Chapter 1.5, Problem 25E

To determine

To find: the standard form of line equation.

Answer to Problem 25E

  $x+5y+15=0$

Explanation of Solution

Given information:

Let the line be A. This line passes through $\left(0,-3\right)$ and perpendicular to $y=5x+12$ .

Calculation:

A line perpendicular to line of equation, $y=5x+12$ . So, the slope (m) is $5$ .

The slope of line A, perpendicular to $y=5x+12$ is opposite reciprocal of the other line. So, the slope of line can be evaluated as follows:

  $\begin{array}{c}\perp m=-\frac{1}{m}\\ 5=-\frac{1}{m}\\ m=-\frac{1}{5}\end{array}$

Put the coordinates and slope value of line A, in the slope intercept equation and solve as follows:

  $\begin{array}{c}y=mx+b\\ -3=\left(-\frac{1}{5}\right)\left(0\right)+b\\ -3=0+b\\ b=-3\end{array}$

Equations for converting the slope and y-intercept into standard form as follows:

  $\begin{array}{l}m=-\frac{A}{B}\text{ and }b=-\frac{C}{B}\\ -\frac{1}{5}=-\frac{A}{B}\text{ and }-3=-\frac{C}{B}\left[\text{Put values of }m\text{ and }b\right]\\ \frac{1}{5}=\frac{A}{B}\text{ and }3=\frac{C}{B}\left[\text{Simplify}\right]\\ \frac{1}{5}=\frac{A}{B}\text{ and }\frac{15}{5}=\frac{C}{B}\left[\text{Finding common denominator}\right]\end{array}$

The value of $A=1,B=5,\text{ and }C=15$ .

Substitute the value of $A,B\text{ and }C$ in the standard form of line as shown below.

  $\begin{array}{l}Ax+By+C=0\\ 1x+5y+15=0\\ x+5y+15=0\end{array}$

Thus, the standard form of line is $x+5y+15=0$ .