Question

Find the distance between the parallel lines $15x+8y-34=0$ and $15x+8y+31=0$

Answer 1

Converting each of the given equations to the form y = mx + C, we get

$15x+8y-34=0\Rightarrow y=\frac{-15}{8}x+\frac{17}{4}$

$15x+8y+31=0\Rightarrow y=\frac{-15}{8}x-\frac{31}{8}$

Clearly, the slopes of the given lines are equal and so they are parallel.

The given lines are of the form $y=mx+C_1$ and $y=mx+C_2$

where$m=\frac{-15}{8},C_1=\frac{17}{4}$ and $C_2=\frac{-31}{8}$

$\therefore$distance between the given lines

d =$\frac{|C_2-C_1|}{\sqrt{1+m^2}}$, where $m=\frac{-15}{8},C_1=\frac{17}{4}$ and $C_2=\frac{-31}{8}$

=$\frac{\left|\begin{array}{c}\frac{-31}{8}-\frac{17}{4}\end{array}\right|}{\sqrt{1+{\left(\frac{-15}{8}\right)}^2}}=\frac{|-\frac{65}{8}|}{\sqrt{1+\frac{225}{64}}}=\frac{\left(\frac{65}{8}\right)}{\sqrt{\frac{289}{64}}}=(\frac{65}{8}\times \frac{8}{17})=\frac{65}{17}$

Hence, the distance between the given lines is $\frac{65}{17}$ units.