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

Passes through $\left(3,-8\right)\text{and}\left(-3,2\right)$

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 slope m of a line passing through two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is expressed below.

$m=\frac{y_2-y_1}{x_2-x_1}$

Now, compute the slope of the line passing through the points $\left(3,-8\right)$ and $\left(-3,2\right)$.

$\begin{array}{c}m=\frac{2-\left(-8\right)}{-3-3}\\ =\frac{2+8}{-6}\\ =\frac{10}{-6}\\ =-\frac{5}{3}\end{array}$

Therefore, slope of the line passing through the points $\left(3,-8\right)$ and $\left(-3,2\right)$ is $m=-\frac{5}{3}$.

The equation in point-slope form is expressed as $y-y_1=m\left(x-x_1\right)$.

Where m is the slope and $\left(x_1,y_1\right)$ is the point through which the line passes.

Now, compute the equation of line with slope $m=-\frac{5}{3}$ and passing through the point $\left(3,-8\right)$.

$\begin{array}{c}y-\left(-8\right)=\frac{-5}{3}\left(x-3\right)\\ y+8=-\frac{5}{3}x+\frac{5}{3}\left(3\right)\\ y=-\frac{5}{3}x+5-8\\ y=-\frac{5}{3}x-3\end{array}$

Recall that equation of line in slope intercept form is expressed as $y=mx+c$

Now, the equation is in the form $y=mx+c$. Here slope m of the line is $-\frac{5}{3}$ and intercept of y-axis c is $-3$.

Hence, the equation of line passing through the points $\left(3,-8\right)$ and $\left(-3,2\right)$ $y=-\frac{5}{3}x-3$.