Answer:
[tex]y=\frac{4}{3}x+\frac{38}{3}[/tex]
Step-by-step explanation:
So we want to find an equation of a line passing through (-8,2) with a slope of 4/3.
To do so, we can use the point-slope form. This is:
[tex]y-y_1=m(x-x_1)[/tex]
So, substitute (-8,2) for x₁ and y₁, and let m be 4/3:
[tex]y-2=\frac{4}{3}(x+8)[/tex]
Distribute the right:
[tex]y-2=\frac{4}{3}x+\frac{32}{3}[/tex]
Add 2 to both sides. Note that 2 can be written as 6/3. Thus:
[tex]y=\frac{4}{3}x+\frac{32}{3}+\frac{6}{3}[/tex]
Simplify:
[tex]y=\frac{4}{3}x+\frac{38}{3}[/tex]
And this is our equation :)
