Chapter 1, Problem 48SGA

To determine

To find: the standard form of line equation.

Answer to Problem 48SGA

  $x-3y+19=0$

Explanation of Solution

Given information:

A line passes through $\left(-1,6\right)$ and parallel to $y=\frac{1}{3}x-2$ .

Calculation:

For $y=\frac{1}{3}x-2$ , the slope is $\frac{1}{3}$ . Since, slope of two parallel lines are always equal. So, the slope (m) of both the lines is $\frac{1}{3}$ .

Since $\left(-1,6\right)$ is on line, the line equation will be as follows:

  $\begin{array}{c}y-y_1=m\left(x-x_1\right)\\ y-6=\left(\frac{1}{3}\right)\left(x-\left(-1\right)\right)\\ y-6=\frac{x+1}{3}\\ 3\left(y+6\right)=x+1\\ 3y+18=x+1\\ 3y=x+1+18\\ 3y=x+19\\ 0=x-3y+19\end{array}$

Thus, the standard form of line is $x-3y+19=0$ .