Chapter 7.7, Problem 11E

To determine

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

Explanation of Solution

Given information:

  $\begin{array}{l}\left(2,0\right)\\ 3x-4y+15=0\end{array}$

Calculation:

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

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

  $\begin{array}{c}=\frac{\left|3\left(2\right)+\left(-4\right)0+\left(15\right)\right|}{\pm \sqrt{{\left(3\right)}^2+{\left(-4\right)}^2}}\left[\begin{array}{l}\text{Put the values }\\ A=3,B=-4,C=15\\ x_1=2,y_1=0\end{array}\right]\end{array}$

  $\begin{array}{c}=\frac{\left|3\left(2\right)+\left(-4\right)0+\left(15\right)\right|}{-\sqrt{{\left(3\right)}^2+{\left(-4\right)}^2}}\left[\begin{array}{l}\text{Use}-\sqrt{A^2+B^2}\\ \text{because }C\text{ is positive}\text{.}\end{array}\right]\end{array}$

  $\begin{array}{c}=\frac{\left|6+0+15\right|}{-\sqrt{9+16}}\end{array}$

  $\begin{array}{c}=\frac{\left|21\right|}{-\sqrt{25}}\end{array}$

  $\begin{array}{c}=-\frac{21}{5}\end{array}$

Thus the distance from the given point to given line is $-\frac{21}{5}$ , where negative sign indicates that the point and origin are on the same side of the line.