Chapter 1.5, Problem 10CFU

To determine

To find: the standard form of line equation.

Answer to Problem 10CFU

  $5x+6y+80=0$

Explanation of Solution

Given information:

Let the line be A. This line passes through $B\left(-10,-5\right)$ and perpendicular to $6x-5y=24$ .

Calculation:

A line perpendicular to line of equation, $6x-5y=24$ .

Convert the equation in slope intercept form as shown below.

  $\begin{array}{l}6x-5y=24\\ -5y=-6x+24\\ y=\frac{-6}{-5}x+\frac{24}{-5}\\ y=\frac{6}{5}x+\frac{24}{5}\end{array}$

So, the slope (m) is $\frac{6}{5}$ .

The slope of line A, perpendicular to $6x-5y=24$ is opposite reciprocal of the other line. So, the slope of line can be evaluated as follows:

  $\begin{array}{c}\perp m=-\frac{1}{m}\\ \frac{6}{5}=-\frac{1}{m}\\ m=-\frac{5}{6}\end{array}$

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

  $\begin{array}{c}y=mx+b\\ -5=-\left(\frac{5}{6}\right)\left(-10\right)+b\\ -5=\frac{50}{6}+b\\ b=-\frac{50}{6}-5\\ b=\frac{-50-30}{6}\\ =-\frac{80}{6}\end{array}$

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

  $\begin{array}{l}m=-\frac{A}{B}\text{ and }b=-\frac{C}{B}\\ -\left(\frac{5}{6}\right)=-\frac{A}{B}\text{ and }-\frac{80}{6}=-\frac{C}{B}\left[\text{Put values of }m\text{ and }b\right]\\ \frac{5}{6}=\frac{A}{B}\text{ and }\frac{\text{80}}{6}=\frac{C}{B}\left[\text{Simplify}\right]\end{array}$

The value of $A=5,B=6,\text{ and }C=80$ .

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

  $\begin{array}{l}Ax+By+C=0\\ 5x+6y+80=0\end{array}$

Thus, the standard form of line is $5x+6y+80=0$ .