Question

Prove that the lines $2x+3y=19$ and $2x+3y+7=0$ are equidistant from the line $2x+3y=6$

Answer 1

Since the coefficient of x and y in the equations $2x+3y-19=0$

$2x+3y-6=0$ and $2x+3y+7=0$ are same, therefore all the lines are parallel.

Distance between parallel lines is $d=\left|\frac{c_2-c_1}{\sqrt{a^2+b^2}}\right|$, where $ax+by+c_2=0$ are the lines parallel to each other.

Distance between the lines $2x+3y-19=0$ and $2x+3y-6=0$ is

$d_1=\left|\frac{-19+6}{\sqrt{2^2+3^2}}\right|=\left|\frac{13}{\sqrt{13}}\right|=\sqrt{13}$

Distance between the lines $2x+3y+7=0$ and $2x+3y-6=0$

$d_2=\left|\frac{7+6}{\sqrt{2^2+3^2}}\right|=\left|\frac{13}{\sqrt{13}}\right|=\sqrt{13}$

Since the distance of both the lines $2x+3y+7=0$ and $2x+3y-19=0$ from the line $2x+3y-6=0$ are equal, therefore the lines equidistant.