Chapter 10.5, Problem 12E

To determine

To find: The distance between the parallel lines.

Answer to Problem 12E

The distance between the parallel lines is $\frac{4\sqrt{2}}{\sqrt{5}}\text{unit}$ .

Explanation of Solution

Given information:

The given equations of the lines are $y=3x+6\text{and }y=3x-2$ .

The slope of the given lines is $3$ , so the slope of the perpendicular line on the given lines will be $\frac{1}{3}$ .

The intersection points of the perpendicular line on the two parallel line equation are $\left(0,6\right)$ and $\left(\frac{12}{5},\frac{26}{5}\right)$ .

Calculate the distance between the parallel lines as,

  $\begin{array}{c}d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(0-\left(\frac{12}{5}\right)\right)}^2+{\left(6-\frac{26}{5}\right)}^2}\\ =\frac{4\sqrt{2}}{\sqrt{5}}\text{unit}\end{array}$

Thus, the distance between the parallel lines is $\frac{4\sqrt{2}}{\sqrt{5}}\text{unit}$ .