Write an equation in slope-intercept form for the line that satisfies each set of conditions.

38. passes through $\left(-1,2\right)$, parallel to the graph of $x-3y=14$

Equation of line in slope intercept form is expressed below.

$y=mx+c$

Where m is the slope and c is the intercept of y-axis.

The y-intercept of a graph is when the x-coordinate is 0. The point where the graph of the equation cuts the y-axis.

It is provided that line passes through the point $\left(-1,2\right)$ and parallel to the graph of $x-3y=14$.

Parallel lines have same slope. This means slope of line is equivalent to slope of the line $x-3y=14$.

Rewrite the equation in slope-intercept form to find the slope of the line.

Subtract both sides by x.

$\begin{array}{c}x-3y-x=14-x\\ -3y=14-x\end{array}$

Divide both sides by –3.

$\begin{array}{c}y=-\frac{14}{3}+\frac{x}{3}\\ y=\frac{1}{3}x-\frac{14}{3}\end{array}$

Therefore, the slope of the line is $m=\frac{1}{3}$.

Now that line passes through the point $\left(-1,2\right)$ with slope $m=\frac{1}{3}$.

So, to find the y-intercept that is value of c, follow the step provided below.

Substitute x as $-1$, y as 2 and m as $\frac{1}{3}$ in the equation $y=mx+c$.

$\begin{array}{c}2=\frac{1}{3}\left(-1\right)+c\\ 2=-\frac{1}{3}+c\\ 2+\frac{1}{3}=c\\ \frac{7}{2}=c\end{array}$

Therefore, y-intercept is $\frac{7}{2}$.

Recall the equation of line in slope-intercept form that is $y=mx+c$.

Where m is the slope and c is the intercept of y-axis.

Replace m by $\frac{1}{3}$ and c by $\frac{7}{2}$ to obtain equation of line in slope-intercept form.

$y=\frac{1}{3}x+\frac{7}{2}$

Hence, equation of line passing through the point $\left(-1,2\right)$ and parallel to the graph of $x-3y=14$ is $y=\frac{1}{3}x+\frac{7}{2}$.