Chapter 8.2, Problem 70SPR

To determine

The distance between the pair of parallel lines.

Answer to Problem 70SPR

  $\mathrm{d}=\sqrt{17}\text{ units}$

Explanation of Solution

Given:

The given equations are:

  $\begin{array}{l}\mathrm{y}=4\mathrm{x}\\ \mathrm{y}=4\mathrm{x}-17\end{array}$

Calculation:

The slope of line p is the opposite reciprocal of 4 is − ¼. Use the y-intercept of the line m, P(0, 0) as one of the endpoints of the perpendicular segment.

  $\begin{array}{l}\mathrm{y}-{\mathrm{y}}_1=\mathrm{m}\left(\mathrm{x}-{\mathrm{x}}_1\right)\\ \mathrm{y}-0=-\frac{1}{4}\left(\mathrm{x}-0\right)\\ \mathrm{y}=-\frac{1}{4}\mathrm{x}\end{array}$

The point of intersection of lines l and p will be:

The length of the sides will be:

  $\begin{array}{l}4\mathrm{x}-17=-\frac{1}{4}\mathrm{x}\\ 4\mathrm{x}+\frac{1}{4}\mathrm{x}=17\\ \frac{17}{4}\mathrm{x}=17\\ \mathrm{x}=4\\ \\ \mathrm{y}=4\mathrm{x}-17\\ \mathrm{y}=4\left(4\right)-17\\ \mathrm{y}=16-17\\ \mathrm{y}=-1\end{array}$

Using distance formula,

  $\begin{array}{l}\mathrm{d}=\sqrt{{\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}^2+{\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)}^2}\\ \mathrm{d}=\sqrt{{\left(4-0\right)}^2+{\left(-1-0\right)}^2}\\ \mathrm{d}=\sqrt{16+1}\\ \mathrm{d}=\sqrt{17}\text{ units}\end{array}$