{|7+ \lambda|}{\sqrt{117}}\)

and distance between (2) and (3) = \(^\frac{|18 - \lambda|}{\sqrt{9^2+6^2}}\)=\(^\frac{|18 - \lambda|}{\sqrt{117}}\)

Given,

\(^\frac{|7+ \lambda|}{\sqrt{117}}\)=\(^\frac{|18 - \lambda|}{\sqrt{117}}\)

⇒ 7 + \(\lambda\) = 18 – \(\lambda\) ⇒ \(\lambda\) = \(\frac{11}{2}\)

The equation of the required line is

9x + 6y +\(\frac{11}{2}\) = 0

⇒ 18x + 12y + 11 = 0