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 $18x + 12y + 11 = 0.$