Chapter 7.7, Problem 14E

To determine

To find: the distance between the point and line for the given coordinates and equation.

Explanation of Solution

Given information:

  $\begin{array}{l}\left(-2,-3\right)\\ y=4-\frac{2}{3}x\end{array}$

Calculation:

First rewrite the given equation in the standard form of equation $\left(Ax+By+C=0\right)$ as follows:

  $\begin{array}{c}y=4-\frac{2}{3}x\end{array}$

  $\begin{array}{c}\frac{2}{3}x+y-4=0\left[\text{Add }\left(\frac{2}{3}x-4\right)\text{ on each side}\right]\end{array}$

  $\begin{array}{c}2x+3y-12=0\left[\text{Taking common denominator}\right]\end{array}$

Thus the value of $A,B$ and $C$ is 2, 3 and $-12$ respectively.

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

  $\begin{array}{c}d=\frac{\left|A{x}_1+B{y}_1+C\right|}{\pm \sqrt{A^2+B^2}}\end{array}$

  $\begin{array}{c}=\frac{\left|2\left(-2\right)+3\left(-3\right)+\left(-12\right)\right|}{\pm \sqrt{{\left(2\right)}^2+{\left(3\right)}^2}}\left[\begin{array}{l}\text{Put the values }\\ A=2,B=3,C=-12\\ x_1=-2,y_1=-3\end{array}\right]\end{array}$

  $\begin{array}{c}=\frac{\left|-4-9-12\right|}{\sqrt{4+9}}\left[\begin{array}{l}\text{Use }+\sqrt{A^2+B^2}\\ \text{because }C\text{ is negative}\text{.}\end{array}\right]\end{array}$

  $\begin{array}{c}=\frac{\left|-25\right|}{\sqrt{13}}\end{array}$

  $\begin{array}{c}=\frac{\left|-25\right|}{\sqrt{13}}\left(\frac{\sqrt{13}}{\sqrt{13}}\right)\left[\text{Multiply and divide by }\sqrt{13}\right]\end{array}$

  $\begin{array}{c}=\frac{25\sqrt{13}}{13}\end{array}$

Thus the distance from the given point to given line is $\frac{25\sqrt{13}}{13}$ , where positive sign indicates that the point and origin are on the opposite side of the line.