Chapter 3.6, Problem 27PPS

To determine

To find distance between parallel lines.

  $y=\frac{1}{4}x+2$

  $4y-x=-60$

Answer to Problem 27PPS

  $d=\frac{68}{\sqrt{17}}units$

Explanation of Solution

Given information:

Parallel lines

  $y=\frac{1}{4}x+2$

  $4y-x=-60$

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

  $y=\frac{1}{4}x+2$

  $4y-x=-60$

Rewrite equations

  $4y=x+8$

  $4y-x+60=0$

Rewrite equations

  $4y-x-8=0$

  $4y-x+60=0$

Here

  $A=-1$

  $B=4$

  $c_1=-8$

  $c_2=60$

Using distance formula

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

  $d=\frac{|-8-60|}{\sqrt{{(-1)}^2+4^2}}$

On solving

  $d=\frac{|-68|}{\sqrt{1+16}}$

On solving

  $d=\frac{68}{\sqrt{17}}units$

Hence, distance between parallel lines is $\frac{68}{\sqrt{17}}units$