Chapter 3.6, Problem 24PPS

To determine

To find distance between parallel lines.

  $y=\frac{1}{3}x-3$

  $y=\frac{1}{3}x+2$

Answer to Problem 24PPS

  $d=\frac{15}{\sqrt{82}}units$

Explanation of Solution

Given information:

Parallel lines

  $y=\frac{1}{3}x-3$

  $y=\frac{1}{3}x+2$

Formula used:

Distance from point to line

  $d=\frac{|c_1-c_2|}{\sqrt{A^2+B^2}}$

Where

  $c_1$ and $c_2$ are intercept of lines.

Line 1

  $Ax+By+c_1=\mathrm{0.............}(1)$

Line 2

  $Ax+By+c_2=\mathrm{0............}(2)$

  $A$ and $B$ are the coefficient.

Calculation:

Parallel lines

  $y=\frac{1}{3}x-3$

  $y=\frac{1}{3}x+2$

Rewrite equations

  $9y=x-9$

  $3y=x+6$

Rewrite equations

  $x-9y-9=0$

  $x-9y+6=0$

Here

  $A=1$

  $B=-9$

  $c_1=-9$

  $c_2=6$

Using distance formula

Substitute value of $A$ , $B$ , $c_1$ and $c_2$

  $d=\frac{|-9-6|}{\sqrt{1^2+{(-9)}^2}}$

On solving

  $d=\frac{|-15|}{\sqrt{1+81}}$

On solving

  $d=\frac{15}{\sqrt{82}}units$

Hence, distance between parallel lines is $\frac{15}{\sqrt{82}}units$