Chapter 7.7, Problem 21E

To determine

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

Explanation of Solution

Given information:

  $y=-3x+6$    … (1)

  $3x+y=4$    … (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.

  $\begin{array}{c}0=-3x+6\\ 3x=6\left[\text{Add 3}x\text{ on each side}\right]\\ x=\frac{6}{3}\left[\text{Divide both sides by 3}\right]\\ =2\end{array}$

So the point $\left(2,0\right)$ lies on the line.

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

  $\begin{array}{c}3x+y=4\\ 3x+y-4=0\left[\text{Subtract 4 from each side}\right]\end{array}$

Thus the value of $A,B$ and $C$ is $3,1$ and $-4$ respectively.

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

  $\begin{array}{c}d=\frac{\left|A{x}_1+B{y}_1+C\right|}{\sqrt{A^2+B^2}}\left[\text{Take absolute value of }d\text{ as distance cannot be negative}\right]\\ =\frac{\left|3\left(2\right)+\left(1\left(0\right)\right)+\left(-4\right)\right|}{\sqrt{{\left(3\right)}^2+{\left(1\right)}^2}}\left[\text{Put the values of }A,B,C,x_1\text{ and }{y}_1\right]\\ =\frac{\left|6+0-4\right|}{\sqrt{9+1}}\\ =\frac{\left|2\right|}{\sqrt{10}}\\ =\frac{2}{\sqrt{10}}\\ =0.63\text{ units}\end{array}$

Thus the distance from the given point to given line is 0.63 units.