# Find equation of the line which is equidistant from parallel lines

Question:

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

Solution:

The equations of the given lines are

9x + 6y – 7 = 0 … (1)

3x + 2y + 6 = 0 … (2)

Let P (hk) be the arbitrary point that is equidistant from lines (1) and (2). The perpendicular distance of P (hk) from line (1) is given by

$d_{1}=\frac{|9 h+6 k-7|}{(9)^{2}+(6)^{2}}=\frac{|9 h+6 k-7|}{\sqrt{117}}=\frac{|9 h+6 k-7|}{3 \sqrt{13}}$

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

$d_{2}=\frac{|3 h+2 k+6|}{\sqrt{(3)^{2}+(2)^{2}}}=\frac{|3 h+2 k+6|}{\sqrt{13}}$

Since $P(h, k)$ is equidistant from lines $(1)$ and $(2), d_{1}=d_{2}$

$\therefore \frac{|9 h+6 k-7|}{3 \sqrt{13}}=\frac{|3 h+2 k+6|}{\sqrt{13}}$

$\Rightarrow|9 h+6 k-7|=3|3 h+2 k+6|$

$\Rightarrow|9 h+6 k-7|=\pm 3(3 h+2 k+6)$

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

The case $9 h+6 k-7=3(3 h+2 k+6)$ is not possible as

$9 h+6 k-7=3(3 h+2 k+6) \Rightarrow-7=18$ (which is absurd)

$\therefore 9 h+6 k-7=-3(3 h+2 k+6)$

$9 h+6 k-7=-9 h-6 k-18$

$\Rightarrow 18 h+12 k+11=0$

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