Answer:
[tex]\large\boxed{y-5=-\dfrac{2}{3}(x+3)}[/tex]
Step-by-step explanation:
The point-slope form of an equation of a line:
[tex]y-y_1=m(x-x_1)[/tex]
m - slope
(x₁, y₁) - point on a line
We have the slope [tex]m=-\dfrac{2}{3}[/tex] and the point [tex](-3,\ 5)[/tex].
Substitute:
[tex]y-5=-\dfrac{2}{3}(x-(-3))\\\\y-5=-\dfrac{2}{3}(x+3)[/tex]
