Chapter 16.1, Problem 7PS

To determine

To find: Thedistance between the graphs of the given equations.

Answer to Problem 7PS

The distance between the graphs of the given equations is $3.051\text{units}$ .

Explanation of Solution

Given information:

The given equationsare,

  $\begin{array}{l}2x-3y+4=0\\ 2x-3y+15=0\end{array}$

The line $2x-3y+4=0$ pass through the point $\left(-2,0\right)$ .

The distance from the point to the graph of the given equation is calculated as,

  $\begin{array}{c}d=\frac{\left|a{x}_1+b{y}_1+c\right|}{\sqrt{a^2+b^2}}\\ =\frac{\left|2\left(-2\right)+\left(-3\right)\left(0\right)+15\right|}{\sqrt{{\left(2\right)}^2+{\left(-3\right)}^2}}\\ =\frac{11}{\sqrt{13}}\\ =3.051\text{units}\end{array}$

Thus, the distance between the graphs of the given equations is $3.051\text{units}$ .