Chapter 16.1, Problem 1PS

To determine

To find: Thedistance from the origin to the graph of the given equation.

Answer to Problem 1PS

The distance from the origin to the graph of the given equation is $3\text{units}$ .

Explanation of Solution

Given information:

The given equation is,

  $4x-3y+15=0$

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

  $\begin{array}{c}d=\frac{\left|a{x}_1+b{y}_1+c\right|}{\sqrt{a^2+b^2}}\\ =\frac{\left|4\left(0\right)+\left(-3\right)\left(0\right)+15\right|}{\sqrt{{\left(4\right)}^2+{\left(-3\right)}^2}}\\ =\frac{15}{5}\\ =3\text{units}\end{array}$

Thus, the distance from the origin to the graph of the given equation is $3\text{units}$ .