Find the equation of the line containing the points $(-5,-3)$and $(4,-6)$. Write the equation in slope-intercept form.

We are given two points $\left(-5,-3\right),(4,-6)$

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

For this, we will first find the slope of the line and then we will use the slope-point formula.

• When $\left(x_1,y_1\right)$and $\left(x_2,y_2\right)$are two points on a line, the slope of the line is $m=\frac{y_2-y_1}{x_2-x_1}$
• The slope point form of an equation of the line is $y-y_1=m(x-x_1)$where $m$is the slope of the line.

The two given points are $\left(x_1,y_1\right)=(-5,-3)$and $\left(x_2,y_2\right)=(4,-6)$

The slope of the line is,

role="math" localid="1644731554525" $$\begin{gathered} m=\frac{y_2-y_1}{x_2-x_1} \\ m=\frac{-6-(-3)}{4-(-5)} \\ m=\frac{-6+3}{4+5} \\ m=-\frac{3}{9} \\ m=-\frac{1}{3} \end{gathered}$$

To find the equation of the line, we will make use of the slope-point formula.

We know that the slope of the line is $m=-\frac{1}{3}$

Let us choose the point $(-5,-3)$

Using the slope-point formula, we have

$$\begin{gathered} y-y_1=m(x-x_1) \\ y-(-3)=-\frac{1}{3}(x-(-5\left)\right) \\ y+3=-\frac{1}{3}x-\frac{5}{3} \\ y=-\frac{1}{3}x-\frac{5}{3}-3 \\ y=-\frac{1}{3}x-\frac{14}{3} \end{gathered}$$

This is in the form$y=mx+b$which is the slope-intercept form.

The slope-intercept form of the equation of the line containing the points$(-5,-3),(4,-6)$is$y=-\frac{1}{3}x-\frac{14}{3}$