Question

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

Answer 1

The equations of the given lines are 
\(9x + 6y - 7 = 0 … (1) \)
\(3x + 2y + 6 = 0 … (2) \)
Let P (h, k) be the arbitrary point that is equidistant from lines (1) and (2).
The perpendicular distance of P (h, k) from line (1) is given by

\(d_1=\frac{\left|9h+6k-7\right|}{\sqrt{(9)^2+(6)^2}}\)

\(=\frac{\left|9h+6k-7\right|}{\sqrt{117}}\)

\(=\frac{\left|9h+6k-7\right|}{3\sqrt{13}}\)

The perpendicular distance of P (h, k) from line (2) is given by 

\(d_2=\frac{\left|3h+2k+6\right|}{\sqrt{(3)^2+(2)^2}}\)

\(=\frac{\left|3h+2k+6\right|}{\sqrt{13}}\)

Since P (h, k) is equidistant from lines (1) and (2),  \(d_1=d_2\)

\(∴\) \(\frac{\left|9h+6k-7\right|}{3\sqrt{13}}\)\(=\frac{\left|3h+2k+6\right|}{\sqrt{13}}\)

\(⇒ |9h+6k-7|=3|3h+2k+6|\)

\(⇒ |9h+6k-7|=±3(3h+2k+6)\)

\(⇒ 9h+6k-7=3(3h+2k+6)\) or \(9h+6k-7=-3(3h+2k+6) \)

The case \(9h+6k-7=3(3h+2k+6)\) is not possible.
\(9h+6k-7=3(3h+2k+6) ⇒ -7=18\) (which is absurd)

\(∴ 9h+6k-7=-3(3h+2k+6) \)
\(9h + 6k - 7 = - 9h - 6k - 18 \)
\(⇒ 18h + 12k + 11 = 0\)

Thus, the required equation of the line is \(18x + 12y + 11 = 0.\)