Chapter 7.7, Problem 8CFU

To determine

To find: the distance between the parallel lines with the given equations.

Explanation of Solution

Given information:

  $y=-\frac{1}{3}x+3$

… (1)

  $y=-\frac{1}{3}x-7$

… (2)

Calculation:

To solve the given problem, we need a point $\left(x_1,y_1\right)$ and the line in the standard form. So substitute $y=0$ in equation (1) and evaluate $x$ as shown below.

  $\begin{array}{c}0=-\frac{1}{3}x+3\\ \frac{1}{3}x=3\left[\text{Add }\frac{1}{3}\text{ on each side}\right]\\ x=9\left[\text{Multiply both sides by 3}\right]\end{array}$

So the point $\left(9,0\right)$ lies on the line.

Now rearrange the equation (2) in the standard form of line $\left(Ax+By+C=0\right)$ as shown below.

  $\begin{array}{c}y=-\frac{1}{3}x-7\\ 3y=-x-21\left[\text{Multiply each side by 3}\right]\\ x+3y=-21\left[\text{Add }x\text{ on each side}\right]\\ x+3y+21=0\left[\text{Add 21 on each side}\right]\end{array}$

Thus the value of $A,B$ and $C$ is $1,3$ and $21$ respectively.

The distance from a point to a line $\left(d\right)$ will be evaluated as shown below.

  $\begin{array}{c}d=\frac{\left|A{x}_1+B{y}_1+C\right|}{\sqrt{A^2+B^2}}\left[\text{Take absolute value of }d\text{ as distance cannot be negative}\right]\\ =\frac{\left|1\left(9\right)+\left(3\left(0\right)\right)+21\right|}{\sqrt{{\left(1\right)}^2+{\left(3\right)}^2}}\left[\text{Put the values of }A,B,C,x_1\text{ and }{y}_1\right]\\ =\frac{\left|9+0+21\right|}{\sqrt{1+9}}\\ =\frac{\left|30\right|}{\sqrt{10}}\\ =\frac{30}{\sqrt{10}}\\ =9.5\text{ units}\end{array}$

Thus the distance from the given point to given line is 9.5 units.