Question

Determine the distance between the following pair of parallel lines:

(i) $4x-3y-9=0$ and $4x-3y-24=0$

(ii) $8x+15y-34=0$ and $8x+15y+31=0$

(iii) $y=mx+c$ and $y=mx+d$

(iv) $4x+3y-11=0$ and $8x+6y=15$

Answer 1

Determine between parallel lines

$ax+by+c_1=0$ and $ax+by+c_2=0$ is

$\left|\frac{c_2-c_1}{\sqrt{a^2+b^2}}\right|$

(i) $4x-3y-9=0$ and $4x-3y-24=0$

Distance between the two parallel lines is:

$\left|\frac{-24-(-9)}{\sqrt{4^2+3^2}}\right|=\left|\frac{-24+9}{5}\right|$

$=3$ units

(ii) Distance between $8x+15y-34=0$ and $8x+15y+31=0$ is $\left|\frac{-34-31}{\sqrt{8^2+{15}^2}}\right|=\frac{65}{17}$ units

(iii) Distance between $y=mx+c$ and $y=mx+d$ is $\left|\frac{c-d}{\sqrt{m^2+1}}\right|$

(iv) Distance between $4x+3y-11=0$ and $8x+6y=15$ is $\left|\frac{-11-15}{\sqrt{4^2+3^2}}\right|=\frac{7}{10}$ units.