Chapter 1.5, Problem 26E

To determine

To find: the standard form of line equation.

Answer to Problem 26E

  $x+6y+5=0$

Explanation of Solution

Given information:

Let the line be A. This line passes through $\left(7,-2\right)$ and perpendicular to $6x-y=3$ .

Calculation:

Convert the line equation into slope intercept form.

  $\begin{array}{c}6x-y=3\\ y=6x-3\end{array}$

A line perpendicular to line of equation, $6x-y=3$ . So, the slope (m) is $6$ .

The slope of line A, perpendicular to $6x-y=3$ is opposite reciprocal of the other line. So, the slope of line can be evaluated as follows:

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

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

  $\begin{array}{c}y=mx+b\\ -2=\left(-\frac{1}{6}\right)\left(7\right)+b\\ -2=-\frac{7}{6}+b\\ b=-2+\frac{7}{6}\\ =\frac{-12+7}{6}\\ =\frac{-5}{6}\end{array}$

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

  $\begin{array}{l}m=-\frac{A}{B}\text{ and }b=-\frac{C}{B}\\ -\frac{1}{6}=-\frac{A}{B}\text{ and }\frac{-5}{6}=-\frac{C}{B}\left[\text{Put values of }m\text{ and }b\right]\\ \frac{1}{6}=\frac{A}{B}\text{ and }\frac{5}{6}=\frac{C}{B}\left[\text{Simplify}\right]\end{array}$

The value of $A=1,B=6,\text{ and }C=5$ .

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

  $\begin{array}{l}Ax+By+C=0\\ 1x+6y+5=0\\ x+6y+5=0\end{array}$

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