Question

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

Answer 1

The equation of the given lines are
$9x+6y7=$ 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
$d_1=\frac{|9h+6k-7|}{\sqrt{(9{)}^6+(12{)}^2}}=\frac{|9h+6k-7|}{\sqrt{117}}=\frac{|9h+6k-7|}{3\sqrt{13}}$
The perpendicular distance of $p(h,k)$ from line (2) is given by
$d_2=\frac{|3h+2k+6|}{\sqrt{(3{)}^2+(12{)}^2}}=\frac{|3h+2k+6|}{\sqrt{13}}$
Since $p(h,k)$ is equidistant from lines (1) and (2), $d_1=d_2$
$\therefore \frac{|9h+6k-7|}{3\sqrt{13}}=\frac{|3h+2k+6|}{\sqrt{13}}$
$\Rightarrow |9h+6k-7|=\pm 3|3h+2k+6|$
$\Rightarrow 9h+6k-7=3(3h+2k+6)or9h+6k-7=-3(3h+2k+6)$
The case $\Rightarrow 9h+6k-7=3(3h+2k+6)$ is not possible as
$9h+6k-7=3(3h+2k+6)\Rightarrow -7=18$ (which is absurd)
Thus $9h+6k-7=-3(3h+2k+6)$
$\Rightarrow 9h+6k-7=-9h-6k-18$
$\Rightarrow 18h+12k+11=0$
Thus, the required equation of the line is $18x+12y+11=0$