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 parallel lines $9x+6y-7=0$ and $3x+2y+6=0$

Let $A(x_1,y_1)$ be any point which is equidistnt from the parallel lines.

$\therefore \left|\frac{9x_1+6y_1-7}{\sqrt{(9{)}^2+(6{)}^2}}\right|=\left|\frac{3x_1+2y_1+6}{\sqrt{(3{)}^2+(2{)}^2}}\right|$

$\Rightarrow \frac{9x_1+6y_1-7}{3\sqrt{13}}=\pm \frac{3x_1+2y_1+6}{3\sqrt{13}}$

Taking $\frac{9x_1+6y_1-7}{3\sqrt{13}}=-\frac{3x_1+2y_1+6}{\sqrt{13}}$

$\Rightarrow 9x_1+6y_1-7=-9x_1-6y_1-18$

$\Rightarrow 18x_1+12y+11=0$

Thus equation of required line is

$18x+12y+11=0$