Chapter 3.6, Problem 28PPS

To determine

To find distance between parallel lines.

  $3x+y=3$

  $y+17=-3x$

Answer to Problem 28PPS

  $d=\frac{20}{\sqrt{10}}units$

Explanation of Solution

Given information:

Parallel lines

  $3x+y=3$

  $y+17=-3x$

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

  $3x+y=3$

  $y+17=-3x$

Rewrite equations

  $3x+y-3=0$

  $3x+y+17=0$

Here

  $A=3$

  $B=1$

  $c_1=-3$

  $c_2=17$

Using distance formula

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

  $d=\frac{|-3-17|}{\sqrt{3^2+{(1)}^2}}$

On solving

  $d=\frac{|-20|}{\sqrt{9+1}}$

On solving

  $d=\frac{20}{\sqrt{10}}units$

Hence, distance between parallel lines is $\frac{20}{\sqrt{10}}units$