Chapter 3.6, Problem 21PPS

To determine

To find distance between parallel lines.

  $y=-2$

  $y=4$

Answer to Problem 21PPS

  $d=6units$

Explanation of Solution

Given information:

Parallel lines

  $y=-2$

  $y=4$

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

  $y=-2$

  $y=4$

Rewrite

  $y+2=0$

  $y-4=0$

Here

  $A=0$

  $B=1$

  $c_1=2$

  $c_2=-4$

Using distance formula

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

  $d=\frac{|2-(-4)|}{\sqrt{0^2+1^2}}$

On solving

  $d=\frac{\left|6\right|}{\sqrt{1}}$

On solving

  $d=6units$

Hence, distance between parallel lines is $6units$ .