Find the equation of the line containing the points $\left(-3.-4\right)$and $\left(2,-5\right)$. Write the equation in slope-intercept form.

We are given two points on a line; they are $\left(-3,-4\right)$and $(2,-5)$.

We have to find the equation of the line containing these two points in slope-intercept form.

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

The given points are $\left(-3,-4\right)$and $(2,-5)$.

Let $\left(x_1,y_1\right)=(-3,-4)$and $(x_2,y_2)=(2,-5)$

Then the slope of the line is

$$\begin{gathered} m=\frac{y_2-y_1}{x_2-x_1} \\ m=\frac{-5-(-4)}{2-(-3)} \\ m=\frac{-5+4}{2+3} \\ m=-\frac{1}{5} \end{gathered}$$

Thus the slope of the line is$m=-\frac{1}{5}$

We need to find the equation of the line in slope-intercept form i.e in the form$y=mx+b$

Let us choose a point, $(x_1,y_1)=(-3,-4)$

Using the slope $m=-\frac{1}{5}$in the slope-point form, we have

$$\begin{gathered} y-y_1=m(x-x_1) \\ y-(-4)=-\frac{1}{5}(x-(-3\left)\right) \\ y+4=-\frac{1}{5}(x+3) \\ y=-\frac{1}{5}x-\frac{3}{5}-4 \\ y=-\frac{1}{5}x-\frac{23}{5} \end{gathered}$$

The equation of the line containing the points $\left(-3,-4\right)$and $\left(2,-5\right)$in the slope-intercept form is$y=-\frac{1}{5}x-\frac{23}{5}$